Equivalency of Infinite Sets
Ross A. Finlayson
No infinity is countable. It is considerable.
Ross
--
Ross Andrew Finlayson
202/387-8208
http://www.tomco.net/~raf/
"C is the speed of light."
Hal Daume III
talking about.
-----------------------------------------------------------------------
"Men are born ignorant, not stupid; they are made stupid by education."
- Bertrand Russell
-----------------------------------------------------------------------
Hope Hubris
Ross A. Finlayson wrote in message <371E9098...@TOMCO.NET>...
>Each infinite set is equivalent.
>
>No infinity is countable. It is considerable.
>
>Ross
A wonderful statement with little in the way of justification. I will now
counter your argument with one with as much justification:
Not.
Most people here could give you a perfectly good explanation of the theory
of infinite sets, but I doubt you'd care.
Hope Hubris
Tyrant of Jupiter
Zdislav V. Kovarik
Hal Daume III <h...@andrew.cmu.edu> wrote:
:Uh....maybe I'm missing something, but I have no idea what you're
:talking about.
:
Just visit his website, especially the "fun" and the "happy" pages, and
you'll understand a little more.
ZVK(Slavek)
:On Thu, 22 Apr 1999, Ross A. Finlayson wrote:
:
:> Each infinite set is equivalent.
:>
:> No infinity is countable. It is considerable.
:>
:> Ross
:>
:> --
:
Ross A. Finlayson
Any infinite set, that is, a set made up of an infinite number of elements, is
equivalent. Equivalency means having the same number of elements.
So, for example, f(x)=x from zero to infinity has the same number of elements
as f(x)=x+1 from zero to infinity. This number of elements is infinity. {0,
1, 2, 3, ...} and {1, 2, 3, 4, ...} are equivalent.
Any infinite set is equivalent to any other infinite set and each infinite
set. So, for example, the following sets are equivalent in that they have the
same number of elements: R, N, P, Q, etcetera.
Hope Hubris wrote:
> Ross A. Finlayson wrote in message <371E9098...@TOMCO.NET>...
> >Each infinite set is equivalent.
> >
> >No infinity is countable. It is considerable.
> >
> >Ross
>
> A wonderful statement with little in the way of justification. I will now
> counter your argument with one with as much justification:
>
> Not.
>
> Most people here could give you a perfectly good explanation of the theory
> of infinite sets, but I doubt you'd care.
>
> Hope Hubris
> Tyrant of Jupiter
I've read some explanations of set theory of infinite sets, and written some.
I always endeavor to say the repsented sides of an issue, to do otherwise is
ignorant, however, decsiveness requires actually determining which side is
correct.
Back to infinite sets, infinity-10000000=infinity. An infinity minus another
infinity might be infinity, zero, or negative infinity. This is determined by
the underlying function that generates infinities in the first place, and
applies to addition and subtraction.
For example:
lim Sum[0->oo] x= oo
lim Sum[0->oo]-0.99x = -oo
lim Sum[0->oo] x-0.99x = oo
In these examples, if the elements that make up the infinite series are when
correlated are positive, the result is positive or rather diverges to positive
infinity, and if negative to negative infinity, and if the additive sum
converges to zero then zero.
Well, I am going to go find some more definitive set theory.
Ross F.
Brian M. Scott
> It is fair enough for you to say "not", although this is incorrect.
> Any infinite set, that is, a set made up of an infinite number of elements, is
> equivalent. Equivalency means having the same number of elements.
You are wrong, and Hope is right.
> {0,
> 1, 2, 3, ...} and {1, 2, 3, 4, ...} are equivalent.
This is correct.
> Any infinite set is equivalent to any other infinite set and each infinite
> set. So, for example, the following sets are equivalent in that they have the
> same number of elements: R, N, P, Q, etcetera.
This is not. The cardinalities of N and Q are the same, and the
cardinalities of P (the irrationals, I assume) and R are the same, but
despite the numerous cranks who have infested the group over the last
few months, the cardinalities of N and R are not the same. The easy
proof has been posted here numerous times.
Brian M. Scott
Mike Oliver
"Ross A. Finlayson" wrote:
> Any infinite set is equivalent to another infinite set.
True.
Not, however, to *any* other infinite set.
Mike Oliver
"Ross A. Finlayson" wrote:
> I conjecture that any infinite set is equivalent to any other infinite
> set, and to each infinite set,
Conjecture all you want. You know what Popper said about conjectures?
He said that a good conjecture is one that, if false, can in principle
be proven false, and that a conjecture that *has* been proven false
has in this sense already demonstrated its merit.
So give yourself a merit badge.
Hal Daume III
cardinalities of nonfinite sets. the name chosen for the cardinality of
the natural numbers is aleph_nought, while the cardinality of the reals
in 2^aleph_nought = aleph_1. it is easy to show that aleph_nought is
strictly smaller than aleph_1. Yes, both sets are infinite -- but
infinite simply means NOT finite. All non finite sets don't necessarily
have the same cardinality.
"Ross A. Finlayson" wrote:
>
> The cardinality of the natural numbers is infinity. The cardinality of the reals is infinity. The cardinality of natural umbers os less than
> that of the reals for any given, range, but the infinite (unbounded) set of natural numbers has cardinality of infinity, and is thus equivalent
> to the reals or any subdomain of reals.
>
> Hal Daume III wrote:
>
> > Uh...no. The cardinality of the natural numbers is strictly smaller
> > than that of the real numbers. One can show (not so easily) that The
> > powerset of any set is strictly greater than the original set. So P(N)
> > > N. Additionally, one can easily show that the cardinality of the powerset of the naturals is equal to that of the reals. Hence |R| > |N|.
> >
> > "Ross A. Finlayson" wrote:
> > > The cardinalities of N and R are the same, and these two sets are equivalent, where
> > > equivalency means that their cardinalities or quantity of elements is the same.
> > >
> > > Between zero and one, or for that matter, zero and any element of N, there are
> > > infinite number of members of R, as it is dense. Thus, the cardinality of a subset
> > > of R bounded by any two different real numbers is the same as of R and of N.
> > >
> > > Any infinite set is equivalent to another infinite set.
> > >
> > > Ross
> > >
> > > --
> > > Ross Andrew Finlayson
> > > 202/387-8208
> > > http://www.tomco.net/~raf/
> > > "C is the speed of light."
> >
> > --
> > -----------------------------------------------------------------------
> > "Men are born ignorant, not stupid; they are made stupid by education."
> > - Bertrand Russell
> > -----------------------------------------------------------------------
>
> --
> Ross Andrew Finlayson
> 202/387-8208
> http://www.tomco.net/~raf/
> "C is the speed of light."
--
Ross A. Finlayson
Ross A. Finlayson
Mike Oliver wrote:
> "Ross A. Finlayson" wrote:
> > Any infinite set is equivalent to another infinite set.
>
> True.
>
> Not, however, to *any* other infinite set.
I conjecture that any infinite set is equivalent to any other infinite
set, and to each infinite set, there being an infinite number of
infinite sets.
At least, certainly, an infinite set is equivalent to an equivalent
set! (Joke.)
The first statement of these two is absolutely serious.
Ross F.
Ross A. Finlayson
other infinite set.
Mike Oliver wrote:
> "Ross A. Finlayson" wrote:
>
> > I conjecture that any infinite set is equivalent to any other infinite
> > set, and to each infinite set,
>
> Conjecture all you want. You know what Popper said about conjectures?
> He said that a good conjecture is one that, if false, can in principle
> be proven false, and that a conjecture that *has* been proven false
> has in this sense already demonstrated its merit.
>
> So give yourself a merit badge.
My preceding conjecture is true. Let's call it a hypothesis.
Consider R, the domain of real numbers. Also, consider N, the range of
integers of natural or whole numbers. The definition of each of these two
sets is an infinitely bounded (that is, unbounded) series of numbers.
Between zero and one, inclusive, there is an infinite quantity of real
numbers, yet only two integers, zero and one. Between zero and an arbitrary
integer, inclusive, there are an infinite quantity of reals and the integer
plus one integers. As the range goes from zero to infinity, there are an
infinite number of reals demarcated by an infinite number of integers, one
per unit.
For any given range not zero, there are an infinite number of reals, and an
arbitrary umber if integers. Thus, there are an infinity of infinite sets
of real numbers, and an infinite set of integers. Each of these is
equivalent, with cardinality of infinity. Any infinite set has a
cardinality of infinity.
Thus, any infinite set is equivalent to each and any of infinity infinite
sets.
Ross
Richard Carr
:Date: Thu, 22 Apr 1999 23:38:04 -0400
:From: Hal Daume III <h...@andrew.cmu.edu>
:Newsgroups: sci.math
:Subject: Re: Equivalency of Infinite Sets
:
:infinity is too vague a concept to deal with if we want to speak of
:cardinalities of nonfinite sets. the name chosen for the cardinality of
:the natural numbers is aleph_nought, while the cardinality of the reals
:in 2^aleph_nought = aleph_1.
Not when the cardinality of the reals is aleph_2 or aleph_3 or
aleph_{aleph_1} etc.
: it is easy to show that aleph_nought is
:strictly smaller than aleph_1. Yes, both sets are infinite -- but
:infinite simply means NOT finite. All non finite sets don't necessarily
:have the same cardinality.
:
:"Ross A. Finlayson" wrote:
:>
:> The cardinality of the natural numbers is infinity. The cardinality of the reals is infinity. The cardinality of natural umbers os less than
:> that of the reals for any given, range, but the infinite (unbounded) set of natural numbers has cardinality of infinity, and is thus equivalent
:> to the reals or any subdomain of reals.
:>
:> Hal Daume III wrote:
:>
:> > Uh...no. The cardinality of the natural numbers is strictly smaller
:> > than that of the real numbers. One can show (not so easily) that The
:> > powerset of any set is strictly greater than the original set. So P(N)
:> > > N. Additionally, one can easily show that the cardinality of the powerset of the naturals is equal to that of the reals. Hence |R| > |N|.
:> >
:> > "Ross A. Finlayson" wrote:
:> > >
:> > > Brian M. Scott wrote:
:> > >
:> > > > Ross A. Finlayson wrote:
:> > > >
:> > > > > It is fair enough for you to say "not", although this is incorrect.
:> > > >
:> > > > > Any infinite set, that is, a set made up of an infinite number of elements, is
:> > > > > equivalent. Equivalency means having the same number of elements.
:> > > >
:> > > > You are wrong, and Hope is right.
:> > > >
:> > > > > {0,
:> > > > > 1, 2, 3, ...} and {1, 2, 3, 4, ...} are equivalent.
:> > > >
:> > > > This is correct.
:> > > >
:> > > > > Any infinite set is equivalent to any other infinite set and each infinite
:> > > > > set. So, for example, the following sets are equivalent in that they have the
:> > > > > same number of elements: R, N, P, Q, etcetera.
:> > > >
:> > > > This is not. The cardinalities of N and Q are the same, and the
:> > > > cardinalities of P (the irrationals, I assume) and R are the same, but
:> > > > despite the numerous cranks who have infested the group over the last
:> > > > few months, the cardinalities of N and R are not the same. The easy
:> > > > proof has been posted here numerous times.
:> > > >
:> > > > Brian M. Scott
:> > >
:> > > The cardinalities of N and R are the same, and these two sets are equivalent, where
:> > > equivalency means that their cardinalities or quantity of elements is the same.
:> > >
:> > > Between zero and one, or for that matter, zero and any element of N, there are
:> > > infinite number of members of R, as it is dense. Thus, the cardinality of a subset
:> > > of R bounded by any two different real numbers is the same as of R and of N.
:> > >
:> > > Any infinite set is equivalent to another infinite set.
:> > >
:> > > Ross
:> > >
:> > > --
:> > > Ross Andrew Finlayson
:> > > 202/387-8208
:> > > http://www.tomco.net/~raf/
:> > > "C is the speed of light."
:> >
:> > --
:> > -----------------------------------------------------------------------
:> > "Men are born ignorant, not stupid; they are made stupid by education."
:> > - Bertrand Russell
:> > -----------------------------------------------------------------------
:>
:> --
:> Ross Andrew Finlayson
:> 202/387-8208
:> http://www.tomco.net/~raf/
:> "C is the speed of light."
:
:--
:
:
Dave Seaman
Ross A. Finlayson <R...@TOMCO.NET> wrote:
>I conjecture that any infinite set is equivalent to any other infinite
>infinite sets.
Do you conjecture that there is a one-to-one correspondence between N
and the power set P(N)?
Mind you, I haven't even asked you to produce one. I am simply asking
whether you consider it possible for a bijection N <--> P(N) to exist.
--
Dave Seaman dse...@purdue.edu
Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
<http://mojo.calyx.net/~refuse/altindex.html>
Nathan the Great
Dave Seaman wrote:
> Do you conjecture that there is a one-to-one correspondence between N
> and the power set P(N)?
>
> Mind you, I haven't even asked you to produce one. I am simply asking
> whether you consider it possible for a bijection N <--> P(N) to exist.
Dave,
First, are you talking about a STATIC or a DYNAMIC bijection?
Second, is there a bijection between N and the set of all odd numbers (a
subset of N - that has only one member for every two members in N).
hahaha ha hehe he ha haahah... :-)
Nathan the Great
Age 11
Hal Daume III
|R| != |R| (as the statment that |R| = aleph_1 or |R|=aleph_3 would
imply). maybe an example would be helpful, rather than just some blind
faith. (or, alternatively, a proof would be nice, as I can _prove_ that
is |N|=aleph_nought, then |R|=aleph_1 (with neither AC nor CH).)
thanks,
hal
On Fri, 23 Apr 1999, Richard Carr wrote:
> On Fri, 23 Apr 1999, Hal Daume III wrote:
>
> :Date: Fri, 23 Apr 1999 15:46:26 -0400
> :From: Hal Daume III <h...@andrew.cmu.edu>
> :To: Richard Carr <ca...@math.columbia.edu>
> :Subject: Re: Equivalency of Infinite Sets
> :
> :under what circumstances would |R| != aleph_1?
>
> For example, when |R|=aleph_2 or when |R|=aleph_3 or when
> |R|=aleph_{aleph_{aleph_7}} etc.
> Any time CH fails or sometimes when CH holds but AC fails.
>
> :
> :Richard Carr wrote:
> :>
> :> :
> :
> :--
Ross A. Finlayson
Dave Seaman wrote:
> In article <371FC9F0...@TOMCO.NET>,
> Ross A. Finlayson <R...@TOMCO.NET> wrote:
> >I conjecture that any infinite set is equivalent to any other infinite
> >set, and to each infinite set, there being an infinite number of
> >infinite sets.
>
> Do you conjecture that there is a one-to-one correspondence between N
> and the power set P(N)?
>
> Mind you, I haven't even asked you to produce one. I am simply asking
> whether you consider it possible for a bijection N <--> P(N) to exist.
>
> --
> Dave Seaman dse...@purdue.edu
> Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
> <http://mojo.calyx.net/~refuse/altindex.html>
As each is an infinite set, it ispossible to match terms one-to-one from
any subset of N to the a subset of the power set of N. It is not possible
to match one-to-one N to P(N), because each is an infinite set. Bijection
is only valid for finite sets. For any subset of N, it is possible, and
vice versa, and for any infinite set, thus infinite sets are equivalent, as
each infinite set has an infinite number of elements beyond any given
subset.
Dave Seaman
>|R| != |R| (as the statment that |R| = aleph_1 or |R|=aleph_3 would
>imply). maybe an example would be helpful, rather than just some blind
>faith. (or, alternatively, a proof would be nice, as I can _prove_ that
>is |N|=aleph_nought, then |R|=aleph_1 (with neither AC nor CH).)
You can't prove that |R|=aleph_1, without CH.
What you can prove is that |R| = 2^aleph_0. The cardinal 2^aleph_0 is
called c, the cardinality of the continuum. It is known that c >
aleph_0 and also that c >= aleph_1. The statement c = aleph_1 is CH.
Richard Carr
On Fri, 23 Apr 1999, Hal Daume III wrote:
:Date: Fri, 23 Apr 1999 16:34:48 -0400 (EDT)
:From: Hal Daume III <h...@andrew.cmu.edu>
:To: Richard Carr <ca...@math.columbia.edu>, post+s...@andrew.cmu.edu
:Subject: Re: Equivalency of Infinite Sets
:
:that doesn't answer my question. maybe i need to be more clear: how can
:|R| != |R| (as the statment that |R| = aleph_1 or |R|=aleph_3 would
It is not possible tha |R|!=|R| but it is possible that |R|=aleph_3. It is
also possible that |R|=aleph_1 (in which case |R| is not aleph_3). It is
also possible that |R|=aleph_{aleph_{aleph_{aleph_2}}} (in which case |R|
is neither aleph_1 nor aleph_3. It is also possible that |R| is not an
aleph at all. There is a lot of literature on this- you just need to look
up "forcing". Cohen did it first, then there's lots of other people too
(Easton did a more general version). You can get the idea in many texts
(e.g. Jech or Kunen etc.)
:imply). maybe an example would be helpful, rather than just some blind
Examples could be slightly difficult to give but basically if you only
have aleph_1 reals and you want aleph_3 reals you just add aleph_3 more
reals. It can be done. Then |R| will be aleph_3 and not aleph_1.
You may be pleased to know that |R| cannot be aleph_{aleph_0}, though.
:faith. (or, alternatively, a proof would be nice, as I can _prove_ that
:is |N|=aleph_nought, then |R|=aleph_1 (with neither AC nor CH).)
No you can't- you may think you can, but although you can prove
|N|=aleph_0, you cannot prove |R|=aleph_1 (this needs AC & CH, AC to get
that it is an aleph (or at least some weak form of AC and then CH to get
that it is aleph_1). It is not possible to prove |R|=aleph_1 in ZF(C).
:
:thanks,
:> :> :> > > Any infinite set is equivalent to another infinite set.
:> :> :> > >
:> :> :> > > Ross
:> :> :> > >
:> :> :> > > --
:> :> :> > > Ross Andrew Finlayson
:> :> :> > > 202/387-8208
:> :> :> > > http://www.tomco.net/~raf/
:> :> :> > > "C is the speed of light."
:> :> :> >
:> :> :> > --
:> :> :> > -----------------------------------------------------------------------
:> :> :> > "Men are born ignorant, not stupid; they are made stupid by education."
:> :> :> > - Bertrand Russell
:> :> :> > -----------------------------------------------------------------------
:> :> :>
:> :> :> --
:> :> :> Ross Andrew Finlayson
:> :> :> 202/387-8208
:> :> :> http://www.tomco.net/~raf/
:> :> :> "C is the speed of light."
:> :> :
:> :> :--
:
Richard Carr
:Date: 23 Apr 1999 15:55:02 -0500
:From: Dave Seaman <a...@seaman.cc.purdue.edu>
:Newsgroups: sci.math
:Subject: Re: Equivalency of Infinite Sets
:
:In article <Pine.SOL.3.96L.99042...@unix6.andrew.cmu.edu>,
:Hal Daume III <h...@andrew.cmu.edu> wrote:
:>that doesn't answer my question. maybe i need to be more clear: how can
:>|R| != |R| (as the statment that |R| = aleph_1 or |R|=aleph_3 would
:>imply). maybe an example would be helpful, rather than just some blind
:>faith. (or, alternatively, a proof would be nice, as I can _prove_ that
:>is |N|=aleph_nought, then |R|=aleph_1 (with neither AC nor CH).)
:
:You can't prove that |R|=aleph_1, without CH.
:
:What you can prove is that |R| = 2^aleph_0. The cardinal 2^aleph_0 is
:called c, the cardinality of the continuum. It is known that c >
:aleph_0 and also that c >= aleph_1.
(This latter part only with some form of choice.)
:The statement c = aleph_1 is CH.
:
:
Dave Seaman
>> Do you conjecture that there is a one-to-one correspondence between N
>> and the power set P(N)?
>> Mind you, I haven't even asked you to produce one. I am simply asking
>> whether you consider it possible for a bijection N <--> P(N) to exist.
>As each is an infinite set, it ispossible to match terms one-to-one from
>any subset of N to the a subset of the power set of N. It is not possible
>to match one-to-one N to P(N), because each is an infinite set. Bijection
>is only valid for finite sets.
Let f: N -> N be given by f(n) = n for each n in N. Then f is a
bijection between the infinite set N and itself. Therefore, bijection
is perfectly valid for at least some infinite sets.
Do you disagree? For which n in N have I not defined f(n)?
>For any subset of N, it is possible, and
>vice versa,
I don't know what you mean by the "any subset of N" part. Are you
claiming there is a bijection between X and P(X) whenever X is a subset
of N? Cantor proved that this is impossible for *any* set, finite or
infinite.
>and for any infinite set, thus infinite sets are equivalent, as
>each infinite set has an infinite number of elements beyond any given
>subset.
Nonsense. N is an infinite set, and N is a subset of N. The number of
elements in N\N is 0, which is a long way from being infinite.
Besides, if we were to accept your premise that bijection is valid only
for finite sets, the logical consequence would be that no two infinite
sets would be equivalent, since equivalence is defined by the existence
of a bijection. Therefore, your conclusion is contradicted by your own
claims.
Ross A. Finlayson
Dave Seaman wrote:
> In article <3720DF0C...@TOMCO.NET>,
> Ross A. Finlayson <R...@TOMCO.NET> wrote:
> >Dave Seaman wrote:
>
> >> Do you conjecture that there is a one-to-one correspondence between N
> >> and the power set P(N)?
>
> >> Mind you, I haven't even asked you to produce one. I am simply asking
> >> whether you consider it possible for a bijection N <--> P(N) to exist.
>
> >As each is an infinite set, it ispossible to match terms one-to-one from
> >any subset of N to the a subset of the power set of N. It is not possible
> >to match one-to-one N to P(N), because each is an infinite set. Bijection
> >is only valid for finite sets.
>
> Let f: N -> N be given by f(n) = n for each n in N. Then f is a
> bijection between the infinite set N and itself. Therefore, bijection
> is perfectly valid for at least some infinite sets.
>
> Do you disagree? For which n in N have I not defined f(n)?
>
I agree that bijection is valid for identical subsets of infinite sets.
>
> >For any subset of N, it is possible, and
> >vice versa,
>
> I don't know what you mean by the "any subset of N" part. Are you
> claiming there is a bijection between X and P(X) whenever X is a subset
> of N? Cantor proved that this is impossible for *any* set, finite or
> infinite.
>
For any X, subset of N, a subset of P(N) can be selected with one-to-one
mappings, that is, an equivalent subset.
>
> >and for any infinite set, thus infinite sets are equivalent, as
> >each infinite set has an infinite number of elements beyond any given
> >subset.
>
> Nonsense. N is an infinite set, and N is a subset of N. The number of
> elements in N\N is 0, which is a long way from being infinite.
>
N is an infinite set, and N is N, not any subset of N. Any subset of N is not
infinite. Any subset of R is finite, but any range over R represents an
infinite set.
>
> Besides, if we were to accept your premise that bijection is valid only
> for finite sets, the logical consequence would be that no two infinite
> sets would be equivalent, since equivalence is defined by the existence
> of a bijection. Therefore, your conclusion is contradicted by your own
> claims.
>
> --
> Dave Seaman dse...@purdue.edu
> Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
> <http://mojo.calyx.net/~refuse/altindex.html>
Equivalence does not mandate a one-to-one mapping, it is impossible to map,
one-to-one, an infinite set, unless such mappings occur at an infinite rate or
there are an infinite number of mappings.
Each infinite set is equivalent to each infinite set, irregardless of any
subset bijection.
Ross F.
Dave Seaman
Ross A. Finlayson <R...@TOMCO.NET> wrote:
>Dave Seaman wrote:
>> >As each is an infinite set, it ispossible to match terms one-to-one from
>> >any subset of N to the a subset of the power set of N. It is not possible
>> >to match one-to-one N to P(N), because each is an infinite set. Bijection
>> >is only valid for finite sets.
>>
>> Let f: N -> N be given by f(n) = n for each n in N. Then f is a
>> bijection between the infinite set N and itself. Therefore, bijection
>> is perfectly valid for at least some infinite sets.
>>
>> Do you disagree? For which n in N have I not defined f(n)?
>I agree that bijection is valid for identical subsets of infinite sets.
There is also a bijection between N and 2N (the integers and the even
integers) given by f(n) = 2*n for each n. Therefore, the subsets need
not be identical.
>For any X, subset of N, a subset of P(N) can be selected with one-to-one
>mappings, that is, an equivalent subset.
Are you claiming that there is a bijection between X and P(X) for some X? Do
you have any particular X in mind? Would you like to give an example?
>> >and for any infinite set, thus infinite sets are equivalent, as
>> >each infinite set has an infinite number of elements beyond any given
>> >subset.
>>
>> Nonsense. N is an infinite set, and N is a subset of N. The number of
>> elements in N\N is 0, which is a long way from being infinite.
>N is an infinite set, and N is N, not any subset of N. Any subset of N is not
>infinite. Any subset of R is finite, but any range over R represents an
>infinite set.
By definition, a set A is said to be a subset of a set B if for every
x, x in A implies x in B.
Therefore, it follows that every set is a subset of itself. If you
want to talk about subsets of N that differ from N, the correct term is
"proper subsets" of N. If you want to talk about only the finite
subsets of N, then you need to say so. P(N)|finite, the finite subsets
of N, is quite a different set from P(N). It is true that there is a
bijection between N and P(N)|finite, but there is no bijection between
N and P(N).
>Equivalence does not mandate a one-to-one mapping, it is impossible to map,
>one-to-one, an infinite set, unless such mappings occur at an infinite rate or
>there are an infinite number of mappings.
By definition, sets are equivalent if they have the same cardinality. That
means there is a bijection between them.
Mappings do not have "rates" attached to them. For example, there is a
bijection between the interval [0,1] and the interval [0,2] given by f(x) = 2*x
for each x. The mapping exists. If you disagree, then show me an x in [0,1]
for which f(x) is not defined.
>Each infinite set is equivalent to each infinite set, irregardless of any
>subset bijection.
Equivalence, by definitioin, means a bijection exists. I don't know
what you mean by a "subset bijection." A "bijection" between A and B
is a mapping f: A -> B that is both injective and surjective. Note, I
did not mention anything about subsets.
Ross A. Finlayson
Dave Seaman wrote:
Every set is identical to itself. There are an infinite number of subsets for any
infinite set.
>
> >Equivalence does not mandate a one-to-one mapping, it is impossible to map,
> >one-to-one, an infinite set, unless such mappings occur at an infinite rate or
> >there are an infinite number of mappings.
>
> By definition, sets are equivalent if they have the same cardinality. That
> means there is a bijection between them.
>
Two sets are equivalent if there is a cardinally equivalent bijection between them
or they are equivalent. Each infinite set is equivalent. I use "equivalent"
perhaps a bit differently than Cantor.
>
> Mappings do not have "rates" attached to them. For example, there is a
> bijection between the interval [0,1] and the interval [0,2] given by f(x) = 2*x
> for each x. The mapping exists. If you disagree, then show me an x in [0,1]
> for which f(x) is not defined.
>
> >Each infinite set is equivalent to each infinite set, irregardless of any
> >subset bijection.
>
> Equivalence, by definitioin, means a bijection exists. I don't know
> what you mean by a "subset bijection." A "bijection" between A and B
> is a mapping f: A -> B that is both injective and surjective. Note, I
> did not mention anything about subsets.
>
> --
> Dave Seaman dse...@purdue.edu
> Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
> <http://mojo.calyx.net/~refuse/altindex.html>
Equivalence, if it thus states, requires redefinition.
Each of N and R has an infinite number of elements, so for as long as one could
ever try
and match them one-to-one, there is an infinite set of each that remains.
Considering such a mapping, there is an infinite number of reals between zero and
one inclusive. Mapping an infinity of integers to that, leaves the infinite range
of the rest of the integers of reals, which is infinite. Happily, there are still
an infinite number of integers to map to the next unit range, etcetera.
Each infinite set is equivalent in that each infinite set can be considered to have
the same number of elements, infinity.
Hal Daume III
> Two sets are equivalent if there is a cardinally equivalent bijection between them
> or they are equivalent. Each infinite set is equivalent. I use "equivalent"
> perhaps a bit differently than Cantor.
So....lemme see if I got this correctly. "Two sets are equivalent if
... they are equivalent." Yes. That is correct. The verb to be is
reflexive.
> Equivalence, if it thus states, requires redefinition.
I think so. Though, even were you to define a set as equivilent meaning
that it was isomorphic to a purple elephant, your previous statement
would remain true.
> Each of N and R has an infinite number of elements, so for as long as one could
> ever try
> and match them one-to-one, there is an infinite set of each that remains.
Hence, we use induction.
> Considering such a mapping, there is an infinite number of reals between zero and
> one inclusive. Mapping an infinity of integers to that, leaves the infinite range
> of the rest of the integers of reals, which is infinite. Happily, there are still
> an infinite number of integers to map to the next unit range, etcetera.
afaik, there is no mapping from the integers to [0,1]. this is because
of stuff that your next statment incorrectly refutes.
> Each infinite set is equivalent in that each infinite set can be considered to have
> the same number of elements, infinity.
and what, may i ask, is infinity. and if two sets have the same "number
of elements", with what are we counting these elements? natural
numbers? we'll soon run out. count for me, why don't you, the powerset
of the powerset of the powerset of |N.
eh?
- hal
Jeremy Boden
<h...@andrew.cmu.edu> writes
>infinity is too vague a concept to deal with if we want to speak of
>cardinalities of nonfinite sets. the name chosen for the cardinality of
>the natural numbers is aleph_nought, while the cardinality of the reals
>strictly smaller than aleph_1. Yes, both sets are infinite -- but
>infinite simply means NOT finite. All non finite sets don't necessarily
>have the same cardinality.
>
>"Ross A. Finlayson" wrote:
>
The cardinality of the natural numbers is aleph_0.
The cardinality of the reals = 2^aleph_0 and this might be aleph_1.
However, no proof is possible within "normal" set theory.
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Dave Seaman
Ross A. Finlayson <R...@TOMCO.NET> wrote:
>Dave Seaman wrote:
>> By definition, sets are equivalent if they have the same cardinality. That
>> means there is a bijection between them.
>Two sets are equivalent if there is a cardinally equivalent bijection between them
>or they are equivalent. Each infinite set is equivalent. I use "equivalent"
>perhaps a bit differently than Cantor.
Then why did you not specify in your first posting on the subject that
you were using the word "equivalent" in a nonstandard way?
What, exactly, is a "cardinally equivalent bijection?" Is this something
different from an ordinary bijection?
>> Equivalence, by definitioin, means a bijection exists. [ ... ]
>Equivalence, if it thus states, requires redefinition.
In other words, you didn't have any idea what you were talking about when you
made your earlier statements.
>Each of N and R has an infinite number of elements, so for as long as one could
>ever try
>and match them one-to-one, there is an infinite set of each that remains.
Each of N and Q has an infinite number of arguments, but there is a
bijection between N and Q. There is no bijection between N and R.
More precisely, there is no surjection from N onto R. That is what it
means to say R, unlike Q, is uncountable.
>Considering such a mapping, there is an infinite number of reals between zero and
>one inclusive. Mapping an infinity of integers to that, leaves the infinite range
>of the rest of the integers of reals, which is infinite.
I am unable to parse that last sentence.
Ross A. Finlayson
Hal Daume III wrote:
> On Fri, 23 Apr 1999, Ross A. Finlayson wrote:
> > Two sets are equivalent if there is a cardinally equivalent bijection between them
> > or they are equivalent. Each infinite set is equivalent. I use "equivalent"
> > perhaps a bit differently than Cantor.
>
> So....lemme see if I got this correctly. "Two sets are equivalent if
> ... they are equivalent." Yes. That is correct. The verb to be is
> reflexive.
>
> > Equivalence, if it thus states, requires redefinition.
>
> I think so. Though, even were you to define a set as equivilent meaning
> that it was isomorphic to a purple elephant, your previous statement
> would remain true.
>
> > Each of N and R has an infinite number of elements, so for as long as one could
> > ever try
> > and match them one-to-one, there is an infinite set of each that remains.
>
> Hence, we use induction.
>
> > Considering such a mapping, there is an infinite number of reals between zero and
> > one inclusive. Mapping an infinity of integers to that, leaves the infinite range
> > of the rest of the integers of reals, which is infinite. Happily, there are still
> > an infinite number of integers to map to the next unit range, etcetera.
>
> afaik, there is no mapping from the integers to [0,1]. this is because
> of stuff that your next statment incorrectly refutes.
>
> > Each infinite set is equivalent in that each infinite set can be considered to have
> > the same number of elements, infinity.
>
> and what, may i ask, is infinity. and if two sets have the same "number
> of elements", with what are we counting these elements? natural
> numbers? we'll soon run out. count for me, why don't you, the powerset
> of the powerset of the powerset of |N.
>
> eh?
>
> - hal
Infinity.
Domnei
>> of elements", with what are we counting these elements? natural
>> numbers? we'll soon run out. count for me, why don't you, the powerset
>> of the powerset of the powerset of |N.
>>
>> eh?
>>
>> - hal
>
>Infinity.
>
Ross -
How about answering the bloody question? He asked
you what you mean by "infinity". Without attaching some
sort of meaning to that word, your answering of the second
question (what is the cardinality of the powerset of
the powerset of N?) by saying "infinity" is meaningless.
And, by the way, since I can't ferret it out from
your ramblings, what is your point? You seem to be trying
hard to say that all infinite sets have the same cardinality.
If you are, then welcome to crankdom. If you're not
saying that (which seems possible, because you keep
using the word "equivalent" instead of "have the
same cardinality") then please explain what you ARE saying.
Mike Keith
Web site: http://users.aol.com/s6sj7gt/mikehome.htm
(remove post-w letters from e-mail address)
Ross A. Finlayson
Domnei wrote:
> >> and what, may i ask, is infinity. and if two sets have the same "number
> >> of elements", with what are we counting these elements? natural
> >> numbers? we'll soon run out. count for me, why don't you, the powerset
> >> of the powerset of the powerset of |N.
> >>
> >> eh?
> >>
> >> - hal
>
> >
> >Infinity.
> >
>
> Ross -
>
> How about answering the bloody question? He asked
> you what you mean by "infinity". Without attaching some
> sort of meaning to that word, your answering of the second
> question (what is the cardinality of the powerset of
> the powerset of N?) by saying "infinity" is meaningless.
>
Infinity, see http://www.astro.virginia.edu/~eww6n/math/Infinity.html.
The number of elements of the set that is the powerset of the powerset of N is
infinity. The term cardinality is often used to mean the number of elements of
a set, thus the cardinality of N is infinity, as is the cardinality of the
powerset of the powerset of N.
What is infinity? Scalar infinity is a numerical increase without bound,
forever and ever.
>
> And, by the way, since I can't ferret it out from
> your ramblings, what is your point? You seem to be trying
> hard to say that all infinite sets have the same cardinality.
> If you are, then welcome to crankdom. If you're not
> saying that (which seems possible, because you keep
> using the word "equivalent" instead of "have the
> same cardinality") then please explain what you ARE saying.
>
> Mike Keith
> Web site: http://users.aol.com/s6sj7gt/mikehome.htm
> (remove post-w letters from e-mail address)
Each infinite set is equivalent in that each infinite set has an infinite
number of elements. For a given range of a domain, leading to a subset of N or
R, a range of N leads to a finite set, and any range of R is an infinite set.
Thus, any range of R is equivalent to R and to N.
Ross F.
Ross A. Finlayson
Dave Seaman wrote:
> In article <3720FF27...@TOMCO.NET>,
> Ross A. Finlayson <R...@TOMCO.NET> wrote:
>
> >Dave Seaman wrote:
>
> >> By definition, sets are equivalent if they have the same cardinality. That
> >> means there is a bijection between them.
>
> >Two sets are equivalent if there is a cardinally equivalent bijection between them
> >or they are equivalent. Each infinite set is equivalent. I use "equivalent"
> >perhaps a bit differently than Cantor.
>
> Then why did you not specify in your first posting on the subject that
> you were using the word "equivalent" in a nonstandard way?
>
> What, exactly, is a "cardinally equivalent bijection?" Is this something
> different from an ordinary bijection?
>
As possible, I will be quite happy to absolutely avoid any use of the words "bijection"
or "cardinality."
In the phrase "cardinally equivalent bijection", the term "cardinally" is redundant. A
bijection is a one-to-one mapping of terms of two separate sets. The term "equivalent"
is an adjective describing a bijection, so a cardinally equivalent bijection is a
bijection.
I never felt that I was using "equivalent" in a non-standard way, that is, representing
the equality of quantity of elements of a set, and throughout this thread and others, I
have not used the term "equivalent" in a way that violates this defintion
>
> >> Equivalence, by definitioin, means a bijection exists. [ ... ]
>
> >Equivalence, if it thus states, requires redefinition.
>
> In other words, you didn't have any idea what you were talking about when you
> made your earlier statements.
>
This previous statement of yours is invalid. I do have an understanding of what I have
been writing. My standing on these issues has not changed since I started this thread,
nor have I seen what I have written invalidated. Don't presume what I or anyone do or
do not know.
Equivalence requires redefinition.
>
> >Each of N and R has an infinite number of elements, so for as long as one could
> >ever try
> >and match them one-to-one, there is an infinite set of each that remains.
>
> Each of N and Q has an infinite number of arguments, but there is a
> bijection between N and Q. There is no bijection between N and R.
> More precisely, there is no surjection from N onto R. That is what it
> means to say R, unlike Q, is uncountable.
>
> >Considering such a mapping, there is an infinite number of reals between zero and
> >one inclusive. Mapping an infinity of integers to that, leaves the infinite range
> >of the rest of the integers of reals, which is infinite.
>
> I am unable to parse that last sentence.
>
> --
> Dave Seaman dse...@purdue.edu
> Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
> <http://mojo.calyx.net/~refuse/altindex.html>
I can understand that the quantity of elements of a powerset of a set is 2^n, where n
is the number of elements of the set. This is similar to probability result
determination
Any infinite set has an infinite number of elements or arguments. Thus, any infinite
set's (eg N, Q, R) cardinality is infinity, and these sets are equivalent, as are they
to the powersets of N, Q, and R.
Richard Carr
:Date: Sat, 24 Apr 1999 05:10:15 GMT
:From: Ross A. Finlayson <R...@TOMCO.NET>
:Newsgroups: sci.math
:Subject: Re: Equivalency of Infinite Sets
:
:
:
:Domnei wrote:
:
:> >> and what, may i ask, is infinity. and if two sets have the same "number
:> >> of elements", with what are we counting these elements? natural
:> >> numbers? we'll soon run out. count for me, why don't you, the powerset
:> >> of the powerset of the powerset of |N.
:> >>
:> >> eh?
:> >>
:> >> - hal
:>
:> >
:> >Infinity.
:> >
:>
:> Ross -
:>
:> How about answering the bloody question? He asked
:> you what you mean by "infinity". Without attaching some
:> sort of meaning to that word, your answering of the second
:> question (what is the cardinality of the powerset of
:> the powerset of N?) by saying "infinity" is meaningless.
:>
:
:Infinity, see http://www.astro.virginia.edu/~eww6n/math/Infinity.html.
The fact that you are citing a website that requires University of
Virginia name and password indicates that you are thinly veiling the fact
that what you are writing is utter nonsense.
:
:The number of elements of the set that is the powerset of the powerset of N is
:infinity. The term cardinality is often used to mean the number of elements of
:a set, thus the cardinality of N is infinity, as is the cardinality of the
:powerset of the powerset of N.
:
:What is infinity? Scalar infinity is a numerical increase without bound,
:forever and ever.
:
:>
:> And, by the way, since I can't ferret it out from
:> your ramblings, what is your point? You seem to be trying
:> hard to say that all infinite sets have the same cardinality.
:> If you are, then welcome to crankdom. If you're not
:> saying that (which seems possible, because you keep
:> using the word "equivalent" instead of "have the
:> same cardinality") then please explain what you ARE saying.
:>
:> Mike Keith
:> Web site: http://users.aol.com/s6sj7gt/mikehome.htm
:> (remove post-w letters from e-mail address)
:
:Each infinite set is equivalent in that each infinite set has an infinite
:number of elements. For a given range of a domain, leading to a subset of N or
:R, a range of N leads to a finite set, and any range of R is an infinite set.
:Thus, any range of R is equivalent to R and to N.
This is gibberish.
:
:Ross F.
:
:
:
:
Russell Easterly
Dave Seaman <a...@seaman.cc.purdue.edu> wrote in message
`Besides, if we were to accept your premise that bijection is valid only
`for finite sets, the logical consequence would be that no two infinite
`sets would be equivalent, since equivalence is defined by the existence
`of a bijection. Therefore, your conclusion is contradicted by your own
`claims.
I don't think that infinite sets can be ordered.
It is meaningless to ask if an infinite set
is smaller, bigger or equal to another infinite set.
Russell
-2 many 2 count
Ross A. Finlayson
Richard Carr wrote:
> :
> :Infinity, see http://www.astro.virginia.edu/~eww6n/math/Infinity.html.
>
> The fact that you are citing a website that requires University of
> Virginia name and password indicates that you are thinly veiling the fact
> that what you are writing is utter nonsense.
>
The fact that this URL points to the CRC Concise Encyclopedia of Mathematics
illustrates that you are unaware of it, or choose to not acknowledge it.
Let me summarize the explicatory contents of the referenced page as it was
yesterday,
lim 1/oo=0
The information contained there is not static.
Like many of us on this forum, I do not have a password to the CRC, so when page
blocks are rotated, I have no access to them either. I have found it very
educational to browse the CRC online, and personally own a copy of the CRC Handbook
of Chemistry and Physics.
"The limit approximation of one divided by infinity is zero." - Ross A. Finlayson,
sci.math.moderated
(http://www.dejanews.com/[ST_rn=ps]/getdoc.xp?AN=453663311)
Ross A. Finlayson
Andrew Martin
>hahaha ha hehe he ha haahah... :-)
>
>Nathan the Great
>Age 11
>
Is it just me or does every hate arrogant little genius pricks like this
guy? Or is it okay to have an ego the size of china when you're 11?
Zdislav V. Kovarik
[discussing N vs. P(N)]
>As each is an infinite set, it ispossible to match terms one-to-one from
>any subset of N to the a subset of the power set of N. It is not possible
>to match one-to-one N to P(N), because each is an infinite set. Bijection
>vice versa, and for any infinite set, thus infinite sets are equivalent, as
>each infinite set has an infinite number of elements beyond any given
>subset.
This is not mathematics of definitions and proofs, but
rather of beliefs and feelings. Here no distinction is
made between theorems and conjectures, or between proofs
and plausibility arguments. And definitions are not a
requirement because one "knows" what the terms mean. If
a term happens to have more than one meaning, so much the
better; the meanings can be switched, preferably without
warning, at the convenience of the user. The context can
likewise be switched, for example from cardinal numbers
to some vaguely intuitive integers (then you can have
negative infinity as the result of counting the elements
in the difference of two sets).
(I amused my friends by describing the taste of rice
cakes as "negative"; you have to add some flavour to them
to make them taste neutral.)
Most of all, a statement and its negation can coexist
because they were reconciled by the good will of the
proposer. Hence, such a set of statements is immune to
criticism. You can prove the negation of the other guy's
statement without disproving that statement. The other guy
won't tell a proof from an opinion.
Do visit Ross's website, especially the "fun" and "happy"
pages (reprinting):
http://www.tomco.net/~raf/
Have fun, ZVK(Slavek)
Brian David Rothbach
>
> Dave Seaman wrote:
>
> > Do you conjecture that there is a one-to-one correspondence between N
> > and the power set P(N)?
> >
> > Mind you, I haven't even asked you to produce one. I am simply asking
> > whether you consider it possible for a bijection N <--> P(N) to exist.
>
> First, are you talking about a STATIC or a DYNAMIC bijection?
> Second, is there a bijection between N and the set of all odd numbers (a
> subset of N - that has only one member for every two members in N).
>
>
> Nathan the Great
> Age 11
Since one uses a function to compare two sets, a bijection is always
static.
Anticipating your comments, Cantor's diagonal does cause a change in a
bijection, but rather for a fixed attempt at a bijection from N to R,
gives a reason why that fixed attempt is not a bijection.
As for a bijection from N to all the odd numbers, the function
f(x)=2x-1 seems to work easily enough. (If we want to think of this in
terms of odometers, you have two odometers that start at 1 miles, but
the first odometer measures miles and the second measures half miles.
What numbers are on the second odometer when the first odometer reaches
1,2,3,...?)
Brian Rothbach
Dave Seaman
>> >Two sets are equivalent if there is a cardinally equivalent bijection between them
>> >or they are equivalent. Each infinite set is equivalent. I use "equivalent"
>> >perhaps a bit differently than Cantor.
>> Then why did you not specify in your first posting on the subject that
>> you were using the word "equivalent" in a nonstandard way?
>I never felt that I was using "equivalent" in a non-standard way, that is, representing
>the equality of quantity of elements of a set, and throughout this thread and others, I
>have not used the term "equivalent" in a way that violates this defintion
You admitted just above that you use "equivalent" in a different way than
Cantor. Cantor's meaning of "equivalent", meaning the existence of a
bijection, is the accepted meaning of the word in set theory. Therefore, you
are using "equivalent" in a nonstandard way.
>> >> Equivalence, by definitioin, means a bijection exists. [ ... ]
>> >Equivalence, if it thus states, requires redefinition.
In other words, if the rest of the world disagrees with you, then it is
necessary for the rest of the world to change in order to conform to your
definition?
>Equivalence requires redefinition.
Welcome to the wonderful world of crankdom.
>I can understand that the quantity of elements of a powerset of a set is 2^n, where n
>is the number of elements of the set. This is similar to probability result
>determination
>Any infinite set has an infinite number of elements or arguments. Thus, any infinite
>set's (eg N, Q, R) cardinality is infinity, and these sets are equivalent, as are they
>to the powersets of N, Q, and R.
The cardinality of N is aleph_0. The cardinality of the power set of N
is 2^aleph_0. Cantor proved that aleph_0 < 2^aleph_0, and in fact 2^n
< n for every cardinal number n, finite or infinite.
Dave Seaman
By definition, a set A is smaller than a set B (|A| < |B|) if there
exists an injection from A into B, but no injection from B into A.
You need the Axiom of Choice to show that all sets are ordered in this
way (given any two sets A and B, exactly one of the relations |A| <
|B|, |A| = |B|, or |A| > |B| holds). Even without AC, some sets can
still be compared. For example, proving that |N| < |R| does not
require AC.
The fact remains that you have contradicted yourself. If you think
there are no bijections between infinite sets, then it follows from the
definition that you are claiming no two infinite sets can possibly be
equivalent.
Dave Seaman
>:What you can prove is that |R| = 2^aleph_0. The cardinal 2^aleph_0 is
>:called c, the cardinality of the continuum. It is known that c >
>:aleph_0 and also that c >= aleph_1.
>(This latter part only with some form of choice.)
I'm not sure I follow that. Which statement is untrue in ZF?
1. 2^aleph_0 is an aleph (I'm not saying which one).
2. Any two alephs are comparable. (I'm not saying any two
cardinals are comparable.)
3. Aleph_1 is the cardinality of the set of countable ordinals.
4. There are no alephs between aleph_0 and aleph_1. There may be
cardinals "out to the side," but none in the aleph hierarchy
between aleph_0 and aleph_1.
5. Therefore, c >= aleph_1.
Putting it another way, I believe CH is equivalent to the assertion
that every uncountable subset of the reals has cardinality 2^aleph_0,
and this does not require AC.
Dave Seaman
Ross A. Finlayson <R...@TOMCO.NET> wrote:
>The fact that this URL points to the CRC Concise Encyclopedia of Mathematics
>illustrates that you are unaware of it, or choose to not acknowledge it.
>
>Let me summarize the explicatory contents of the referenced page as it was
>yesterday,
>
>lim 1/oo=0
>Like many of us on this forum, I do not have a password to the CRC, so when page
>blocks are rotated, I have no access to them either. I have found it very
>educational to browse the CRC online, and personally own a copy of the CRC Handbook
>of Chemistry and Physics.
>
>"The limit approximation of one divided by infinity is zero." - Ross A. Finlayson,
You may find it instructive to look up "Cantor's Theorem" in that same
volume.
Cantor's Theorem
----------------
The CARDINAL NUMBER of any set is lower than the CARDINAL
NUMBER of the set of all its subsets. A COROLLARY is that there
is no highest aleph.
-- CRC Concise Encyclopedia of Mathematics
Richard Carr
:Date: 24 Apr 1999 08:05:10 -0500
:From: Dave Seaman <a...@seaman.cc.purdue.edu>
:Newsgroups: sci.math
:Subject: Re: Equivalency of Infinite Sets
:
:In article <Pine.LNX.4.10.990423...@cpw.math.columbia.edu>,
:Richard Carr <ca...@math.columbia.edu> wrote:
:>On 23 Apr 1999, Dave Seaman wrote:
:
:>:What you can prove is that |R| = 2^aleph_0. The cardinal 2^aleph_0 is
:>:called c, the cardinality of the continuum. It is known that c >
:>:aleph_0 and also that c >= aleph_1.
:>(This latter part only with some form of choice.)
:
:I'm not sure I follow that. Which statement is untrue in ZF?
:
: 1. 2^aleph_0 is an aleph (I'm not saying which one).
This one. Actually, it's not outright untrue in ZF, but it is not provable
in ZF. Statements 2-4 are true.
: 2. Any two alephs are comparable. (I'm not saying any two
: cardinals are comparable.)
: 3. Aleph_1 is the cardinality of the set of countable ordinals.
: 4. There are no alephs between aleph_0 and aleph_1. There may be
: cardinals "out to the side," but none in the aleph hierarchy
: between aleph_0 and aleph_1.
: 5. Therefore, c >= aleph_1.
This one needs statement 1 which does not follow from ZF.
:
:Putting it another way, I believe CH is equivalent to the assertion
:that every uncountable subset of the reals has cardinality 2^aleph_0,
Yes, that is correct.
:and this does not require AC.
So is that, but one does need AC to get c>=aleph_1.
:
:--
:
:
Brian M. Scott
wrote:
>Dave Seaman wrote:
>> Ross A. Finlayson <R...@TOMCO.NET> wrote:
>> >Dave Seaman wrote:
>> >> By definition, sets are equivalent if they have the same cardinality. That
>> >> means there is a bijection between them.
>> >Two sets are equivalent if there is a cardinally equivalent bijection between them
>> >or they are equivalent. Each infinite set is equivalent. I use "equivalent"
>> >perhaps a bit differently than Cantor.
>> Then why did you not specify in your first posting on the subject that
>> you were using the word "equivalent" in a nonstandard way?
>> What, exactly, is a "cardinally equivalent bijection?" Is this something
>> different from an ordinary bijection?
>As possible, I will be quite happy to absolutely avoid any use of the words "bijection"
>or "cardinality."
>In the phrase "cardinally equivalent bijection", the term "cardinally" is redundant. A
>bijection is a one-to-one mapping of terms of two separate sets. The term "equivalent"
>is an adjective describing a bijection, so a cardinally equivalent bijection is a
>bijection.
>I never felt that I was using "equivalent" in a non-standard way, that is, representing
>the equality of quantity of elements of a set, and throughout this thread and others, I
>have not used the term "equivalent" in a way that violates this defintion
You appear to be contradicting yourself. A bit earlier you said that
your use of 'equivalent' is perhaps a bit different from Cantor's.
That definition is that two sets are equivalent if and only if there
is a bijection between them. You then said that you were using the
term 'equivalent' to mean that there is a 'cardinally equivalent
bijection' between the two sets. It now appears that what you mean by
'cardinally equivalent bijection' is indeed just the ordinary notion
of a bijection. Thus, your definition of equivalence is no different
from anyone else's.
>> >> Equivalence, by definitioin, means a bijection exists. [ ... ]
>> >Equivalence, if it thus states, requires redefinition.
>> In other words, you didn't have any idea what you were talking about when you
>> made your earlier statements.
>This previous statement of yours is invalid. I do have an understanding of what I have
>been writing.
This is not apparent from your statement. The phrase 'if it thus
states' clearly suggests that you did not previously know the
definition of equivalence that Dave had just given you. Had you known
it, you would presumably have said something like 'I know, but
equivalence requires redefinition because ...'. Moreover, the
definition of equivalence that you've offered in this post is *not* a
redefinition: it's exactly the same as the standard one given by Dave.
That you don't appear to recognize this strongly suggests a
fundamental misunderstanding on your part.
>Equivalence requires redefinition.
Then why have you given the same old definition?
[...]
>Any infinite set has an infinite number of elements or arguments. Thus, any infinite
>set's (eg N, Q, R) cardinality is infinity, and these sets are equivalent, as are they
>to the powersets of N, Q, and R.
You'll have to offer more than a statement of blind faith of you wish
to convince a mathematician. In fact N and Q have cardinality w
(omega); R, P(N), and P(Q) have cardinality 2^w > w; and P(R) has
cardinality 2^(2^w) > 2^w.
Brian M. Scott
Brian M. Scott
<logi...@wolfenet.com> wrote:
>It is meaningless to ask if an infinite set
>is smaller, bigger or equal to another infinite set.
It is meaningless only until one defines what |A| < |B| is to mean.
Once a definition is in place, it is entirely meaningful to use it.
Mathematicians have a definition of this relationship and can easily
prove that |N| < |P(N)|. You are free to argue that the definition
doesn't match your private intuitions about notions of relative size,
but that doesn't render meaningless either the definition or
statements based on it.
Brian M. Scott
Tapio Hurme
denominator. The division result equals to pi/2. How those two infinite
products can be meaningless? The numerator is clearly pi times bigger than
the denominator. Both are infinite sets with bijection of each members: one
bijective member in numerator and one in denominator.
The answer: Make a difference between Archimedean absolute infinity and
relative infinity. "Give me the biggest (=infinity) number and I add always
one" (Euclides).
The equivalency of infinite sets is based on Archimedean infinity concept
that was canceled already by Euclides.
Tapio
Russell Easterly wrote in message <7frpq8$rao$1...@sparky.wolfe.net>...
|
|Dave Seaman <a...@seaman.cc.purdue.edu> wrote in message
|`Besides, if we were to accept your premise that bijection is valid only
|`for finite sets, the logical consequence would be that no two infinite
|`sets would be equivalent, since equivalence is defined by the existence
|`of a bijection. Therefore, your conclusion is contradicted by your own
|`claims.
|
|I don't think that infinite sets can be ordered.
|It is meaningless to ask if an infinite set
|is smaller, bigger or equal to another infinite set.
|
Ross A. Finlayson
Dave Seaman wrote:
> In article <372155BF...@TOMCO.NET>,
> Ross A. Finlayson <R...@TOMCO.NET> wrote:
> >Dave Seaman wrote:
>
> >> >Two sets are equivalent if there is a cardinally equivalent bijection between them
> >> >or they are equivalent. Each infinite set is equivalent. I use "equivalent"
> >> >perhaps a bit differently than Cantor.
>
> >> Then why did you not specify in your first posting on the subject that
> >> you were using the word "equivalent" in a nonstandard way?
>
> >I never felt that I was using "equivalent" in a non-standard way, that is, representing
> >the equality of quantity of elements of a set, and throughout this thread and others, I
> >have not used the term "equivalent" in a way that violates this defintion
>
> You admitted just above that you use "equivalent" in a different way than
> Cantor. Cantor's meaning of "equivalent", meaning the existence of a
> bijection, is the accepted meaning of the word in set theory. Therefore, you
> are using "equivalent" in a nonstandard way.
>
> >> >> Equivalence, by definitioin, means a bijection exists. [ ... ]
>
> >> >Equivalence, if it thus states, requires redefinition.
>
> In other words, if the rest of the world disagrees with you, then it is
> necessary for the rest of the world to change in order to conform to your
> definition?
>
> >Equivalence requires redefinition.
>
> Welcome to the wonderful world of crankdom.
>
Thank you for your warm welcome.
>
> >I can understand that the quantity of elements of a powerset of a set is 2^n, where n
> >is the number of elements of the set. This is similar to probability result
> >determination
>
> >Any infinite set has an infinite number of elements or arguments. Thus, any infinite
> >set's (eg N, Q, R) cardinality is infinity, and these sets are equivalent, as are they
> >to the powersets of N, Q, and R.
>
> The cardinality of N is aleph_0. The cardinality of the power set of N
> is 2^aleph_0. Cantor proved that aleph_0 < 2^aleph_0, and in fact 2^n
> < n for every cardinal number n, finite or infinite.
>
> --
> Dave Seaman dse...@purdue.edu
> Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
> <http://mojo.calyx.net/~refuse/altindex.html>
Could you provide a reference to Cantor's proof? That would help me to establish the
veracity of these claims. I am finding several good references about this, and CH. I would
like to see the actual proof.
If it is comparing one-to-one bijections of N to the powerset of N by subsets of N and the
powerset of N, thus for any subset of N generating a larger number of elements than exist in
the subset of N as the powerset, then it is not considering the actual range and number of
elements of N and E(N), which for each of those two cases is infinite.
For any subset of N, the powerset of said subset of N has a larger number of elements. For
the complete set N, with an infinity of elements, the corresponding powerset also has an
infinity of elements.
The powerset of a set is all the permutations of the elements of the set, each as an element
of the powerset. Consider an augmented set where for each element there is added an element
of the element and zero. In this case, the cardinality would fall between aleph_0 and
aleph_1, using the diagonal method. There are an infinite number of alternate permutations
to the primary set that could produce an infinite number of cardinalities of permutations of
any set.
Consider X={a, b, c}. The powerset P(X) is {a, b, c, ab, ac, bc, abc}. The cardinality of
X, |X|, is 3. The cardinality of P(X), |P(X)|, is 7. Considering a four element X, {a, b,
c, d}, the powerset is {a, b, c, d, ab, ac, ad, bc, bd, cd, abc, acd, bcd, abcd}.
Why was the powerset notation chosen to represent an aleph_1 as compared to an aleph_0, the
"cardinality" of N? What is particular about the powerset notation as the permutation of an
infinite set as opposed to any other permutation of an infinite set that is still infinite?
Powerset is arbitrary, to the extent that an infinite number of alternate permutations can
be considered.
Now considering that aleph_0 is noted as the cardinality of N, and aleph_1 that of R, that
is a bit different, perhaps. For each unit in N, there is not 2^N elements of R, but
infinity elements of R, |N|. To say aleph_0, the cardinality of N, is related to the
cardinality of R by aleph_0=aleph_1=2^aleph0, one might as well say that aleph_1 is
10^aleph_0, 13^aleph_0, or aleph_0^aleph_0. If 2 exponentiated is abitrary, then
aleph_0^aleph_0 is closer to representation of the cardinality of the reals over the domain
of N.
At this point I will note this observation:
aleph_0=aleph_0^aleph_0 = oo
I need to do more research because I do not feel satisfied either way at this point. I have
been learning some more about what is called omega and whatnot.
Infinity in one dimension is infinitely large, and nothing can be larger, and a comparative
infinity in that same dimension is also infinitely large.
I still do not feel that my conjecture or hypothesis has been disproved, and and still am in
search of further enlightenment. I know that some of my statements on this thread have not
been rigorously mathematically correct, that was not the intent at the time, although I do
not write to sci.math with the intention of not being rigorously mathematically correct. I
have read all the messages on this thread to date. At some point I will gather their
represented contents and summarize.
More later, ciao, auf Wiedersehen,
Ross
Brian M. Scott
> Could you provide a reference to Cantor's proof?
It's been given here any number of times recently. Here it is yet
again.
Let X be any set. For any function f : X --> P(X) define
S(f) = {x in X : x is not in f(x)}. Clearly S(f) is a
subset of X. But for each x in X, x is in S(f) if and only
if x is not in f(x), so S(f) is not one of the sets in the
range of f. Hence f does not map X *onto* P(X). Since f
was arbitrary, there is no map of X onto P(X). There clearly
is a map of X *into* P(X), e.g., f(x) = {x}, so it must be
that |P(X)| > |X|.
> If it is comparing one-to-one bijections of N to the powerset of N by subsets of N and the
> powerset of N, thus for any subset of N generating a larger number of elements than exist in
> the subset of N as the powerset, then it is not considering the actual range and number of
> elements of N and E(N), which for each of those two cases is infinite.
This is incoherent.
> For any subset of N, the powerset of said subset of N has a larger number of elements. For
> the complete set N, with an infinity of elements, the corresponding powerset also has an
> infinity of elements.
N *is* a subset of N. Did you perhaps mean your first
statement to refer to only to proper subsets of N? Then
take E = {2n : n in N}, the set of even natural numbers.
It's clearly the same size as N, since the map f(n) = 2n
from N to E is a bijection. It's obviously a proper
subset of N, so apparently you agree that it has more
subsets than it has elements. But there's an obvious
bijection F between P(N) and P(E) given by F(S) =
{f(n) : n in S}, so P(N) is the same size as P(E). That
is, |N| = |E| < |P(E) = |P(N)|, and hence |N| < |P(N)|.
Or did your 'any subset of N' in fact mean only any *finite*
subset of N?
> The powerset of a set is all the permutations of the elements of the set, each as an element
> of the powerset.
No, it isn't. A permutation of a set X is a bijection
from X to X; this isn't at all the same thing as a
subset of X.
> Consider an augmented set where for each element there is added an element
> of the element and zero. In this case, the cardinality would fall between aleph_0 and
> aleph_1, using the diagonal method.
Your first sentence is incomprehensible: an element of
*what* element? There is no cardinality strictly between
aleph_0 and aleph_1.
> Consider X={a, b, c}. The powerset P(X) is {a, b, c, ab, ac, bc, abc}.
You forgot the empty set, which is also a member of P(X).
You have also failed to distinguish between strings (or
finite sequences) and sets; the actual members of P(X) are
0 (empty set), {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and
{a,b,c}. This is not a trivial distinction.
> The cardinality of
> X, |X|, is 3. The cardinality of P(X), |P(X)|, is 7.
No, it's 8.
> Considering a four element X, {a, b,
> c, d}, the powerset is {a, b, c, d, ab, ac, ad, bc, bd, cd, abc, acd, bcd, abcd}.
Same errors.
[...]
> Now considering that aleph_0 is noted as the cardinality of N, and aleph_1 that of R,
No. Aleph_1 is the first cardinal number larger than
aleph_0. Whether |R| = aleph_1 is consistent with but
also independent of ZFC (or any other customary
axiomatization of set theory). The statement that they
are equal is the Continuum Hypothesis.
[snip more confusion in this vein]
> aleph_0=aleph_0^aleph_0 = oo
The first '=' is false, and the second is meaningless
until you define 'oo'.
Brian M. Scott
Bennett Standeven
On Fri, 23 Apr 1999, Ross A. Finlayson wrote:
> There is an infinity of infinite sets, each being equivalent to each and any
> other infinite set.
>
> Mike Oliver wrote:
>
> > "Ross A. Finlayson" wrote:
> >
> > > I conjecture that any infinite set is equivalent to any other infinite
> > > set, and to each infinite set,
> >
> > Conjecture all you want. You know what Popper said about conjectures?
> > He said that a good conjecture is one that, if false, can in principle
> > be proven false, and that a conjecture that *has* been proven false
> > has in this sense already demonstrated its merit.
> >
> > So give yourself a merit badge.
>
> My preceding conjecture is true. Let's call it a hypothesis.
>
> Consider R, the domain of real numbers. Also, consider N, the range of
> integers of natural or whole numbers. The definition of each of these two
> sets is an infinitely bounded (that is, unbounded) series of numbers.
> Between zero and one, inclusive, there is an infinite quantity of real
> numbers, yet only two integers, zero and one.
Correct. However, these two sets contain the same number of elements.
Bennett Standeven
On 24 Apr 1999, Dave Seaman wrote:
> In article <Pine.LNX.4.10.990423...@cpw.math.columbia.edu>,
> Richard Carr <ca...@math.columbia.edu> wrote:
> >On 23 Apr 1999, Dave Seaman wrote:
>
> >:What you can prove is that |R| = 2^aleph_0. The cardinal 2^aleph_0 is
> >:called c, the cardinality of the continuum. It is known that c >
> >:aleph_0 and also that c >= aleph_1.
> >(This latter part only with some form of choice.)
>
> I'm not sure I follow that. Which statement is untrue in ZF?
>
> 1. 2^aleph_0 is an aleph (I'm not saying which one).
This one requires AC. Otherwise the continuum may not be well ordered.
> 2. Any two alephs are comparable. (I'm not saying any two
> cardinals are comparable.)
> 3. Aleph_1 is the cardinality of the set of countable ordinals.
> 4. There are no alephs between aleph_0 and aleph_1. There may be
> cardinals "out to the side," but none in the aleph hierarchy
> between aleph_0 and aleph_1.
> 5. Therefore, c >= aleph_1.
>
> Putting it another way, I believe CH is equivalent to the assertion
> that every uncountable subset of the reals has cardinality 2^aleph_0,
> and this does not require AC.
Without AC, these are not equivalent. (But adding two more cases of GCH
for c and 2^c would allow proof of the usual form, as I recall.)
Hal Daume III
am bored of trying to explain things over and over again just to get a
"blind faith" response. If you truly are interested in this topic, I
would highly recommend the book:
Introduction to Set Theory
by K. Hrbacek and T. Jech
Published by Marcel Dekker, Inc.
It's quite a good introduction to the topic. The chapters on cardinals,
ordinals, natural numbers, reals, and the topology of the real line, in
addition to the final chapter on AC and CH and forcing might interest
you.
- Hal
On Sat, 24 Apr 1999, Ross A. Finlayson wrote:
> If it is comparing one-to-one bijections of N to the powerset of N by subsets of N and the
> powerset of N, thus for any subset of N generating a larger number of elements than exist in
> the subset of N as the powerset, then it is not considering the actual range and number of
> elements of N and E(N), which for each of those two cases is infinite.
>
> For any subset of N, the powerset of said subset of N has a larger number of elements. For
> the complete set N, with an infinity of elements, the corresponding powerset also has an
> infinity of elements.
>
> The powerset of a set is all the permutations of the elements of the set, each as an element
> of the powerset. Consider an augmented set where for each element there is added an element
> of the element and zero. In this case, the cardinality would fall between aleph_0 and
> aleph_1, using the diagonal method. There are an infinite number of alternate permutations
> to the primary set that could produce an infinite number of cardinalities of permutations of
> any set.
>
> Consider X={a, b, c}. The powerset P(X) is {a, b, c, ab, ac, bc, abc}. The cardinality of
> X, |X|, is 3. The cardinality of P(X), |P(X)|, is 7. Considering a four element X, {a, b,
> c, d}, the powerset is {a, b, c, d, ab, ac, ad, bc, bd, cd, abc, acd, bcd, abcd}.
>
-----------------------------------------------------------------------
"Men are born ignorant, not stupid; they are made stupid by education."
- Bertrand Russell
-----------------------------------------------------------------------
Mike Oliver
"Ross A. Finlayson" wrote:
> What I am finding instructive is this paradox:
>
> http://www.astro.virginia.edu/~eww6n/math/CantorsParadox.html
>
> "The Set of all Sets is its own Power Set. Therefore, the Cardinality of the Set of all
> Sets must be bigger than itself. "
>
> This paradox does not exist if each infinite set is equivalent.
Let's get the grammar flame out of the way: "each infinite set is
equivalent" to what? Yes, I know what you mean; you ought to say
it right anyway.
Second and more important, you don't eliminate this paradox simply by choosing not
to look at it. Your notion of "equivalent," defined as
A equivalent to B iff either (A,B finite and |A| = |B| ) or (A,B both infinite)
is a perfectly good equivalence relation, and I'm willing to call it
"equivalence" for the purposes of this discussion. (Most mathematicians
don't use the word "equivalent" to mean "of the same cardinality" anyway;
if they want a single word they say "equinumerous" or (archaically) "equipollent.")
But it *doesn't* eliminate the paradox, because we can just put some
other word to the notion "there exists a bijection between A and B",
and if you assume that a set of all sets exists and that it satisfies
certain natural properties, then you get the same contradiction back
again.
If you really insist on having a set of all sets, then you need to
find some way around the paradox. None of the ones that exist seem
particularly motivated. But you can look up "New Foundations" or
"Modern Logic" (both due to Quine) if you're interested in this
sort of thing.
Mike Oliver
"Brian M. Scott" wrote:
> Ross A. Finlayson wrote:
>> Brian M. Scott wrote:
>>> It's not a good idea to use the word 'range' here. In the standard
>>> standard mathematical sense of the word R, not being a relation, has no
>>> range.
>> Instead of range, domain.
> Same problem. It would really help if you would learn basic standard
> terminology.
Actually, I do think it is *fairly* standard to use range(R) to mean
{ y | (exists x) xRy } and domain(R) for { x | (exists y) xRy }. For
example I believe "range" in this sense is one of the Goedel operations.
I haven't bothered to check whether that's the way Ross is using the
words.
Ross A. Finlayson
Brian M. Scott wrote:
> Ross A. Finlayson wrote:
>
> > Could you provide a reference to Cantor's proof?
>
> It's been given here any number of times recently. Here it is yet
> again.
>
> Let X be any set. For any function f : X --> P(X) define
> S(f) = {x in X : x is not in f(x)}. Clearly S(f) is a
> subset of X. But for each x in X, x is in S(f) if and only
> if x is not in f(x), so S(f) is not one of the sets in the
> range of f. Hence f does not map X *onto* P(X). Since f
> was arbitrary, there is no map of X onto P(X). There clearly
> is a map of X *into* P(X), e.g., f(x) = {x}, so it must be
> that |P(X)| > |X|.
>
> > If it is comparing one-to-one bijections of N to the powerset of N by subsets of N and the
> > powerset of N, thus for any subset of N generating a larger number of elements than exist in
> > the subset of N as the powerset, then it is not considering the actual range and number of
> > elements of N and E(N), which for each of those two cases is infinite.
>
> This is incoherent.
>
Replace E(N) with P(N).
>
> > For any subset of N, the powerset of said subset of N has a larger number of elements. For
> > the complete set N, with an infinity of elements, the corresponding powerset also has an
> > infinity of elements.
>
> N *is* a subset of N. Did you perhaps mean your first
> statement to refer to only to proper subsets of N? Then
> take E = {2n : n in N}, the set of even natural numbers.
> It's clearly the same size as N, since the map f(n) = 2n
> from N to E is a bijection. It's obviously a proper
> subset of N, so apparently you agree that it has more
> subsets than it has elements. But there's an obvious
> bijection F between P(N) and P(E) given by F(S) =
> {f(n) : n in S}, so P(N) is the same size as P(E). That
> is, |N| = |E| < |P(E) = |P(N)|, and hence |N| < |P(N)|.
>
> Or did your 'any subset of N' in fact mean only any *finite*
> subset of N?
>
For any finite subset of N, the powerset of this finite subset of N is larger than the given
finite subset of N.
>
> > The powerset of a set is all the permutations of the elements of the set, each as an element
> > of the powerset.
>
> No, it isn't. A permutation of a set X is a bijection
> from X to X; this isn't at all the same thing as a
> subset of X.
>
> > Consider an augmented set where for each element there is added an element
> > of the element and zero. In this case, the cardinality would fall between aleph_0 and
> > aleph_1, using the diagonal method.
>
> Your first sentence is incomprehensible: an element of
> *what* element? There is no cardinality strictly between
> aleph_0 and aleph_1.
>
> > Consider X={a, b, c}. The powerset P(X) is {a, b, c, ab, ac, bc, abc}.
>
> You forgot the empty set, which is also a member of P(X).
> You have also failed to distinguish between strings (or
> finite sequences) and sets; the actual members of P(X) are
> 0 (empty set), {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, and
> {a,b,c}. This is not a trivial distinction.
>
My mistake. Including the empty set leads to the cardinality of a powerset of a three element to
be 2^3, 8.
>
> > The cardinality of
> > X, |X|, is 3. The cardinality of P(X), |P(X)|, is 7.
>
> No, it's 8.
>
> > Considering a four element X, {a, b,
> > c, d}, the powerset is {a, b, c, d, ab, ac, ad, bc, bd, cd, abc, acd, bcd, abcd}.
>
> Same errors.
>
Adding the empty set, and changing notation, P(X) where |X|=4 is { {}, {a}, {b}, {c}, {d}, {a, b},
{a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}
}, the cardinality of which is 16, not 4^2.
>
> [...]
>
> > Now considering that aleph_0 is noted as the cardinality of N, and aleph_1 that of R,
>
> No. Aleph_1 is the first cardinal number larger than
> aleph_0. Whether |R| = aleph_1 is consistent with but
> also independent of ZFC (or any other customary
> axiomatization of set theory). The statement that they
> are equal is the Continuum Hypothesis.
>
> [snip more confusion in this vein]
>
> > aleph_0=aleph_0^aleph_0 = oo
>
> The first '=' is false, and the second is meaningless
> until you define 'oo'.
>
> Brian M. Scott
Removing from R the open intervals for each (n, n-1) for each n in N leaves N.
The removed intervals are each infinite sets of ranges of R, the remainder
is the infinite set N.
Placing each element of N over infinity gives a unit range.
limit N0/x {x:oo->0} = 0
limit Noo/x {x:0->oo} = 1
Thus, N maps as a function to the real number line from zero to one.
Thus, N is surjective onto the reals.
Ross A. Finlayson
> }, the cardinality of which is 16, not 4^2.
>
16, 2^4
Ross A. Finlayson
Dave Seaman wrote:
> In article <37216DF8...@TOMCO.NET>,
> Ross A. Finlayson <R...@TOMCO.NET> wrote:
>
> >The fact that this URL points to the CRC Concise Encyclopedia of Mathematics
> >illustrates that you are unaware of it, or choose to not acknowledge it.
> >
> >Let me summarize the explicatory contents of the referenced page as it was
> >yesterday,
> >
> >lim 1/oo=0
>
> >Like many of us on this forum, I do not have a password to the CRC, so when page
> >blocks are rotated, I have no access to them either. I have found it very
> >educational to browse the CRC online, and personally own a copy of the CRC Handbook
> >of Chemistry and Physics.
> >
> >"The limit approximation of one divided by infinity is zero." - Ross A. Finlayson,
>
> You may find it instructive to look up "Cantor's Theorem" in that same
> volume.
>
> Cantor's Theorem
> ----------------
> The CARDINAL NUMBER of any set is lower than the CARDINAL
> NUMBER of the set of all its subsets. A COROLLARY is that there
> is no highest aleph.
>
> -- CRC Concise Encyclopedia of Mathematics
>
> --
> Dave Seaman dse...@purdue.edu
> Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
> <http://mojo.calyx.net/~refuse/altindex.html>
What I am finding instructive is this paradox:
http://www.astro.virginia.edu/~eww6n/math/CantorsParadox.html
"The Set of all Sets is its own Power Set. Therefore, the Cardinality of the Set of all
Sets must be bigger than itself. "
This paradox does not exist if each infinite set is equivalent.
Ross
Brian M. Scott
> Brian M. Scott wrote:
> > Ross A. Finlayson wrote:
> > > If it is comparing one-to-one bijections of N to the powerset of N by subsets of N and the
> > > powerset of N, thus for any subset of N generating a larger number of elements than exist in
> > > the subset of N as the powerset, then it is not considering the actual range and number of
> > > elements of N and E(N), which for each of those two cases is infinite.
> > This is incoherent.
> Replace E(N) with P(N).
If I had merely wondered what you meant by E(N), I'd have asked; in fact
I assumed that you meant P(N). The *sentence* is incoherent; please
rewrite it in English.
> > > For any subset of N, the powerset of said subset of N has a larger number of elements. For
> > > the complete set N, with an infinity of elements, the corresponding powerset also has an
> > > infinity of elements.
[...]
> > Or did your 'any subset of N' in fact mean only any *finite*
> > subset of N?
> For any finite subset of N, the powerset of this finite subset of N is larger than the given
> finite subset of N.
I certainly won't argue, since the power set of any set at all is larger
than the given set. But please try to be more careful: we can't
reasonably be expected to guess that 'any subset' means 'any finite
subset'.
[...]
> > > aleph_0=aleph_0^aleph_0 = oo
> > The first '=' is false, and the second is meaningless
> > until you define 'oo'.
> Removing from R the open intervals for each (n, n-1) for each n in N leaves N.
Actually it doesn't, since it also leaves all of the negative real
numbers, but you could repair matters by using Z instead of N in both
places.
> The removed intervals are each infinite sets of ranges of R, the remainder
> is the infinite set N.
It's not a good idea to use the word 'range' here. In the standard
standard mathematical sense of the word R, not being a relation, has no
range.
> Placing each element of N over infinity gives a unit range.
You still haven't defined infinity or explained what n/infinity (for n
in N) means. Until you give these definitions, it's all just
meaningless marks on paper.
> limit N0/x {x:oo->0} = 0
> limit Noo/x {x:0->oo} = 1
You'll have to explain your notation; these two lines make no sense as
they stand.
> Thus, N maps as a function to the real number line from zero to one.
> Thus, N is surjective onto the reals.
This is utter nonsense: N isn't a function in the first place.
Brian M. Scott
Ross A. Finlayson
Brian M. Scott wrote:
> Ross A. Finlayson wrote:
>
> > Brian M. Scott wrote:
>
> > > Ross A. Finlayson wrote:
>
> > > > If it is comparing one-to-one bijections of N to the powerset of N by subsets of N and the
> > > > powerset of N, thus for any subset of N generating a larger number of elements than exist in
> > > > the subset of N as the powerset, then it is not considering the actual range and number of
> > > > elements of N and E(N), which for each of those two cases is infinite.
>
> > > This is incoherent.
>
> > Replace E(N) with P(N).
>
> If I had merely wondered what you meant by E(N), I'd have asked; in fact
> I assumed that you meant P(N). The *sentence* is incoherent; please
> rewrite it in English.
>
> > > > For any subset of N, the powerset of said subset of N has a larger number of elements. For
> > > > the complete set N, with an infinity of elements, the corresponding powerset also has an
> > > > infinity of elements.
>
> [...]
>
> > > Or did your 'any subset of N' in fact mean only any *finite*
> > > subset of N?
>
> > For any finite subset of N, the powerset of this finite subset of N is larger than the given
> > finite subset of N.
>
> I certainly won't argue, since the power set of any set at all is larger
> than the given set. But please try to be more careful: we can't
> reasonably be expected to guess that 'any subset' means 'any finite
> subset'.
>
> [...]
>
> > > > aleph_0=aleph_0^aleph_0 = oo
>
> > > The first '=' is false, and the second is meaningless
> > > until you define 'oo'.
>
> > Removing from R the open intervals for each (n, n-1) for each n in N leaves N.
>
> Actually it doesn't, since it also leaves all of the negative real
> numbers, but you could repair matters by using Z instead of N in both
> places.
>
You are correct, I was only considering R from zero positively, and N as integers from zero upwards.
Removing from non-negative R the open intervals (n, n-1) for each n in N leaves N.
>
> > The removed intervals are each infinite sets of ranges of R, the remainder
> > is the infinite set N.
>
> It's not a good idea to use the word 'range' here. In the standard
> standard mathematical sense of the word R, not being a relation, has no
> range.
>
Instead of range, domain.
>
> > Placing each element of N over infinity gives a unit range.
>
> You still haven't defined infinity or explained what n/infinity (for n
> in N) means. Until you give these definitions, it's all just
> meaningless marks on paper.
>
> > limit N0/x {x:oo->0} = 0
> > limit Noo/x {x:0->oo} = 1
>
> You'll have to explain your notation; these two lines make no sense as
> they stand.
>
N0 is the zeroeth element of N, 0. Noo is the infinity element of N.
N0/x {x:oo->0+} = 0
Noo/x {x:0+->oo} = (oo,1]
Infinity is increase without bound.
>
> > Thus, N maps as a function to the real number line from zero to one.
> > Thus, N is surjective onto the reals.
>
> This is utter nonsense: N isn't a function in the first place.
>
> Brian M. Scott
N is a step function of R, floor R. N is also N by itself. As N goes from zero to infinity, it can be
mapped to all of the infinite real numbers from zero to one, and thus as the reals from zero to one
can be mapped to R, as can N.
The function that describes R from 0 to 1 using N is 1/n in N as n goes from infinity to one.
For these statements, I have been including zero in N. The set of Z from zero upwards would be the
same.
I'm still in the process of collating information and trying to find logical truths. I'm somewhat of
a skeptic and cynic. At the same time , I do not deny valid proof. Also, if I am wrong I do not have
a problem accepting that, of course, I prefer being right.
I'm still considering many statements and counter-statements of this thread, I know that I have not
been exactly accurate in all cases, and do plan to establish rigorous proofs of any summary materials.
Ciao,
Ross F.
Brian M. Scott
> Brian M. Scott wrote:
> > Ross A. Finlayson wrote:
[After removing from R the intervals of the form (n, n+1):]
> > > The removed intervals are each infinite sets of ranges of R, the remainder
> > > is the infinite set N.
> > It's not a good idea to use the word 'range' here. In the standard
> > standard mathematical sense of the word R, not being a relation, has no
> > range.
> Instead of range, domain.
Same problem. It would really help if you would learn basic standard
terminology.
> > > Placing each element of N over infinity gives a unit range.
> > You still haven't defined infinity or explained what n/infinity (for n
> > in N) means. Until you give these definitions, it's all just
> > meaningless marks on paper.
> > > limit N0/x {x:oo->0} = 0
> > > limit Noo/x {x:0->oo} = 1
> > You'll have to explain your notation; these two lines make no sense as
> > they stand.
> N0 is the zeroeth element of N, 0. Noo is the infinity element of N.
If N0 is 0, why not just write '0', like everyone else? In its normal
ordering (0, 1, 2, 3, ...) the set N has no 'infinity element' (or
'infinitieth element').
> N0/x {x:oo->0+} = 0
> Noo/x {x:0+->oo} = (oo,1]
You still haven't explained notations '{x:oo->0+}' and '{x:0+->oo}'.
You've also introduced the unexplained '(oo,1]', and you haven't
explained how a quotient (or the limit of a quotient?) is a whole
interval. And since Noo remains undefined, so does Noo/x.
> Infinity is increase without bound.
So Noo is the 'increase without bound element of N'? Does that make any
sense to you?
> > > Thus, N maps as a function to the real number line from zero to one.
> > > Thus, N is surjective onto the reals.
> > This is utter nonsense: N isn't a function in the first place.
> N is a step function of R, floor R.
No, it isn't; it's a subset of R. But it's no wonder that you're having
trouble if you're confusing the set N of natural numbers with a certain
function whose domain is R and whose range is N.
> N is also N by itself. As N goes from zero to infinity,
So now N is a subset of R, a function on R, and something that 'goes'.
Oy, veh.
> it can be
> mapped to all of the infinite real numbers from zero to one,
There are no infinite real numbers; I expect that you mean 'the infinite
collection of real numbers between zero and one'. In fact N cannot be
mapped onto [0,1], and throwing around undefined symbols won't change
the fact.
> and thus as the reals from zero to one
> can be mapped to R, as can N.
> The function that describes R from 0 to 1 using N is 1/n in N as n goes from infinity to one.
This gives you the set {1, 1/2, 1/3, 1/4, 1/5, ...}; please show me what
value of n gives the real number 2/3.
> I'm still in the process of collating information and trying to find logical truths.
Frankly, you're wasting your time trying to run before you can crawl.
Someone recommended the book by Hrbacek and Jech; I'm not at all sure
that you're ready for it, but it's an excellent book if you *can* handle
it (and if you can find it). There are certainly other (and easier)
books available as well, though I'm afraid that I haven't any handy to
recommend.
Brian M. Scott
Dave Seaman
>http://www.astro.virginia.edu/~eww6n/math/CantorsParadox.html
The paradox also does not exist if there is no set of all sets, which
there isn't in ZF(C). This resolution has the advantage of not being
contradictory, in light of Cantor's Theorem.
Ross A. Finlayson
Brian M. Scott wrote:
> Ross A. Finlayson wrote:
>
> > Brian M. Scott wrote:
>
> > > Ross A. Finlayson wrote:
>
> [After removing from R the intervals of the form (n, n+1):]
>
> > > > The removed intervals are each infinite sets of ranges of R, the remainder
> > > > is the infinite set N.
>
> > > It's not a good idea to use the word 'range' here. In the standard
> > > standard mathematical sense of the word R, not being a relation, has no
> > > range.
>
> > Instead of range, domain.
>
> Same problem. It would really help if you would learn basic standard
> terminology.
>
> > > > Placing each element of N over infinity gives a unit range.
>
> > > You still haven't defined infinity or explained what n/infinity (for n
> > > in N) means. Until you give these definitions, it's all just
> > > meaningless marks on paper.
>
> > > > limit N0/x {x:oo->0} = 0
> > > > limit Noo/x {x:0->oo} = 1
>
> > > You'll have to explain your notation; these two lines make no sense as
> > > they stand.
>
> > N0 is the zeroeth element of N, 0. Noo is the infinity element of N.
>
> If N0 is 0, why not just write '0', like everyone else? In its normal
> ordering (0, 1, 2, 3, ...) the set N has no 'infinity element' (or
> 'infinitieth element').
>
N has infinity elements, Noo is infinity.
>
> > N0/x {x:oo->0+} = 0
> > Noo/x {x:0+->oo} = (oo,1]
>
> You still haven't explained notations '{x:oo->0+}' and '{x:0+->oo}'.
> You've also introduced the unexplained '(oo,1]', and you haven't
> explained how a quotient (or the limit of a quotient?) is a whole
> interval. And since Noo remains undefined, so does Noo/x.
>
The notation {x:00->0} means x goes from infinity to approaching approaching zero.
The notation (oo,1] is [1, oo).
>
> > Infinity is increase without bound.
>
> So Noo is the 'increase without bound element of N'? Does that make any
> sense to you?
>
> > > > Thus, N maps as a function to the real number line from zero to one.
> > > > Thus, N is surjective onto the reals.
>
> > > This is utter nonsense: N isn't a function in the first place.
>
> > N is a step function of R, floor R.
>
> No, it isn't; it's a subset of R. But it's no wonder that you're having
> trouble if you're confusing the set N of natural numbers with a certain
> function whose domain is R and whose range is N.
>
The floor function applied to non-negative R is N.
>
> > N is also N by itself. As N goes from zero to infinity,
>
> So now N is a subset of R, a function on R, and something that 'goes'.
> Oy, veh.
>
I should have stated as n in N goes from zero to infinity.
>
> > it can be
> > mapped to all of the infinite real numbers from zero to one,
>
> There are no infinite real numbers; I expect that you mean 'the infinite
> collection of real numbers between zero and one'. In fact N cannot be
> mapped onto [0,1], and throwing around undefined symbols won't change
> the fact.
>
> > and thus as the reals from zero to one
> > can be mapped to R, as can N.
>
> > The function that describes R from 0 to 1 using N is 1/n in N as n goes from infinity to one.
>
> This gives you the set {1, 1/2, 1/3, 1/4, 1/5, ...}; please show me what
> value of n gives the real number 2/3.
>
You're correct, that is an obvious fault that I should have caught.
This might be better, j and k in N, as j goes from zero to infinity and k from infinity to zero the
union of each j/k.
>
> > I'm still in the process of collating information and trying to find logical truths.
>
> Frankly, you're wasting your time trying to run before you can crawl.
> Someone recommended the book by Hrbacek and Jech; I'm not at all sure
> that you're ready for it, but it's an excellent book if you *can* handle
> it (and if you can find it). There are certainly other (and easier)
> books available as well, though I'm afraid that I haven't any handy to
> recommend.
>
> Brian M. Scott
I'm always amenable to reading more.
Ross A. Finlayson
> ...>
> > > N is also N by itself. As N goes from zero to infinity,
> >
> > So now N is a subset of R, a function on R, and something that 'goes'.
> > Oy, veh.
> >
>
> I should have stated as n in N goes from zero to infinity.
>
> >
> > > it can be
> > > mapped to all of the infinite real numbers from zero to one,
> >
> > There are no infinite real numbers; I expect that you mean 'the infinite
> > collection of real numbers between zero and one'. In fact N cannot be
> > mapped onto [0,1], and throwing around undefined symbols won't change
> > the fact.
> >
> > > and thus as the reals from zero to one
> > > can be mapped to R, as can N.
> >
> > > The function that describes R from 0 to 1 using N is 1/n in N as n goes from infinity to one.
> >
> > This gives you the set {1, 1/2, 1/3, 1/4, 1/5, ...}; please show me what
> > value of n gives the real number 2/3.
> >
>
> You're correct, that is an obvious fault that I should have caught.
>
> This might be better, j and k in N, as j goes from zero to infinity and k from infinity to zero the
> union of each j/k.
>
This is still a problem, as that set is only containing the rational numbers, as well as an infinite
number of reducible rational numbers, in fact, this construct repeats each rational number infinitely.
To construct the irrational numbers as well and thus the reals requires a method to construct an
infinite quantity of irrational numbers between each pair of rational numbers, which being infinite, is
infinity.
Consider 1/x as x goes from one to infinity in N. This produces 1, 1/2, 1/3, ..1/oo. 1/oo is not real,
if it is irrational, then it can be used as the basis for irrationality. THen, this quantity could be
used additively to create all rational and irrational numbers from 0 to 1.
That is, the set made up of the elements of each n in N of ( n * 1/oo). This set is the reals from
zero to one.
>
> >
> > > I'm still in the process of collating information and trying to find logical truths.
> >
> > Frankly, you're wasting your time trying to run before you can crawl.
> > Someone recommended the book by Hrbacek and Jech; I'm not at all sure
> > that you're ready for it, but it's an excellent book if you *can* handle
> > it (and if you can find it). There are certainly other (and easier)
> > books available as well, though I'm afraid that I haven't any handy to
> > recommend.
> >
> > Brian M. Scott
>
> I'm always amenable to reading more.
>
Russell Easterly
'For any subset of N, the powerset of said subset of N has a larger number
of elements.
Not completely true.
The powerset of null is null.
Thomas Womack
>'Brian M. Scott wrote:
>'For any subset of N, the powerset of said subset of N has a larger
number
>of elements.
>
>Not completely true.
>The powerset of null is null.
Um, isn't P(\emptyset) = \{\emptyset\}, with cardinality 1?
Tom
Dave Seaman
>of elements.
>Not completely true.
>The powerset of null is null.
Since the empty set is a subset of itself, we get P(0) = {0}, which
(having one element) is a larger set than 0. If fact, if we apply the
Cantor diagonal argument to your empty mapping L: 0 -> P(0), we get the
empty set as a member of P(0) that is not in the range of L.
In general, if X has n elements, then P(X) has 2^n elements, and Cantor
proved that n < 2^n for every n (even if n is infinite).
David C. Ullrich
a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
> In article <7fuqou$13c$1...@sparky.wolfe.net>,
> Russell Easterly <logi...@wolfenet.com> wrote:
> >'Brian M. Scott wrote:
> >of elements.
>
> >Not completely true.
> >The powerset of null is null.
>
> Since the empty set is a subset of itself, we get P(0) = {0}, which
> (having one element) is a larger set than 0. If fact, if we apply the
> Cantor diagonal argument to your empty mapping L: 0 -> P(0), we get the
> empty set as a member of P(0) that is not in the range of L.
>
> In general, if X has n elements, then P(X) has 2^n elements, and Cantor
> proved that n < 2^n for every n (even if n is infinite).
Right. Hence {} has more elements than P({}) does: P({}) has
only one element, while {} has _zero_ elements!
Maybe you're forgetting that zero is everything?
David C. "sorry, the weather's very strange here lately" Ullrich
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Brian M. Scott
wrote:
>"Brian M. Scott" wrote:
>> Ross A. Finlayson wrote:
>>> Brian M. Scott wrote:
>>>> It's not a good idea to use the word 'range' here. In the standard
>>>> standard mathematical sense of the word R, not being a relation, has no
>>>> range.
>>> Instead of range, domain.
>> Same problem. It would really help if you would learn basic standard
>> terminology.
>Actually, I do think it is *fairly* standard to use range(R) to mean
>{ y | (exists x) xRy } and domain(R) for { x | (exists y) xRy }. For
>example I believe "range" in this sense is one of the Goedel operations.
That's exactly the way I had in mind, and I'd have called it
bog-standard.
>I haven't bothered to check whether that's the way Ross is using the
>words.
It isn't; that was the problem.
Brian M. Scott
Brian M. Scott
wrote:
Ross:
>> > > > Placing each element of N over infinity gives a unit range.
Brian:
>> > > You still haven't defined infinity or explained what n/infinity (for n
>> > > in N) means. Until you give these definitions, it's all just
>> > > meaningless marks on paper.
Ross:
>> > > > limit N0/x {x:oo->0} = 0
>> > > > limit Noo/x {x:0->oo} = 1
Brian:
>> > > You'll have to explain your notation; these two lines make no sense as
>> > > they stand.
Ross:
>> > N0 is the zeroeth element of N, 0. Noo is the infinity element of N.
Brian:
>> If N0 is 0, why not just write '0', like everyone else? In its normal
>> ordering (0, 1, 2, 3, ...) the set N has no 'infinity element' (or
>> 'infinitieth element').
>N has infinity elements, Noo is infinity.
And {0, 1, 2} has three elements, but none of them is 3. What element
of N -- i.e., what natural number -- is Noo? Or if Noo isn't a
natural number, what is it? (English usage correction: 'N has an
infinite number of elements', not 'N has infinity elements'.)
>> > N0/x {x:oo->0+} = 0
>> > Noo/x {x:0+->oo} = (oo,1]
>> You still haven't explained notations '{x:oo->0+}' and '{x:0+->oo}'.
>> You've also introduced the unexplained '(oo,1]', and you haven't
>> explained how a quotient (or the limit of a quotient?) is a whole
>> interval. And since Noo remains undefined, so does Noo/x.
>The notation {x:00->0} means x goes from infinity to approaching approaching zero.
Is your first line an attempt at 'the limit of 0/x as x approaches 0
from the right is 0'? If so, you can drop the 'oo', since limits
depend only on what happens for x near the 'target', in this case 0
(with x constrained to be greater than 0). You should also forget the
notion of motion implicit in the usual language of 'approaching' and
think in terms of the rigorous definition of limit. The picture of an
x sweeping down the number line is very misleading.
>The notation (oo,1] is [1, oo).
That still doesn't explain what Noo is, what it means to divide it by
a positive real number, or how the limit of a quotient can be a whole
ray and not a single number.
>> > Infinity is increase without bound.
>> So Noo is the 'increase without bound element of N'? Does that make any
>> sense to you?
You didn't answer this.
>> > > > Thus, N maps as a function to the real number line from zero to one.
>> > > > Thus, N is surjective onto the reals.
>> > > This is utter nonsense: N isn't a function in the first place.
>> > N is a step function of R, floor R.
>> No, it isn't; it's a subset of R. But it's no wonder that you're having
>> trouble if you're confusing the set N of natural numbers with a certain
>> function whose domain is R and whose range is N.
>The floor function applied to non-negative R is N.
No. The RANGE of the floor function on R+ is N, which is a different
statement entirely. Your statement is analogous to 'a coffee-grinder
is ground coffee', or 'this assembly line is the set of Ford Escorts'.
>> > The function that describes R from 0 to 1 using N is 1/n in N as n goes from infinity to one.
>> This gives you the set {1, 1/2, 1/3, 1/4, 1/5, ...}; please show me what
>> value of n gives the real number 2/3.
>You're correct, that is an obvious fault that I should have caught.
>This might be better, j and k in N, as j goes from zero to infinity and k from infinity to zero the
>union of each j/k.
Not union: you want the set of all such j/k. This is better, but it
only gives you the rational numbers. You won't get sqrt(2), pi, or
any other irrational number.
Brian M. Scott
Brian M. Scott
wrote:
>This is still a problem, as that set is only containing the rational numbers, as well as an infinite
>number of reducible rational numbers, in fact, this construct repeats each rational number infinitely.
Ah, good: you caught this.
>To construct the irrational numbers as well and thus the reals requires a method to construct an
>infinite quantity of irrational numbers between each pair of rational numbers, which being infinite, is
>infinity.
See Walter Rudin's :Principles of Mathematical Analysis: for a
construction using Dedekind cuts.
>Consider 1/x as x goes from one to infinity in N. This produces 1, 1/2, 1/3, ..1/oo. 1/oo is not real,
>if it is irrational,
If it's not real, a fortiori it's not irrational. In fact it's
undefined, since you still haven't defined oo or explained how to
extend the definition of division to this new object.
>That is, the set made up of the elements of each n in N of ( n * 1/oo). This set is the reals from
>zero to one.
Let's suppose that your 1/oo, whatever it is, is some real number. If
it's negative, obviously you won't get [0,1] in this way. If it's 0,
each n * 1/oo = 0 as well, so you'll get only 0. Suppose, then, that
1/oo is some positive real, say p; for which n is n * 1/oo equal to
the positive real number p/2?
Brian M. Scott
Brian M. Scott
<logi...@wolfenet.com> wrote:
>'Brian M. Scott wrote:
>'For any subset of N, the powerset of said subset of N has a larger number
>of elements.
>Not completely true.
>The powerset of null is null.
On the contrary, it *is* completely true. If 0 stands for the empty
set, then P(0) = {0}, a set with 1 element. (You've overlooked the
fact that 0 is a subset of itself.)
Brian M. Scott
Brian M. Scott
> Ross A. Finlayson wrote:
> > That is, the set made up of the elements of each n in N of ( n * 1/oo). This set is the reals from
> > zero to one.
> I have been an idiot. The set made up of elements of for each n in N, Sum{i:0->n}(1/oo) is the reals from
> zero to one.
How is Sum{i:0->n}(1/oo) any different from (n+1)*(1/oo)? And you still
haven't defined oo or told us how the operation of division, defined on
R, applies to oo, which apparently isn't in R.
Brian M. Scott
Ross A. Finlayson
Brian M. Scott wrote:
> On Sun, 25 Apr 1999 07:35:37 GMT, "Ross A. Finlayson" <R...@TOMCO.NET>
> wrote:
>
> >This is still a problem, as that set is only containing the rational numbers, as well as an infinite
> >number of reducible rational numbers, in fact, this construct repeats each rational number infinitely.
>
> Ah, good: you caught this.
>
> >To construct the irrational numbers as well and thus the reals requires a method to construct an
> >infinite quantity of irrational numbers between each pair of rational numbers, which being infinite, is
> >infinity.
>
> See Walter Rudin's :Principles of Mathematical Analysis: for a
> construction using Dedekind cuts.
>
> >Consider 1/x as x goes from one to infinity in N. This produces 1, 1/2, 1/3, ..1/oo. 1/oo is not real,
> >if it is irrational,
>
> If it's not real, a fortiori it's not irrational. In fact it's
> undefined, since you still haven't defined oo or explained how to
> extend the definition of division to this new object.
>
If 1/oo is real, it is irrational, and infinitely small. In this case, it would fulfill the goal of
replicating the reals from zero to one.
Consider this, the limit approximation of 0 * 1/oo is zero. The limit approximation of oo times 1/oo is
one. The function x/oo for each n in N is continuous, so it covers all reals from zero to one.
Whether 1/oo is real or surreal, this is so.
>
> >That is, the set made up of the elements of each n in N of ( n * 1/oo). This set is the reals from
> >zero to one.
>
> Let's suppose that your 1/oo, whatever it is, is some real number. If
> it's negative, obviously you won't get [0,1] in this way. If it's 0,
> each n * 1/oo = 0 as well, so you'll get only 0. Suppose, then, that
> 1/oo is some positive real, say p; for which n is n * 1/oo equal to
> the positive real number p/2?
>
>
> Brian M. Scott
The quantity 1/oo is not negative, and not zero. It's not real.
Ross A. Finlayson
Brian M. Scott wrote:
> Ross A. Finlayson wrote:
>
> > Ross A. Finlayson wrote:
>
> > > That is, the set made up of the elements of each n in N of ( n * 1/oo). This set is the reals from
> > > zero to one.
>
> > I have been an idiot. The set made up of elements of for each n in N, Sum{i:0->n}(1/oo) is the reals from
> > zero to one.
>
> How is Sum{i:0->n}(1/oo) any different from (n+1)*(1/oo)? And you still
> haven't defined oo or told us how the operation of division, defined on
> R, applies to oo, which apparently isn't in R.
>
> Brian M. Scott
It's not any different.
Brian M. Scott
> Brian M. Scott wrote:
> > On Sun, 25 Apr 1999 07:35:37 GMT, "Ross A. Finlayson" <R...@TOMCO.NET>
> > wrote:
[...]
> > >Consider 1/x as x goes from one to infinity in N. This produces 1, 1/2, 1/3, ..1/oo. 1/oo is not real,
> > >if it is irrational,
> > If it's not real, a fortiori it's not irrational. In fact it's
> > undefined, since you still haven't defined oo or explained how to
> > extend the definition of division to this new object.
> If 1/oo is real, it is irrational, and infinitely small. In this case, it would fulfill the goal of
> replicating the reals from zero to one.
There are no infinitely small, non-zero real numbers; if 1/oo is
infinitely small and not zero, then it is not a real number, and we
don't have any idea how you mean to do arithmetic with it. You still
haven't defined oo or explained how to extend the addition defined on R
to apply also to this new object.
> Consider this, the limit approximation of 0 * 1/oo is zero. The limit approximation of oo times 1/oo is
> one.
More private terminology, I'm afraid; what do you mean by 'limit
approximation'? If your 1/oo is an actual fixed object of some kind, 0
* 1/oo should be a single number of some kind; which one? And of course
both of your statements remain meaningless, since you still haven't
defined the symbol 'oo' and explained how to do arithmetic with it.
> The function x/oo for each n in N is continuous, so it covers all reals from zero to one.
What is the relationship between x and n? What is oo? How do you
divide by it? What is your function, and what is its domain?
> Whether 1/oo is real or surreal, this is so.
At the moment no one can tell whether it is so, because it doesn't say
anything. And surreal numbers have nothing to do with the original
question of the relative cardinalities of N and R: neither of these
standard mathematical objects contains any surreal elements.
Brian M. Scott
Brian M. Scott
[some reformatting]
Ross:
> > > > That is, the set made up of the elements of each
> > > > n in N of ( n * 1/oo). This set is the reals from
> > > > zero to one.
Ross:
> > > I have been an idiot. The set made up of elements of
> > > for each n in N, Sum{i:0->n}(1/oo) is the reals from
> > > zero to one.
Brian:
> > How is Sum{i:0->n}(1/oo) any different from (n+1)*(1/oo)? And you still
> > haven't defined oo or told us how the operation of division, defined on
> > R, applies to oo, which apparently isn't in R.
> It's not any different.
Then why did you feel that you needed to correct yourself?! And you
still haven't answered the important questions.
Brian M. Scott
Mike Oliver
"Brian M. Scott" wrote:
>
> On Sat, 24 Apr 1999 23:08:19 -0700, Mike Oliver <oli...@math.ucla.edu>
> wrote:
>> Actually, I do think it is *fairly* standard to use range(R) to mean
>> { y | (exists x) xRy } and domain(R) for { x | (exists y) xRy }. For
>> example I believe "range" in this sense is one of the Goedel operations.
> That's exactly the way I had in mind, and I'd have called it
> bog-standard.
Now that's a lovely word! I'm curious what typing style you use that
would have you hit first 'b' and then 'g' for 'n'. But that's not
important right now.
I've noticed other instances where one can use a term such that
o People will ask you what it means, but
o They'll make a guess, and
o The guess will always be right.
The other one that pops to mind is "Beth" notation for cardinalities
of iterated powersets.
I propose that henceforward we call such terminologies and notiations
"bogstandard" -- perhaps, standard in the bog of our collective mathematical
unconscious?
What makes this irresistible is that the first use of "bog-standard" was
itself bogstandard.
--
Disclaimer: I could be wrong -- but I'm not. (Eagles, "Victim of Love")
Finger for PGP public key, or visit http://www.math.ucla.edu/~oliver.
1500 bits, fingerprint AE AE 4F F8 EA EA A6 FB E9 36 5F 9E EA D0 F8 B9
Brian M. Scott
> "Brian M. Scott" wrote:
> > On Sat, 24 Apr 1999 23:08:19 -0700, Mike Oliver <oli...@math.ucla.edu>
> > wrote:
> >> Actually, I do think it is *fairly* standard to use range(R) to mean
> >> { y | (exists x) xRy } and domain(R) for { x | (exists y) xRy }. For
> >> example I believe "range" in this sense is one of the Goedel operations.
> > That's exactly the way I had in mind, and I'd have called it
> > bog-standard.
> Now that's a lovely word! I'm curious what typing style you use that
> would have you hit first 'b' and then 'g' for 'n'. But that's not
> important right now.
I typed exactly what I meant to type: 'bog-standard' is an English
(British) idiom for 'common as mud'.
Brian M. Scott
Mike Oliver
"Brian M. Scott" wrote:
> I typed exactly what I meant to type: 'bog-standard' is an English
> (British) idiom for 'common as mud'.
Ah. Well, then it seems that we don't have the "guessed meaning is
always right" test of my proposed definition, which I withdraw.
Richard Carr
:Date: Sun, 25 Apr 1999 17:20:02 -0400
:From: Brian M. Scott <BMS...@stratos.net>
:Newsgroups: sci.math
:Subject: Re: Bogstandard notation
:
:Mike Oliver wrote:
:
:> "Brian M. Scott" wrote:
:
:> > On Sat, 24 Apr 1999 23:08:19 -0700, Mike Oliver <oli...@math.ucla.edu>
:> > wrote:
:
:> >> Actually, I do think it is *fairly* standard to use range(R) to mean
:> >> { y | (exists x) xRy } and domain(R) for { x | (exists y) xRy }. For
:> >> example I believe "range" in this sense is one of the Goedel operations.
:
:> > That's exactly the way I had in mind, and I'd have called it
:> > bog-standard.
:
:> Now that's a lovely word! I'm curious what typing style you use that
:> would have you hit first 'b' and then 'g' for 'n'. But that's not
:> important right now.
:
:I typed exactly what I meant to type: 'bog-standard' is an English
:(British) idiom for 'common as mud'.
:
:Brian M. Scott
:
:
Can't say that I've heard of it and I can't find it in the OED. Which
dialect is it? 'Common as mud' is more familiar to me or 'common as muck'.
Is it from the south?
Brian M. Scott
<ca...@math.columbia.edu> wrote:
>:I typed exactly what I meant to type: 'bog-standard' is an English
>:(British) idiom for 'common as mud'.
>Can't say that I've heard of it and I can't find it in the OED. Which
>dialect is it? 'Common as mud' is more familiar to me or 'common as muck'.
>Is it from the south?
I picked it up from a friend who's been doing research in Aberdeen for
several years, so I really should have said 'Scottish idiom'. I don't
know how widespread it is, though I'm pretty sure that she's not my
only source for it. In fact I thought that I'd run across it in
England, but I'm not actually certain.
Brian M. Scott
Michel Hack
equivalence relation (two sets are RF-equivalent iff they are both finite
or they are both infinite). It is easy to verify that this is a genuine
equivalence relation. Unless we exclude the axiom of inifinity, we can prove
that there are exactly two equivalence classes.
Ross clearly states that his equivalence relation is not necessarily the
same one as Cantor's equipollence.
We know the word "necessarily" is not necessary in the previous sentence.
Michel.
Jeremy Boden
<oli...@math.ucla.edu> writes
...
>
>I've noticed other instances where one can use a term such that
>
> o People will ask you what it means, but
> o They'll make a guess, and
> o The guess will always be right.
>
>The other one that pops to mind is "Beth" notation for cardinalities
>of iterated powersets.
>
>I propose that henceforward we call such terminologies and notiations
>"bogstandard" -- perhaps, standard in the bog of our collective mathematical
>unconscious?
>
>What makes this irresistible is that the first use of "bog-standard" was
>itself bogstandard.
>
standard "Beth" notation for cardinalities of iterated power sets?
I would hazard a guess that since aleph is the first letter of the
Hebrew alphabet that Beth is the second letter. However whilst that
guess might be right, it doesn't tell me much about power sets etc.
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Jeremy Boden
<sc...@math.csuohio.edu> writes
I think bog-standard is indeed countrywide as it is common in the
Midlands and in London. However, I would give "bog-standard" a slightly
different meaning than "common as muck" - it has rather more of a
meaning of something that's pretty ordinary, i.e. nothing special.
In computer programming terms a piece of bog-standard code might be
something that's not particularly elegant or efficient but is reasonably
readable and actually works.
Brian M. Scott
> I don't understand what the fuss is about. Ross Finlayson defines a simple
> equivalence relation (two sets are RF-equivalent iff they are both finite
> or they are both infinite).
Except that he doesn't. In at least one post he defined two sets to be
equivalent if there was a 'cardinally equivalent bijection' between
them; this subsequently turned out to be an ordinary bijection.
Brian M. Scott
Ross A. Finlayson
Brian M. Scott wrote:
Two finite sets are equivalent if they have the same count of elements.
Bijection, using the diagonal argument, is a simple method, essentially saying
that two sets have the same number of elements if they have the same number of
elements, which is obvious and apparent, for finite sets. For two finite sets A
and B: if |A|=|B|, then |A|=|B|. An infinite set, and any infinite set, is
equivalent to any other infinite set in that they each have a quantity of
elements that can never be completely counted, and is infinite. Thus lim |N|=lim
|R|=oo.
Two sets are equivalent if they are both finite and have the same number of
elements or if they are both infinite.
The process of mapping through a one-to-one bijection to determine equivalency is
invalid for an infinite set, because for whatever one-to-one mapping between
elements is described, there still exists in each set an infinite quantity of
unmapped elements.
Mathematical infinity exists despite any axioms, like any number.
The limit approximation of one divided by infinity is zero.
Ross F.
Mike Oliver
Jeremy Boden wrote:
> I know this rather spoils your argument - but what exactly is the bog-
> standard "Beth" notation for cardinalities of iterated power sets?
Beth_0 = Aleph_0
Beth_{alpha+1} = 2^Beth_alpha
Beth_lambda = sup{beta<lambda} Beth_beta, for lambda limit
> I would hazard a guess that since aleph is the first letter of the
> Hebrew alphabet that Beth is the second letter.
True.
Brian M. Scott
> Two finite sets are equivalent if they have the same count of elements.
> Bijection, using the diagonal argument, is a simple method, essentially saying
> that two sets have the same number of elements if they have the same number of
> elements, which is obvious and apparent, for finite sets.
The diagonal argument has nothing to do with finding bijections between
two finite sets of the same cardinality.
> For two finite sets A
> and B: if |A|=|B|, then |A|=|B|. An infinite set, and any infinite set, is
> equivalent to any other infinite set in that they each have a quantity of
> elements that can never be completely counted, and is infinite. Thus lim |N|=lim
> |R|=oo.
By any reasonable definition, standard or otherwise, |N| and |R| are
constant if they are defined at all; what is the purpose of 'lim'?
> Two sets are equivalent if they are both finite and have the same number of
> elements or if they are both infinite.
Fine; if you'd said that before, we could have saved a lot of bandwidth.
> The process of mapping through a one-to-one bijection to determine equivalency is
> invalid for an infinite set, because for whatever one-to-one mapping between
> elements is described, there still exists in each set an infinite quantity of
> unmapped elements.
Nonsense. Let N and E be respectively the set of
non-negative integers and the set of even non-negative
integers. Define f : N --> E by f(n) = 2n. What
element of N or of E isn't paired up by this mapping?
> Mathematical infinity exists despite any axioms, like any number.
> The limit approximation of one divided by infinity is zero.
You still haven't defined infinity or told us how to extend to it the
operation of division of real numbers.
Brian M. Scott
Dave Seaman
Ross A. Finlayson <R...@TOMCO.NET> wrote:
>Two sets are equivalent if they are both finite and have the same number of
>elements or if they are both infinite.
>
>The process of mapping through a one-to-one bijection to determine equivalency is
>invalid for an infinite set, because for whatever one-to-one mapping between
>elements is described, there still exists in each set an infinite quantity of
>unmapped elements.
Let A = [0,1] and B = [0,2]. Then A and B are both infinite sets, and
they are not identical. Nevertheless, there is a bijection f: A <--> B
defined by f(x) = 2*x for each x.
Note that dom(f) = A. That is, f is defined for each member of A,
despite your claims that this is impossible.
>Mathematical infinity exists despite any axioms, like any number.
There is no axiom in ZFC saying that infinity does not exist; quite the
reverse. There is an Axiom of Infinity that guarantees the existence
of a set that contains 0 and is closed under the successor operation,
and there is also the Power Set Axiom which guarantees that additional
infinite sets exist (which turn out to be of larger cardinality because
of Cantor's theorem).
From the other axioms we can deduce the existence of the Cartesian
products of sets and also subsets of given sets that are specified by a
given property (the principle of Separation). That means, given the
sets A = [0,1] and B = [0,2], we can form the set
S = { (a,b) in A x B : b = 2 * a }
which is the graph of the bijection f described above.
>The limit approximation of one divided by infinity is zero.
What does that have to do with cardinality?
Jake Wildstrom
Easily said. Not so easily done.
>Bijection, using the diagonal argument, is a simple method, essentially saying
>that two sets have the same number of elements if they have the same number of
>elements, which is obvious and apparent, for finite sets. For two finite sets
>A and B: if |A|=|B|, then |A|=|B|.
Well, you're somewhat trivializing the nature of bijection there. I'll get back
to this point.
>An infinite set, and any infinite set, is
>equivalent to any other infinite set in that they each have a quantity of
>elements that can never be completely counted, and is infinite. Thus lim
>|N|=lim |R|=oo.
This is handwaving. None of what you say here is supported by mathematical
definitions. An infinite set is simply defined as one that can be paired
bijectively with at least one proper subset of itself. How does your claim
follow from this definition?
>Two sets are equivalent if they are both finite and have the same number of
>elements or if they are both infinite.
Wrong. Keep reading.
>The process of mapping through a one-to-one bijection to determine equivalency
>is invalid for an infinite set, because for whatever one-to-one mapping
>between elements is described, there still exists in each set an infinite
>quantity of unmapped elements.
You are, like many others, confused about the nature of bijection. If I define
the trivial bijection f:N->N as f(n)=n, then _every_ element of n is
bijectively mapped onto itself. You claim that an infinite number of the
naturals should be unmapped. Name one of them for me please.
The thing is, you seem to be under the impression that a bijection is a sort of
process which considers first f(1), then f(2), and so on, so in any finite time
it'll never reach infinity. This does not correspond with the mathematical
definition thereof. Time plays no role in mathematical calculations except
where explicitly defined as a variable (e.g. in calculus). A bijection is
simply an infinite set of ordered pairs. If you can believe that the infinite
set of natural numbers contains all natural numbers (which I hope you can),
there's no reason why an infinite set of ordered pairs shouldn't contain every
natural number as an abscissa somewhere. Think on it.
Ultimately, dealing with the infinite is a matter of faith. You might obejct
that "all the natural numbers can never be created". It doesn't matter. We do
not have to "create" sets to deal with them. All we need to know is that they
_do_ exist and to know a little about their behavior. The set of natural
numbers contains all natural numbers. It seems like a truism, but it can be
hard to accept. A better point might be "Why can/would/should the set of all
natural numbers contain all natural nubers?" It does because we define it
that way. Similarly, we define the bijection f(n)=n as the set of all ordered
pairs of the form (x,x). Such a construction is valid under both the Peano
postulates and the ZFC set theory. Until such time as both of these theories
are shown to have internal inconsistencies, the set of natural numbers exists.
>Mathematical infinity exists despite any axioms, like any number.
Well, we're descending into philosophy here, but from a purely mathematical
standpoint, you'd be wrong. "Infinity" is a meaningless concept without a
system of rules governing its behavior. The ZFC definition of an infinite set
characterizes infinity one way. Theories of cardinals or surreal numbers
characterize it a completely different way. Projective geometry has a third
interpretation. The idea of the infinite, a philosophical logos if you will,
does not have a place in mathematics because the concept of the infinite is one
which can and must be treated differently in different contexts.
>The limit approximation of one divided by infinity is zero.
True, but irrelevant. And more accurately stated (avoiding unnecessary
reference to infinity):
"As x increases without bound, 1/x approaches 0."
Or, if you want to be really ultraprecise about semantics and use only simple
arithmetic and not the somewhat alarming phrases "without bound" and
"approaches":
"For every real number e, there is a real number d such that if x>d, 0<1/x<e."
which is actually quite easy to prove.
+--First Church of Briantology--Order of the Holy Quaternion--+
| A mathematician is a device for turning coffee into |
| theorems. -Paul Erdos |
+-------------------------------------------------------------+
| Jake Wildstrom |
+-------------------------------------------------------------+
Ross A. Finlayson
Brian M. Scott wrote:
> Ross A. Finlayson wrote:
>
> > Two finite sets are equivalent if they have the same count of elements.
> > Bijection, using the diagonal argument, is a simple method, essentially saying
> > that two sets have the same number of elements if they have the same number of
> > elements, which is obvious and apparent, for finite sets.
>
> The diagonal argument has nothing to do with finding bijections between
> two finite sets of the same cardinality.
>
> > For two finite sets A
> > and B: if |A|=|B|, then |A|=|B|. An infinite set, and any infinite set, is
> > equivalent to any other infinite set in that they each have a quantity of
> > elements that can never be completely counted, and is infinite. Thus lim |N|=lim
> > |R|=oo.
>
The limit of the count of the elements of N is equal to the limit of the count of the
elements of R which is infinity.
>
> By any reasonable definition, standard or otherwise, |N| and |R| are
> constant if they are defined at all; what is the purpose of 'lim'?
>
> > Two sets are equivalent if they are both finite and have the same number of
> > elements or if they are both infinite.
>
> Fine; if you'd said that before, we could have saved a lot of bandwidth.
>
> > The process of mapping through a one-to-one bijection to determine equivalency is
> > invalid for an infinite set, because for whatever one-to-one mapping between
> > elements is described, there still exists in each set an infinite quantity of
> > unmapped elements.
>
> Nonsense. Let N and E be respectively the set of
> non-negative integers and the set of even non-negative
> integers. Define f : N --> E by f(n) = 2n. What
> element of N or of E isn't paired up by this mapping?
>
An n in N has a corresponding element in f(n), to map them all at once requires an
infinite set of mappings, or in this case, a function over N. Note that as n goes to
infinity f(n) goes to infinity, not 2*infinity.
Any r in R(0,1) can be mapped to an element of n in N of n/oo.
Consider the map of N to the powerset of N. What element of N or of P(N) is not paired
by this mapping?
>
> > Mathematical infinity exists despite any axioms, like any number.
>
> > The limit approximation of one divided by infinity is zero.
>
> You still haven't defined infinity or told us how to extend to it the
> operation of division of real numbers.
>
> Brian M. Scott
Infinity is the limit approximation of one divided by zero approached from the positive
side. That is one expression that produces infinity, of course, there are an infinite
quantity of expressions that produce infinity.
Ken Cox
> Dave Seaman wrote:
> > Do you conjecture that there is a one-to-one correspondence between N
> > and the power set P(N)?
> First, are you talking about a STATIC or a DYNAMIC bijection?
He is talking about a bijection, a well-defined mathematical
object with certain properties. If you want to define new types
of mathematical objects and call them STATIC and DYNAMIC bijections
you should feel free to do so. You can even investigate their
properties and, perhaps, publish the results.
Do not, however, think that any properties of these objects are
properties of bijections just because they have similar names.
Also, as a matter of politeness and clarity, when you discuss
these objects please state immediately that they are not the
standard sort of bijections.
> Second, is there a bijection between N and the set of all odd numbers (a
> subset of N - that has only one member for every two members in N).
Yes, of course. If O is the set of odd integers, f(n) = 2n+1
is one such bijection. This demonstrates that N and O have the
same cardinality.
--
Ken Cox k...@research.bell-labs.com
Ken Cox
> I don't think that infinite sets can be ordered.
I would tend to agree. However, their cardinalities *can* be
ordered, and the result of that ordering is that your next
sentence is false:
> It is meaningless to ask if an infinite set
> is smaller, bigger or equal to another infinite set.
N is smaller than R; the cardinality of N is less than that of R.
--
Ken Cox k...@research.bell-labs.com
Ken Cox
> >Nathan the Great
> >Age 11
> Is it just me or does every hate arrogant little genius pricks like this
> guy? Or is it okay to have an ego the size of china when you're 11?
It's fine to have a huge ego at any age, providing it's merited.
In this case it would not be.
--
Ken Cox k...@research.bell-labs.com
Domnei
>not paired
>by this mapping?
>
Boy is that a vacuous question. You didn't specify what the
mapping is! (And yes, I did look for it in your post. I could
not find it.)
Of course, no matter what mapping you specify it will
be easy to answer your question (probably using - you
guessed it - diagonalization), because N and P(N) do
not have the same cardinality.
Mike Keith
Web site: http://users.aol.com/s6sj7gt/mikehome.htm
(remove post-w letters from e-mail address)
Ross A. Finlayson
Domnei wrote:
> >Consider the map of N to the powerset of N. What element of N or of P(N) is
> >not paired
> >by this mapping?
> >
>
> Boy is that a vacuous question. You didn't specify what the
> mapping is! (And yes, I did look for it in your post. I could
> not find it.)
>
Consider any map of N to P(N) where each element of N is used to map to one
element of P(N). As the mapping progresses, or considered all at once, for any
unmapped element in P(N) there exist an infinite quantity of elements in N that
can be mapped exclusively to it, and vice versa.
>
> Of course, no matter what mapping you specify it will
> be easy to answer your question (probably using - you
> guessed it - diagonalization), because N and P(N) do
> not have the same cardinality.
>
> Mike Keith
> Web site: http://users.aol.com/s6sj7gt/mikehome.htm
> (remove post-w letters from e-mail address)
Any finite subset of n in N and its corresponding powerset P(n) do not have the
same cardinality.
--
Virgil
>The process of mapping through a one-to-one bijection to determine
equivalency is
>invalid for an infinite set, because for whatever one-to-one mapping between
>elements is described, there still exists in each set an infinite quantity of
>unmapped elements.
If each pairing takes an equal amount of time, however small, sets can be
found large enough so that a pairing off of elements could not be
completed before the extinction of humanity. Thus large enough finite sets
have the same problem and infinite sets, that the one-to-one mappings
cannot be completed in time to be of any use.
On the other hand, if each pairing takes half the time as the previous
pairing, then if any pairing can be done in finite time, so can a
countably infinite number of pairings be compoete in finite time.
Once you outlaw the infinite, you will have to answer to Zeno of Elea
about how the world wags. He asked some hard questions.
--
Virgil
vm...@frii.com
Bill Dubuque
Given a square matrix F one may construct a row-vector different from
those in F by simply taking the diagonal of F and changing each element.
In detail: suppose matrix F(i,j) has entries from a set B with two or
more elements (so there is a "change" map ~ on B: ~b differs from b).
Let ~F(i,i) denote the "changed diagonal" of F, i.e. the row-vector
obtained by applying ~ to each element of the diagonal F(i,i) of F.
The changed diagonal of F differs from each row-vector of F because
it differs on diagonal elements: the i'th row-vector f_i of F has
i'th entry = F(i,i) versus ~F(i,i) = i'th entry of changed diagonal:
[ f1(1) f1(2) f1(3) . . . ] row i = f_i(j) maps A to B
| |
| f2(1) f2(2) f2(3) |
| | --> [~f1(1) ~f2(2) ~f3(3) ... ]
| f3(1) f3(2) f3(3) |
| . . | = "changed diagonal" of F
| . . |
[ . . ] where ~x differs from x
Notice that the above proof required no hypotheses on the index
set of the rows and columns of F. Although classically one thinks
of a square matrix as indexed by a finite ordered set {1 2 ... n},
in fact the above proof holds for any abstract index set A; then
the rows are simply functions f(j) from A to B, and the matrix F
is just a set of such row-functions {f_i(j)}, indexed by i in A.
This matrix viewpoint is adopted in the proof given below that
nontrivial exponentiation B^A increases the cardinality |A| of A.
THEOREM Let A, B be sets, and B^A = set of functions from A to B.
If |B| > 1 then |B^A| > |A|, i.e. B^A has more than |A| functions.
Proof: Given |A| functions f_i : A -> B, construct the |A| x |A|
"matrix" F(i,j) = f_i(j). The proof presented above shows that
the changed diagonal function ~F(i,i) differs from the given |A|
functions (= row-vectors of F); hence |B^A| > |A| when |B| > 1.
In particular, for B = 2 = {0 1} the theorem implies |2^A| > |A|,
i.e. the cardinality of the powerset of A exceeds that of A; here
we've employed the well-known fact that the powerset of A is the
same as 2^A (subsets S of A are in 1-1 correspondence with functions
f : A -> 2 via: i in S <=> f(i) = 1; in other words consider f
as the characteristic function (bit-vector) for the subset S of A).
A close examination of the classical proof [1] will show that it
is essentially identical to the above "matrix-theoretic" proof.
By the way, it's a historical mistake that the diagonal construct
is attributed to Cantor. In fact it dates to at least 1871 when
du Bois-Reymond diagonalized on the growth rates of functions in
his pioneering studies on "orders of infinity" (see my prior post
[2]); in fact such a diagonalization on growth rates is a simple
way of proving that Ackermann function is not primitive recursive
(this problem was not considered until almost 60 years later [3]).
-Bill Dubuque
[1] http://www.dejanews.com/getdoc.xp?AN=470526822
[2] http://www.dejanews.com/getdoc.xp?AN=255882969
[3] http://www.dejanews.com/getdoc.xp?AN=271857318
David Kastrup
> Andrew Martin wrote:
> > >Nathan the Great
> > >Age 11
>
> > Is it just me or does every hate arrogant little genius pricks like this
> > guy? Or is it okay to have an ego the size of china when you're 11?
>
> It's fine to have a huge ego at any age, providing it's merited.
I would tend to disagree. A merited ego would be no excuse for being
obnoxious, getting on everyone's nerves, calling people names and
making stupid jokes about them.
Personality disorders might be *tolerated* if the person has some
other redeeming qualities, but they are never something that is
"fine". There are some people that are, well, let's say kind of
strange and without too good a feeling for proper human interaction,
but they quite often make an honest effort of compensation.
> In this case it would not be.
I have to agree.
--
David Kastrup Phone: +49-234-700-5570
Email: d...@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany
Brian M. Scott
wrote:
>Brian M. Scott wrote:
>> Ross A. Finlayson wrote:
>> > For two finite sets A
>> > and B: if |A|=|B|, then |A|=|B|. An infinite set, and any infinite set, is
>> > equivalent to any other infinite set in that they each have a quantity of
>> > elements that can never be completely counted, and is infinite. Thus lim |N|=lim
>> > |R|=oo.
>The limit of the count of the elements of N is equal to the limit of the count of the
>elements of R which is infinity.
You're still talking neither English nor mathematics. If it makes
sense to talk about the 'count of the elements of N', that count is a
single object, not a function or sequence that has a limit. If you
are talking about the limit of some process of countING, you will have
to define what the process is and what it means for something to be
its limit.
If you're simply trying to say that N and R have the same cardinality,
something that you choose to call 'infinity', you're just wrong. You
must know by now that as (thousands of) mathematicians have defined
and used the notion of cardinality, |N| < |R|. This is a fact, and an
easily provable one, irrespective of whether you like it or understand
the proof. As others have pointed out, you may define a new concept -
call it RAF-cardinality, say - that is equivalent to the usual one for
finite sets and assigns to all infinite sets the same cardinality. It
isn't a very useful notion, so far as I can see, but it could
certainly be defined and studied. But as long as you continue to
insist that cardinality is (or somehow 'ought' to be) RAF-cardinality,
you will continue to write nonsense.
>> > The process of mapping through a one-to-one bijection to determine equivalency is
>> > invalid for an infinite set, because for whatever one-to-one mapping between
>> > elements is described, there still exists in each set an infinite quantity of
>> > unmapped elements.
>> Nonsense. Let N and E be respectively the set of
>> non-negative integers and the set of even non-negative
>> integers. Define f : N --> E by f(n) = 2n. What
>> element of N or of E isn't paired up by this mapping?
>An n in N has a corresponding element in f(n), to map them all at once requires an
>infinite set of mappings, or in this case, a function over N.
A function *is* a mapping; *one* mapping.
> Note that as n goes to
>infinity f(n) goes to infinity, not 2*infinity.
This is an altogether different notion of infinity; hadn't you
realized that yet? The infinity of limit statements like these is a
verbal shorthand for 'increases without bound'. It isn't really a
number at all, let alone the first infinite cardinal number (aleph_0)
or the first infinite ordinal number (omega_0).
>Any r in R(0,1) can be mapped to an element of n in N of n/oo.
This is incoherent. And you STILL haven't defined oo or explained how
the operation of division of real numbers is supposed to apply to it.
>Consider the map of N to the powerset of N. What element of N or of P(N) is not paired
>by this mapping?
Your question is meaningless: 'the map of N to the powerset of N' is
an empty phrase, since you have specified no map. If you specify one,
I will of course apply diagonalization to produce a member of P(N)
that is not paired with any element of N.
>> You still haven't defined infinity or told us how to extend to it the
>> operation of division of real numbers.
>Infinity is the limit approximation of one divided by zero approached from the positive
>side.
That's not a definition, even after the mangled language is sorted
out. What you're trying to say, I assume, is that you wish to define
infinity to be the limit as x approaches 0 from the right of 1/x. But
you haven't shown that there is any such object. A good calculus text
will tell you that in that context 'infinity' is simply an
abbreviation for 'increases without bound', not an actual number of
some kind. In a more advanced setting one might consider adding a
point at infinity and calling it the limit of that function, but it
still won't behave like a number: you can't extend the usual
arithmetic operations to it without losing some of their usual
properties.
You also haven't explained what connection there is between this
notion of infinite limit and the notion of infinite set.
Brian M. Scott
Ken Cox
> Ken Cox <k...@research.bell-labs.com> writes:
> > It's fine to have a huge ego at any age, providing it's merited.
> I would tend to disagree. A merited ego would be no excuse for being
> obnoxious, getting on everyone's nerves, calling people names and
> making stupid jokes about them.
Let me clarify. Part of what merits a huge ego (IMHO) is the
humility to avoid doing the above. In other words, it's fine
to have a large sense of one's own importance, as long as you
don't let it show.
--
Ken Cox k...@research.bell-labs.com
Torkel Franzen
>I would tend to disagree. A merited ego would be no excuse for being
>obnoxious, getting on everyone's nerves, calling people names and
>making stupid jokes about them.
Nathan, who by the way is 42, is by no means to be blamed for any of
the endless crapola that his postings have generated in the group.