Millionaire Mathematics
Koe Nic
some of you have done the same thing. I know I've had fun with it
several times.
I would be in a conversation with someone I know. Eventually I would
make a comment about seeing if they would give me a penny today and
then 2 cents tomorrow, 4 cents the next day, and then just continue to
double it each day for a month. Most people will agree to it since it
only seems to them that they will be giving me not much more than a
few cents.
There was this one friend I would see on a daily basis and he actually
did start doing this. However, after it reached $2.56 he finally
started thinking and decided he didn't want to continue.
The typical person might not stop and think that if he would do this
for a month he would end up making me a millionaire, since it starts
with such a small amount, but he would.
If you know a few millionaires, this might be a good trick to play on
them, or just with some friends who might not be aware of how much
this would actually cost them.
--
Almost Anything You Want To Know
http://www.weeuniverse.com
Nico Benschop
>
> Here is a little trick that I like to try on people at times. Perhaps
> some of you have done the same thing. I know I've had fun with it
> several times.
>
> I would be in a conversation with someone I know. Eventually I would
> make a comment about seeing if they would give me a penny today and
> then 2 cents tomorrow, 4 cents the next day, and then just continue to
> double it each day for a month. Most people will agree to it since it
> only seems to them that they will be giving me not much more than a
Anything that grows exponentially, is uncountable
after countably infinite time (isn't it?-)
PS: This trick is a few thousand years old
(re: Persia & the chess game, with 64 fields, each filled with 2^n
[n=1..64] grains; the sultan agreed to pay the wizzard, but could'nt
keep his promise - and those are only *finite* # fields;-) -- NB
Henry
wrote:
>Anything that grows exponentially, is uncountable
> after countably infinite time (isn't it?-)
That depends. 2^n increases to the same infinity as n when n tends to
infinity. But the power set of a infinite set is a higher infinity.
(The difference is between ordinal and cardinal infinities.) It all
depends on what you might mean by the meaningless phrase "after
countably infinite time".
Daniel McLaury
This was a well-known trick at my elementary school. However, when I
tried to explain it to some people, they didn't believe me, even after
I showed them the sum of the series, and even after I printed out a
table of doubles. One girl told me, "But you can't keep doubling it
every time!", even though I made very clear that was what I intended
to do. Some people won't let truth get in the way of "common
sense"...
Wade Ramey
Nico Benschop <n.ben...@chello.nl> wrote:
> Anything that grows exponentially, is uncountable
> after countably infinite time (isn't it?-)
??? I think this makes no sense.
--W.
Adam Russell
"Wade Ramey" <wrame...@attbi.remove13.com> wrote in message
news:wrameyxiii-6D52B...@netnews.attbi.com...
Had me reaching for my glasses.
Ahmed
This reminds me the story about the Harvard professor that after 30
years meets an old friend that dropped out from high school because he
couldn't do mathematics .
The professor is surprised to hear that this old friend is now a heavy
millionair .
He ask him : well how did you earn all that money ?
and the friend replies : it is very simple,
I buy a product at $10 and sell it in $20 , this gives me my 10% profit !
A.F.
Ugly Dude 42
1-4 came out (the ones to multiply by 7).
Nico Benschop
I just mean lim n-->inf. (as you guessed).
I'm no specialist on completed infinities, obviously.
But since you mentioned 'powerset', I thought of that too, which
for finite set S with |S|=n can be considered as the Boolean lattice
of all its 2^n subsets, under intersection and union. I must admit
having a bit of trouble with a 'completed' infinite, as you say:
'powerset of a infinite set'.
But I could get used to it after a while, if you let me think of it
as a Boolean lattice generated under union over the 'countably inf.
layer of singletons' (ordered just above the empty set) in the infinite
Boolean lattice 2^N (Peano's naturals set N). To sketch what I've in
mind, the lattice elements, as subsets of N, have unique binary code
(membership n \in N yields in binary string_place n a '1', else '0')
The lattice top would be N, with binary code the all_1 string 1*,
while the lattice bottom element is the all_0 string 0*.
The mentioned singleton layer of the naturals, ordered just above 0*,
would consist of all countably infinite strings with just one '1' and
rest zero. Does *that* make sense? As I was told 2^N has uncountable
cardinality, I thought |2^N| = lim [n-->inf] 2^n. So I figured that
exponential growth is 'equivalent' (necess & suff) for the concept
of some 'uncountable' infinity, while polynomial growth n^k for
any finite k (n --> inf) is not. That's all, really. -- NB
Dave Seaman
Nico Benschop <n.ben...@chello.nl> wrote:
>As I was told 2^N has uncountable
>cardinality, I thought |2^N| = lim [n-->inf] 2^n.
You are confusing cardinal exponentiation with ordinal exponentiation.
According to cardinal exponentiation, |2^N| = 2^aleph_0 is the
cardinality of the power set of N, which is an uncountable cardinal.
According to ordinal exponentiation, 2^N = 2^w = lim_{n->w} 2^n, which is
a countable ordinal.
--
Dave Seaman dse...@purdue.edu
Amnesty International says Mumia Abu-Jamal decision falls short of justice.
<http://www.amnestyusa.org/news/2001/usa12192001.html>
Simon Nickerson
<koe...@usa.com> writes
>The typical person might not stop and think that if he would do this
>for a month he would end up making me a millionaire, since it starts
>with such a small amount, but he would.
A similar story appeared as an anecdote on the BBC Radio 4 programme
"Home Truths" a couple of weeks ago. Specifically, a couple and their
two children were on holiday in Germany (before everyone started using
Euros, presumably). The father, wanting to instil his children with the
idea of the benefits of saving, made a deal with them. He would give
them 1 Deutschmark each day, but in addition, if they had saved any of
their money, he would give the same amount to them again (e.g. if after
the first day they had saved 35 pfennigs, he would give them DM 1.35).
To his pleasure, though eventual horror, neither of his children spent
any of their holiday money. Unfortunately, I was laughing so hard by
this point that I failed to learn whether the father really gave each of
his children 32,767 Deutschmarks in pocket money.
--
Simon Nickerson
"It means," said Aslan, "that though the Witch knew the Deep
Magic, there is a magic deeper still which she did not know."
C.S.Lewis, The Lion, the Witch and the Wardrobe
Nico Benschop
>
> In article <3CB94B00...@chello.nl>,
> Nico Benschop <n.ben...@chello.nl> wrote:
>
> >As I was told 2^N has uncountable
> >cardinality, I thought |2^N| = lim [n-->inf] 2^n.
>
> You are confusing cardinal exponentiation with ordinal exponentiation.
>
> According to cardinal exponentiation, |2^N| = 2^aleph_0 is the
> cardinality of the power set of N, which is an uncountable cardinal.
>
> According to ordinal exponentiation, 2^N = 2^w = lim_{n->w} 2^n,
> which is a countable ordinal. -- Dave Seaman
OK Dave, thanks (some more reading is required;-) -- NB
Nico Benschop
Consider alphabet B of two symbols, say B={a,b} or {0,1} if preferred.
Then there are 2^n binary strings of finite length n over B,
"growing exponentially" with n (the above "countably inf. time").
Set B* = 2^N, of all binary strings of countably inf. length over B,
has the same cardinality as the reals R01 on interval (0,1) - AFAIK,
just taking the arithmetic semantics of binary coded reals 0 < r < 1.
There are only countably many rationals in R01, so qua cardinality
there is no difference between |B*| and |R01| .
And doesn't Cantors diagonal argument hold for both: set B* of
binary strings without arithmetic semantics, and R01 of reals ?
Cantor talks of strings in a w x w binary marix,
with 'bitwise' complemented diagonal, also of length w.
If so, please tell me what is wrong with |B*| = |R01| both being
uncountable due to their "exponential growth rate" (in lim n --> w,
with Peanos naturals |N| = w being 'countable') -- NB
Nico Benschop
Consider alphabet B of two symbols, say B={a,b} or {0,1} if preferred.
Then there are 2^n binary strings of finite length n over B,
"growing exponentially" with n (the above "countably inf. time").
Set B* = 2^N, of all binary strings of countably inf. length over B,
has the same cardinality as the reals R01 on interval (0,1) - AFAIK,
just taking the arithmetic semantics of binary coded reals 0 < r < 1.
There are only countably many rationals in R01, so qua cardinality
there is no difference between |B*| and |R01| .
And doesn't Cantors diagonal argument hold for both: set B* of
binary strings without arithmetic semantics, and R01 of reals ?
Cantor talks of strings in a w x w binary marix,
with 'bitwise' complemented diagonal, also of length w.
If so, please tell me what is wrong with |B*| = |R01| both being
uncountable due to their "exponential growth rate" (in lim n --> w,
with Peanos naturals |N| = w being 'countable'; for comparison:
set N = A* = the set of all strings over one_symbol alphabet |A|=1
)
Robin Chapman
news:3CBA9AE7...@chello.nl...
> > "Wade Ramey" <wrame...@attbi.remove13.com> wrote in message
> > news:wrameyxiii-6D52B...@netnews.attbi.com...
> > > In article <3CB810CC...@chello.nl>,
> > > Nico Benschop <n.ben...@chello.nl> wrote:
> > >
> > > > Anything that grows exponentially, is uncountable
> > > > after countably infinite time (isn't it?-)
> > >
> > > ??? I think this makes no sense.
Wise words.
> Consider alphabet B of two symbols, say B={a,b} or {0,1} if preferred.
> Then there are 2^n binary strings of finite length n over B,
> "growing exponentially" with n (the above "countably inf. time").
OK
> Set B* = 2^N, of all binary strings of countably inf. length over B,
> has the same cardinality as the reals R01 on interval (0,1) - AFAIK,
Yes, easy enough to prove.
> just taking the arithmetic semantics of binary coded reals 0 < r < 1.
"arithmetic semantics"?
> And doesn't Cantors diagonal argument hold for both: set B* of
> binary strings without arithmetic semantics, and R01 of reals ?
The diagonal argument shows straightforwardly that {0,1}^N
is uncountable. For the reals, it's better to use a Baire
Category Theorem argument than to mess around with digital
representations.
> If so, please tell me what is wrong with |B*| = |R01| both being
> uncountable due to their "exponential growth rate" (in lim n --> w,
> with Peanos naturals |N| = w being 'countable'; for comparison:
> set N = A* = the set of all strings over one_symbol alphabet |A|=1
> )
It's a non-sequitur. {0,1}^N and (0,1) are uncountable because
they are uncountable, not because of any "exponential growth
rate".
Robin Chapman
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
Nico Benschop
>
> "Nico Benschop" <n.ben...@chello.nl> wrote
> > > "Wade Ramey" <wrame...@attbi.remove13.com> wrote
> > > >
> > > > > Anything that grows exponentially, is uncountable
> > > > > after countably infinite time (isn't it?-)
> > > >
> > > > ??? I think this makes no sense.
>
> Wise words.
>
> > Consider alphabet B of two symbols, say B={a,b} or {0,1}
> > "growing exponentially" with n (the above "countably inf. time").
>
> OK
>
> > Set B* = 2^N, of all binary strings of countably inf. length
> > over B, has the same cardinality as the reals R01 on interval
> > (0,1) - AFAIK,
>
> Yes, easy enough to prove.
>
>
> "arithmetic semantics"?
Well, real r = \sum r_i.2^{-i} for i \in N,
with binary coefficients r_i \in {0,1}
> > And doesn't Cantors diagonal argument hold for both: set B* of
> > binary strings without arithmetic semantics, and R01 of reals ?
>
> The diagonal argument shows straightforwardly that {0,1}^N
> is uncountable. For the reals, it's better to use a Baire
> Category Theorem argument than to mess around with digital
> representations.
>
> > If so, please tell me what is wrong with |B*| = |R01| both being
> > uncountable due to their "exponential growth rate" in lim n --> w,
> > with Peanos naturals |N| = w being 'countable'. For comparison:
> > set N= A* : the set of all strings over one_symbol alphabet |A|=1.
>
> It's a non-sequitur.
??
> {0,1}^N and (0,1) are uncountable because they are uncountable, ..[*]
> not because of any "exponential growth rate". -- Robin Chapman
You mean by definition, or axiom?
Sorry, I did'nt realize that the exponential origin
of the uncountable was also unspeakable.
(a non-generated infinity does have its fascination, doesn't it?-)
- http://www.iae.nl/users/benschop/cantor.htm [1]
http://www.iae.nl/users/benschop/ism.htm [2]
[2] gives a ref to an AMM-aug99 paper on an infinite semigroup A*
of integer 2 x 2 matrices generated by two of them, A={a,b} :
with a = 0 1 b = 1 1 Notice that aa = a (idempotent).
0 1 0 1
The set A^n of distinct such matrices represented by strings
of length n over A, has order |A^n| which, for n --> inf,
grows faster than any polynomial, and slower than exponential,
thus with essentially an 'in-between' growth rate.
This appears to defy the Continuum Hypothesis, doesn't it?
I suspect many (most?) integer state machines (state set N) with 2
inputs A={a,b} and inf.closure (semigroup) S= A*/N have this property.
-- NB
Robin Chapman
> Robin Chapman wrote:
> >
> > "Nico Benschop" <n.ben...@chello.nl> wrote
> > > If so, please tell me what is wrong with |B*| = |R01| both being
> > > uncountable due to their "exponential growth rate" in lim n --> w,
> > > with Peanos naturals |N| = w being 'countable'. For comparison:
> > > set N= A* : the set of all strings over one_symbol alphabet |A|=1.
> >
> > It's a non-sequitur.
>
> ??
>
> > {0,1}^N and (0,1) are uncountable because they are uncountable, ..[*]
> > not because of any "exponential growth rate". -- Robin Chapman
>
> You mean by definition, or axiom?
No, in each case there is a proof. Nothing to do with
any "exponential growth rate".
> Sorry, I did'nt realize that the exponential origin
"didn't".
> of the uncountable was also unspeakable.
The word is not "unspeakable" but "meaningless".
> http://www.iae.nl/users/benschop/ism.htm [2]
>
> [2] gives a ref to an AMM-aug99 paper on an infinite semigroup A*
> of integer 2 x 2 matrices generated by two of them, A={a,b} :
> with a = 0 1 b = 1 1 Notice that aa = a (idempotent).
> 0 1 0 1
> The set A^n of distinct such matrices represented by strings
> of length n over A, has order |A^n| which, for n --> inf,
> grows faster than any polynomial, and slower than exponential,
> thus with essentially an 'in-between' growth rate.
Fascinating.
> This appears to defy the Continuum Hypothesis, doesn't it?
No. It actually appears to have nothing to do with the
continuum hypothesis.
> I suspect many (most?) integer state machines (state set N) with 2
> inputs A={a,b} and inf.closure (semigroup) S= A*/N have this property.
What property? That they "appear to defy the continuum
hypothesis"?
Nico Benschop
>
> "Nico Benschop" <n.ben...@chello.nl> wrote
> > Robin Chapman wrote:
> > >
> > > "Nico Benschop" <n.ben...@chello.nl> wrote
> > > > uncountable due to their "exponential growth rate" for n -> w,
> > > > with Peanos naturals |N| = w being 'countable'. By comparison:
> > > > set N= A* of all strings over one_symbol alphabet |A|=1.
> > >
> > > It's a non-sequitur.
> >
> > ??
again ??
> >
> > > {0,1}^N and (0,1) are uncountable because they are uncountable,
> > > not because of any "exponential growth rate". -- Robin Chapman
> >
> > You mean by definition, or axiom?
>
> No, in each case there is a proof.
> Nothing to do with any "exponential growth rate".
With Peano generating by the successor function S(n)=n+1 the
'countably infinite' set N of all naturals, or equivalently the
countable infinite set A* of all strings over an alphabet A of
one symbol, say A={1}, then what is so strange about generating an
uncountable infinite set B* of all strings over B={0,1} - which is
precisely what Cantor considered (at least: any countable subset
thereof, by his diagonal of a w x w binary table showing some
string remains missing, independent of which table is taken;-)
> > http://www.iae.nl/users/benschop/ism.htm [2]
> >
> > [2] gives a ref to an AMM-aug99 paper on an infinite semigroup A*
> > of integer 2 x 2 matrices generated by two of them, A={a,b} :
> > with a = 0 1 b = 1 1 Notice that aa = a (idempotent).
> > 0 1 0 1
> > The set A^n of distinct such matrices represented by strings
> > of length n over A, has order |A^n| which, for n --> inf,
> > grows faster than any polynomial, and slower than exponential,
> > thus with essentially an 'in-between' growth rate.
>
> Fascinating.
>
> > This appears to defy the Continuum Hypothesis, doesn't it?
>
> No. It actually appears to have nothing to do with the
> continuum hypothesis.
How come? CH conjectures there is no infinite set 'between' Cantors
uncountable~ and Peanos countable infinity. And unless you specify
how some infinite set is _generated_ , as Peano and Cantor did,
you just don't know what kind of infinity you are talking about.
> > I suspect many (most?) integer state machines (state set N) with
> > 2 inputs A={a,b} and inf.closure (semigroup) S= A*/N have this
> > property.
>
> What property? That they "appear to defy the continuum hypothesis"?
> -- Robin Chapman
No, the property of generating an infinity between countable
and uncountable, thus between A* and B* where |A|=1 and |B|=2,
having a growth rate of | (B^n)/N | for n --> w=|N|, which is
faster than polynomial but slower than exponential.
(re: the machine M(N,B): N x B --> N induces equivalences in B*
that are not countable, yet the set of remaining non-equiv.
strings '(mod M)' is also not countable).
http://home.iae.nl/users/benschop/ism.htm
PS: surely sci.math can yield a response less prejudiced than RC's ?
Nico Benschop
>
> "Nico Benschop" <n.ben...@chello.nl> wrote
> > Robin Chapman wrote:
> > >
> > > "Nico Benschop" <n.ben...@chello.nl> wrote
> > > > uncountable due to their "exponential growth rate" for n -> w,
> > > > with Peanos naturals |N| = w being 'countable'. By comparison:
> > >
> > > It's a non-sequitur.
> >
> > ??
again ??
> >
> > > {0,1}^N and (0,1) are uncountable because they are uncountable,
> > > not because of any "exponential growth rate". -- Robin Chapman
> >
> > You mean by definition, or axiom?
>
> No, in each case there is a proof.
> Nothing to do with any "exponential growth rate".
With Peano generating by the successor function S(n)=n+1 the
'countably infinite' set N of all naturals, or equivalently the
countable infinite set A* of all strings over an alphabet A of
one symbol, say A={1}, then what is so strange about generating an
uncountable infinite set B* of all strings over B={0,1} - which is
precisely what Cantor considered (at least: any countable subset
thereof, by his diagonal of a w x w binary table showing some
string remains missing, independent of which table is taken;-)
> > http://www.iae.nl/users/benschop/ism.htm [2]
> >
> > [2] gives a ref to an AMM-aug99 paper on an infinite semigroup A*
> > of integer 2 x 2 matrices generated by two of them, A={a,b} :
> > with a = 0 1 b = 1 1 Notice that aa = a (idempotent).
> > 0 1 0 1
> > The set A^n of distinct such matrices represented by strings
> > of length n over A, has order |A^n| which, for n --> inf,
> > grows faster than any polynomial, and slower than exponential,
> > thus with essentially an 'in-between' growth rate.
>
> Fascinating.
>
> > This appears to defy the Continuum Hypothesis, doesn't it?
>
> No. It actually appears to have nothing to do with the
> continuum hypothesis.
How come? CH conjectures there is no infinite set 'between' Cantors
uncountable~ and Peanos countable infinity. And unless you specify
how some infinite set is _generated_ , as Peano and Cantor did,
you just don't know what you are talking about.
> > I suspect many (most?) integer state machines (state set N) with
> > 2 inputs A={a,b} and inf.closure (semigroup) S= A*/N have this
> > property.
>
> What property? That they "appear to defy the continuum hypothesis"?
> -- Robin Chapman
No, the property of generating an infinity between countable
and uncountable, thus between A* and B* where |A|=1 and |B|=2,
being neither of polynomial- nor of exponential growth rate of
|A^n| resp. |B^n| for n --> w. -- NB
Robin Chapman
> > >
> > > > {0,1}^N and (0,1) are uncountable because they are uncountable,
> > > > not because of any "exponential growth rate". -- Robin Chapman
> > >
> > > You mean by definition, or axiom?
> >
> > No, in each case there is a proof.
> > Nothing to do with any "exponential growth rate".
>
> With Peano generating by the successor function S(n)=n+1 the
> 'countably infinite' set N of all naturals, or equivalently the
> countable infinite set A* of all strings over an alphabet A of
> one symbol, say A={1}, then what is so strange about generating an
> uncountable infinite set B* of all strings over B={0,1}
You are talking about "generating" again. I seem to recall
a dismal thread a little while ago, where you insisted
that the set of all permutations of Z were "generated"
by two or three particular permutations. However this
use of "generated" seemed to come (like many other of
your terms) from your own private lexicon, and you steadfastly
refused to explain what on earth you meant by it (certainly
not the conventional usage in group theory, where a
countable set of generators must generate a countable group).
> > > The set A^n of distinct such matrices represented by strings
> > > of length n over A, has order |A^n| which, for n --> inf,
> > > grows faster than any polynomial, and slower than exponential,
> > > thus with essentially an 'in-between' growth rate.
> >
> > Fascinating.
> >
> > > This appears to defy the Continuum Hypothesis, doesn't it?
> >
> > No. It actually appears to have nothing to do with the
> > continuum hypothesis.
>
> How come? CH conjectures there is no infinite set 'between' Cantors
> uncountable~ and Peanos countable infinity.
OK, no set in cardinality between aleph_0 and 2^(aleph_0),
but what the fuck has your example to do with that? Where
is the putative set that might have inequality. Let me
stress it again: CH has to do with SETs. All I see is a
sequence of numbers |A^n| which tend to infinity,
superpolynomially but subexponentially. No set at all.
Anyway the existence of such a sequence ain't news: n^(log n)
tends to infinity superpolynomially and subexponentially.
The point about the Monthly article was that it's an amusing
fact that one particular sequence can be proved to
have this property.
> And unless you specify
> how some infinite set is _generated_ , as Peano and Cantor did,
> you just don't know what you are talking about.
Actually no, it's you who do not know what you are talking
about Mr Benschop.
> >
> > What property? That they "appear to defy the continuum hypothesis"?
> > -- Robin Chapman
>
> No, the property of generating an infinity between countable
> and uncountable, thus between A* and B* where |A|=1 and |B|=2,
What the fuck are you talking about here? What does it
mean for a sequence of numbers ro "generate an infinity".
> being neither of polynomial- nor of exponential growth rate of
> |A^n| resp. |B^n| for n --> w. -- NB
As I said, such sequences are not news, and have nothing
to do with infinite cardinalities.
> PS: surely sci.math can yield a response less prejudiced than RC's ?
Yes I am prejudiced. I hate idiocy and I can scarcely
tolerate morons. I just have to tell them how stupid they
are. I know I should be unprejudiced and just accept that
they are different; I just can't do it though.
David C. Ullrich
Nobody but you has any idea what "the exponential origin of
the uncountable" _means_.
Look. Say S_n is the set of all strings of 0's and 1's of length n.
Then S_n has 2^n elements - this grows exponentially. But the
union of the S_n is countable.
>(a non-generated infinity does have its fascination, doesn't it?-)
>
> - http://www.iae.nl/users/benschop/cantor.htm [1]
> http://www.iae.nl/users/benschop/ism.htm [2]
>
>[2] gives a ref to an AMM-aug99 paper on an infinite semigroup A*
> of integer 2 x 2 matrices generated by two of them, A={a,b} :
> with a = 0 1 b = 1 1 Notice that aa = a (idempotent).
> 0 1 0 1
>The set A^n of distinct such matrices represented by strings
> of length n over A, has order |A^n| which, for n --> inf,
>grows faster than any polynomial, and slower than exponential,
>thus with essentially an 'in-between' growth rate.
>
>This appears to defy the Continuum Hypothesis, doesn't it?
This is nonsense. Not _false_, it makes no sense. To defy CH
you have to give us a _set_ with cardinality larger than
aleph_0 but smaller than c. Exactly what set with this
property are you referring to here?
>I suspect many (most?) integer state machines (state set N) with 2
>inputs A={a,b} and inf.closure (semigroup) S= A*/N have this property.
>
>-- NB
David C. Ullrich
David C. Ullrich
<n.ben...@chello.nl> wrote:
That is exactly correct. Except I don't know if you've noticed,
but it's _you_ making assertions about CG, not RC. If you claim,
as you have, that you're "defying" CH then it's up to _you_ to
say which set you're talking about that has cardinality larger
than aleph_0 and smaller than c. You haven't done that.
>> > I suspect many (most?) integer state machines (state set N) with
>> > 2 inputs A={a,b} and inf.closure (semigroup) S= A*/N have this
>> > property.
>>
>> What property? That they "appear to defy the continuum hypothesis"?
>> -- Robin Chapman
>
>No, the property of generating an infinity between countable
>and uncountable, thus between A* and B* where |A|=1 and |B|=2,
>being neither of polynomial- nor of exponential growth rate of
>|A^n| resp. |B^n| for n --> w. -- NB
>
>PS: surely sci.math can yield a response less prejudiced than RC's ?
What does his supposed prejudice have to do with the validity of
his or your arguments?
Never mind. I'm not RC. When you claim to be defying CH, exactly
what set is it that you are asserting has cardinality strictly
between aleph_0 and c?
David C. Ullrich
Nico Benschop
>
> "Nico Benschop" <n.ben...@chello.nl> wrote in message
> news:3CBC1612...@chello.nl...
>
> > > >
> > > > > {0,1}^N and (0,1) are uncountable because they are
> > > > > uncountable, not because of any "exponential growth rate".
> > > > > -- Robin Chapman
> > > >
> > > > You mean by definition, or axiom?
> > >
> > > No, in each case there is a proof.
> > > Nothing to do with any "exponential growth rate".
> >
> > With Peano generating by the successor function S(n)=n+1 the
> > 'countably infinite' set N of all naturals, or equivalently the
> > countable infinite set A* of all strings over an alphabet A of
> > one symbol, say A={1}, then what is so strange about generating
> > an uncountable infinite set B* of all strings over B={0,1}
>
> You are talking about "generating" again. I seem to recall
> a dismal thread a little while ago, where you insisted
> that the set of all permutations of Z were "generated"
> by two or three particular permutations. However this
> use of "generated" seemed to come (like many other of
> your terms) from your own private lexicon, and you steadfastly
> refused to explain what on earth you meant by it (certainly
> not the conventional usage in group theory, where a
> countable set of generators must generate a countable group).
Yeah, just as countable (in fact binary) alphabet B={0,1} must
necessarily generate the countable [%] set of reals on (0,1)
in binary code of countable_length strings in B* over B. ??
([%]: I think not~ ;-)
Well, now that you bring this up: in the meantime I found a simpler
construction, using a special form of Cantor table, to show the
generative use of his diagonals, under table-row permutations,
mapping lattice 2^N one-one into group N!
Consider the first n naturals in 'tally' code, filling a Cantor
type binary table, for instance n=6 (as I've shown earlier):
1. 1 0 0 0 0 0 Cantor diagonal CD = ~(1 1 1 1 1 1) = 0 0 0 0 0 0
2. 1 1 0 0 0 0
3. 1 1 1 0 0 0
4. 1 1 1 1 0 0
5. 1 1 1 1 1 0
6. 1 1 1 1 1 1
Cyclic permute rows (1,2,3):
3. 1 1 1 0 0 0
1. 1 0 0 0 0 0
2. 1 1 0 0 0 0 yields CD = -(1 0 0 1 1 1) = 0 1 1 0 0 0
4. 1 1 1 1 0 0
5. 1 1 1 1 1 0
6. 1 1 1 1 1 1
Thus any k-cycle permutation (m, m+1, .. , m+k-1) produces as CD
the binary code of the order k-1 subset {m+1, ..m+k-1}. And each
subset consists of a set of such disjoint intervals, which procedure
can be extended for n --> w (set N of rows in w x w table).
This maps each element (= subset of N) of Boolean lattice 2^N
(coded by countably long binary strings) 1-1 into group N!
of all permutations of N, generated by two permutations
- as given in http://home.iae.nl/users/benschop/ism.htm
> > > > The set A^n of distinct such matrices represented by strings
> > > > of length n over A, has order |A^n| which, for n --> inf,
> > > > grows faster than any polynomial, and slower than exponential,
> > > > thus with essentially an 'in-between' growth rate.
> > >
> > > Fascinating.
> > >
> > > > This appears to defy the Continuum Hypothesis, doesn't it?
> > >
> > > No. It actually appears to have nothing to do with the
> > > continuum hypothesis.
> >
> > How come? CH conjectures there is no infinite set 'between'
> > Cantors uncountable~ and Peanos countable infinity.
>
> OK, no set in cardinality between aleph_0 and 2^(aleph_0),
> but what the fuck has your example to do with that?
> Where is the putative set that might have inequality.
> Let me stress it again: CH has to do with SETs. All I see is a
> sequence of numbers |A^n| which tend to infinity,
NO, there is the set of non-equivalent strings (B*)/N which are
functions N --> N, implemented as strings over input alphabet B
of a sequential machine M (integer state machine):
M: N x B --> N, with state set N (the naturals),
and equivalence defined as: x ~ y in B* iff qx == qy for all q in N.
> superpolynomially but subexponentially. No set at all.
> Anyway the existence of such a sequence ain't news: n^(log n)
> tends to infinity superpolynomially and subexponentially.
> The point about the Monthly article was that it's an amusing fact
> that one particular sequence can be proved to have this property.
>
> > And unless you specify how some infinite set is _generated_, as
> > Peano and Cantor did, you just don't know what you are talking
> > about.
>
> Actually no, it's you who do not know what you are talking
> about Mr Benschop.
(re: unless generation procedure is defined...)
I adjusted that already to:
...you don't know what kind of infinity you are talking about.
^^^^^^^^^^^^^^^^
> > >
> > > What property? That they "appear to defy the continuum
> > > hypothesis"? -- Robin Chapman
> >
> > No, the property of generating an infinity between countable
> > and uncountable, thus between A* and B* where |A|=1 and |B|=2,
>
> What the fuck are you talking about here? What does it
> mean for a sequence of numbers ro "generate an infinity".
>
> > being neither of polynomial- nor of exponential growth rate
> > of |A^n| resp. |B^n| for n --> w. -- NB
>
> As I said, such sequences are not news, and have nothing
> to do with infinite cardinalities.
>
> > PS: surely sci.math can yield a response less prejudiced than RC's ?
>
> Yes I am prejudiced. I hate idiocy and I can scarcely
> tolerate morons. I just have to tell them how stupid they
> are. I know I should be unprejudiced and just accept that
> they are different; I just can't do it though. -- Robin Chapman
It requires a lot patience from me to keep on a discussion with
someone who cannot refrain from using 'fuck' - 'idiocy' - 'moron' -
'stupid' and God knows what other terms (from a supposedly academic
person;-( So unless you shape up, and contribute something useful,
and in a palatable manner, I bid you farewell...
I wonder, does using such terms mean that you're losing grip, again ?
-- NB
David C. Ullrich
And now where is the proof that this set is uncountable but has
cardinality less than c?
I'd be willing to bet a large sum that it is either easy to
show this set is countable or easy to show it has cardinality c.
I can't quite prove that right now because I haven't been paying
attention to the notation. I think I know what you mean by
"alphabet" - two questions about the meaning of the rest of it:
Q1: Does B* consist of the finite strings over B or does it include
infinite strings?
Q2: What does qx mean, for q in N and x in B*?
David C. Ullrich
Robin Chapman
> Robin Chapman wrote:
> >
>
> > You are talking about "generating" again. I seem to recall
> > a dismal thread a little while ago, where you insisted
> > that the set of all permutations of Z were "generated"
> > by two or three particular permutations. However this
> > use of "generated" seemed to come (like many other of
> > your terms) from your own private lexicon, and you steadfastly
> > refused to explain what on earth you meant by it (certainly
> > not the conventional usage in group theory, where a
> > countable set of generators must generate a countable group).
>
> Yeah, just as countable (in fact binary) alphabet B={0,1} must
> necessarily generate the countable [%] set of reals on (0,1)
> in binary code of countable_length strings in B* over B. ??
> ([%]: I think not~ ;-)
You think not. I agree. I see no sign of thought from you
at all.
Are you *ever* going to divulge to use what you *mean*
by "generate"? As evidenced by the previous paragraph
you seem to be in some doubt whether the set of functions
from N to {0,1} is countable or not. Lack of thought indeed!
> Well, now that you bring this up: in the meantime I found a simpler
> construction, using a special form of Cantor table, to show the
> generative use of his diagonals, under table-row permutations,
> mapping lattice 2^N one-one into group N!
That's not hard to do.
> Consider the first n naturals in 'tally' code, filling a Cantor
> type binary table, for instance n=6 (as I've shown earlier):
>
> 1. 1 0 0 0 0 0 Cantor diagonal CD = ~(1 1 1 1 1 1) = 0 0 0 0 0 0
> 2. 1 1 0 0 0 0
> 3. 1 1 1 0 0 0
> 4. 1 1 1 1 0 0
> 5. 1 1 1 1 1 0
> 6. 1 1 1 1 1 1
>
> Cyclic permute rows (1,2,3):
>
> 3. 1 1 1 0 0 0
> 1. 1 0 0 0 0 0
> 2. 1 1 0 0 0 0 yields CD = -(1 0 0 1 1 1) = 0 1 1 0 0 0
> 4. 1 1 1 1 0 0
> 5. 1 1 1 1 1 0
> 6. 1 1 1 1 1 1
>
> Thus any k-cycle permutation (m, m+1, .. , m+k-1) produces as CD
> the binary code of the order k-1 subset {m+1, ..m+k-1}. And each
> subset consists of a set of such disjoint intervals, which procedure
> can be extended for n --> w (set N of rows in w x w table).
Now what does "can be extended for n --> w" mean?
> This maps each element (= subset of N) of Boolean lattice 2^N
> (coded by countably long binary strings) 1-1 into group N!
So what permutation gives the set {1,3}?
The set {n in N: n odd}? The set {n in N: n prime}.
And anyway, you said above that you would show
"mapping lattice 2^N one-one into group N!".
You bloody didn't!
> of all permutations of N, generated by two permutations
> - as given in http://home.iae.nl/users/benschop/ism.htm
No, a countable set of permutations only generates
a countable group. The set of all permutations of an infinite
set, N for example, is not countable. You are wrong, when
you maintain that the set of all permutations of N
are generated by two permutations. I have pointed this out
to you on many occasions, so why do you continue to maintain it?
Do you really want the whole world to see how thick you are?
>
> NO, there is the set of non-equivalent strings (B*)/N which are
> functions N --> N, implemented as strings over input alphabet B
> of a sequential machine M (integer state machine):
> M: N x B --> N, with state set N (the naturals),
> and equivalence defined as: x ~ y in B* iff qx == qy for all q in N.
What has this to do with your putative intermediate
cardinality set?
> >
> > Actually no, it's you who do not know what you are talking
> > about Mr Benschop.
>
> (re: unless generation procedure is defined...)
So then Mr Benschop, define your notion of "generation".
> > >
> > > No, the property of generating an infinity between countable
> > > and uncountable, thus between A* and B* where |A|=1 and |B|=2,
> >
> > What the fuck are you talking about here? What does it
> > mean for a sequence of numbers ro "generate an infinity".
> >
> > > being neither of polynomial- nor of exponential growth rate
> > > of |A^n| resp. |B^n| for n --> w. -- NB
So again I ask, what the fuck were you talking about?
You have a sequence of sets whose sizes increase
subexponetially and superpolynomially? I ask again,
what the fuck has that to do with CH?
And I ask anew, why the fuck are you avoiding these questions?
> It requires a lot patience from me to keep on a discussion with
> someone who cannot refrain from using 'fuck' - 'idiocy' - 'moron' -
> 'stupid' and God knows what other terms (from a supposedly academic
> person;-( So unless you shape up, and contribute something useful,
Tu quoque Mr Benschop. It is you who has failed to contribute
anything useful to this discussion. I have attempted to
usefully point out your errors and confusions, but it is you
have stubbornly refused to move this thread into the realm
of useful and productive dialogue.
> and in a palatable manner, I bid you farewell...
> I wonder, does using such terms mean that you're losing grip, again ?
And is this another excuse for you to avoid admitting
that you are talking a load of bollocks yet again?
Nico Benschop
>
> Nico Benschop <n.ben...@chello.nl> wrote:
>
> >Robin Chapman wrote:
> >>
> >> "Nico Benschop" <n.ben...@chello.nl> wrote
> >> > > "Wade Ramey" <wrame...@attbi.remove13.com> wrote
> >> > > > Nico Benschop <n.ben...@chello.nl> wrote:
> >> > > >
> >> > > > > Anything that grows exponentially, is uncountable
> >> > > > > after countably infinite time (isn't it?-)
> >> > > >
> >> > > > ??? I think this makes no sense.
> >>
> >> Wise words.
> >>
> >> > Consider alphabet B of two symbols, say B={a,b} or {0,1}
> >> > Then there are 2^n binary strings of finite length n over B,
> >>
> >> OK
> >>
> >> > Set B* = 2^N, of all binary strings of countably inf. length
> >> > over B, has the same cardinality as the reals R01 on interval
> >> > (0,1) - AFAIK,
> >>
> >> Yes, easy enough to prove.
> >>
> >> > taking the arithmetic semantics of binary coded reals 0 < r < 1.
> >>
> >> "arithmetic semantics"?
> >
> > Well, real r = \sum r_i.2^{-i} for i \in N,
> > with binary coefficients r_i \in {0,1}
> >
> >> > And doesn't Cantors diagonal argument hold for both: set B* of
> >> > binary strings without arithmetic semantics, and R01 of reals ?
> >>
> >> The diagonal argument shows straightforwardly that {0,1}^N
> >> is uncountable. For the reals, it's better to use a Baire
> >> Category Theorem argument than to mess around with digital
> >> representations.
> >>
> >> > If so, please tell me what is wrong with |B*| = |R01| both
> >> > being uncountable due to their "exponential growth rate" in
> >> > lim n --> w, with Peanos naturals |N| = w being 'countable'.
> >> > For comparison: set N= A* : the set of all strings over
> >> > one_symbol alphabet |A|=1.
like the tally way of representation:
1. 1
2. 11
3. 111
4. 1111
5. 11111
6. 111111
etc...
> >>
> >> It's a non-sequitur.
> >
> > ??
> >
> >> {0,1}^N and (0,1) are uncountable because they are uncountable,
> >> not because of any "exponential growth rate". -- Robin Chapman
> >
> > You mean by definition, or axiom?
> > Sorry, I did'nt realize that the exponential origin
> > of the uncountable was also unspeakable.
>
> Nobody but you has any idea what "the exponential origin of
> the uncountable" _means_.
>
> Look. Say S_n is the set of all strings of 0's and 1's of length n.
> Then S_n has 2^n elements - this grows exponentially. But the
> union of the S_n is countable. [...] -- David C. Ullrich
Indeed for finite n. Now let n --> w, yielding strings of
countably infinite length having the above given arithmetic
semantics as reals on (0,1). Then the union of countably many S_n :
Union S_n (n --> w), which in my case is the set B* of all binary
strings (without arithmetic interpretation, including countably
many equivalences of type ..x999... == ..y000... where x<9 and
y= x+1)
: precisely the set of reals on (0,1) thus uncountable, isn't it?
-- NB
David C. Ullrich
<n.ben...@chello.nl> wrote:
>"David C. Ullrich" wrote:
>>
>> Nico Benschop <n.ben...@chello.nl> wrote:
>>
>> >Robin Chapman wrote:
>> >>
>> >> "Nico Benschop" <n.ben...@chello.nl> wrote
>[...]
>> >
>> >> {0,1}^N and (0,1) are uncountable because they are uncountable,
>> >> not because of any "exponential growth rate". -- Robin Chapman
>> >
>> > You mean by definition, or axiom?
>> > Sorry, I did'nt realize that the exponential origin
>> > of the uncountable was also unspeakable.
>>
>> Nobody but you has any idea what "the exponential origin of
>> the uncountable" _means_.
>>
>> Look. Say S_n is the set of all strings of 0's and 1's of length n.
>> Then S_n has 2^n elements - this grows exponentially. But the
>> union of the S_n is countable. [...] -- David C. Ullrich
>
>Indeed for finite n. Now let n --> w, yielding strings of
> countably infinite length having the above given arithmetic
> semantics as reals on (0,1). Then the union of countably many S_n :
> Union S_n (n --> w), which in my case is the set B* of all binary
>strings
I can't make much sense of this. I don't know what you mean by
"Now let n -> w". If you mean "let n tend to w" nothing has
changed - S_n has cardinality 2^n, which grows exponentially,
and the union of the S_n is countable. (And in this case
the union of the S_n does _not_ contain all binary strings.)
If otoh you mean something more like "let n = w" (taking the
"-->" to be an assignment) then you're talking about S_w.
S_w has cardinality c to begin with - there's no "exponential
growth" in sight.
Probably you just haven't seen it yet - do reply to my question
about the cardinality of B*/N in that other post (as well as
to my two questions about what you meant by B* and qx.)
>(without arithmetic interpretation, including countably
>many equivalences of type ..x999... == ..y000... where x<9 and
>y= x+1)
>: precisely the set of reals on (0,1) thus uncountable, isn't it?
>
>-- NB
David C. Ullrich
Dik T. Winter
> Robin Chapman wrote:
> > However this
> > use of "generated" seemed to come (like many other of
> > your terms) from your own private lexicon, and you steadfastly
> > refused to explain what on earth you meant by it (certainly
> > not the conventional usage in group theory, where a
> > countable set of generators must generate a countable group).
>
> Yeah, just as countable (in fact binary) alphabet B={0,1} must
> necessarily generate the countable [%] set of reals on (0,1)
> in binary code of countable_length strings in B* over B. ??
> ([%]: I think not~ ;-)
Yes, you are using your private lexicon, thanks that you now have
proven it. Your "alphabet" does not generate the set of reals in
the conventional meaning of generate. It generates the set of
rationals with denominators of the form 2^p with p some integer.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> "David C. Ullrich" wrote:
> > Look. Say S_n is the set of all strings of 0's and 1's of length n.
> > Then S_n has 2^n elements - this grows exponentially. But the
> > union of the S_n is countable. [...] -- David C. Ullrich
>
> Indeed for finite n. Now let n --> w, yielding strings of
> countably infinite length having the above given arithmetic
> semantics as reals on (0,1).
At no point will the union contain any string of infinite length, because
there is no single set that contains a string of infinite length. So also
in the limit the union will not contain any string of infinite length.
ma...@mimosa.csv.warwick.ac.uk
>> Then S_n has 2^n elements - this grows exponentially. But the
>> union of the S_n is countable. [...] -- David C. Ullrich
>
>Indeed for finite n. Now let n --> w, yielding strings of
I do not understand what 'let n --> w' means.
In fact what do you mean by w? Limits only make sense in the context
of a topology. What is your topology, and what is the underlying set
on which it is defined?
One of the basic principals adhered to by mathemticians is that they do
not use notation without being able to define it absolutely precisely.
I know that mathematicians frequently do use expressions like 'now let
n -> oo', but that is actually an informal use of language, and if
challenged they should always be able to replace it by a more precise
and formal statement.
> countably infinite length having the above given arithmetic
> semantics as reals on (0,1). Then the union of countably many S_n :
> Union S_n (n --> w),
What does "Union S_n (n --> w)" mean?
I do not understand this notation at all, so please define it precisely.
I do not even know what (n --> w) means.
You can take a union of a set of sets. If N is the set {0,1,2,3,... } of
non-negative integers, then
Union S_n (n \in N)
makes sense, and is a countably infinite set consisting of all finite
strings of 0's and 1's. But
I cannot see any meaningful way in which you can 'generate' countably infinite
strings starting from finite strings.
Derek Holt.
Nico Benschop
>
> Nico Benschop <n.ben...@chello.nl> writes:
> > "David C. Ullrich" wrote:
> >> Look. Say S_n is the set of all strings of 0's and 1's of
> >> length n. Then S_n has 2^n elements - this grows exponentially.
> >> But the union of the S_n is countable. [...] -- David C. Ullrich
> >
> > Indeed for finite n. Now let n --> w, yielding strings of
> > countably infinite length having the above given arithmetic
> > semantics as reals on (0,1).
>
> At no point will the union contain any string of infinite length,
> because there is no single set that contains a string of infinite
> length. So also in the limit the union will not contain any string
> of infinite length. -- dik t. winter
Right you are, I have to get used to the difference. -- NB
Even 1/3 in binary code has an infinite representation:
1/3 = .010101010101... thus .(01)* in repetition notation
2/3 = .101010101010... thus .(10)* with '...' : limit
+ -------------------------
1 = .111111111111... thus .(1)*
I get your point: no finite binary string suffices,
since it ends in 0* for any finite length n.
Natural Binary 'NB' (Cantor) table, up to n=6 :
\ \
1. 1 0 0 0 0 0 1 1 0 0 0 0 2.
2. 1 1 0 0 0 0 1 0 0 0 0 0 1.
3. 1 1 1 0 0 0 1 1 1 1 0 0 4.
4. 1 1 1 1 0 0 1 1 1 0 0 0 3.
5. 1 1 1 1 1 0 1 1 1 1 1 1 6.
6. 1 1 1 1 1 1 1 1 1 1 1 0 5.
\ \
CD= 0 0 0 0 0 0 CD= 0 1 0 1 0 1 <-> 1/3 <-> even n \in N.
In terms of Cantor Diagonals (CD), or Complemented Diagonal,
in binary code representing subsets of naturals set N,
with place 1 just right of the binary point, etc.:
1/3 <-> all even n \in N <-> CD of all C2 (swap) perm's (2i-1,2i) i>0
2/3 <-> all odd n \in N <-> CD of all C2 (swap) perm's (2i-2,2i-1) i>0
1 <-> all n \in N <-> uncomplemented Diagonal of Nat_Bin table.
For infinite subsets, resp. reals in (0,1) - this representation
by special permutations of the NB_table rows (disjoint cycles of
adjacent rows) suffices, doesn't it?
Robin Chapman
> "Dik T. Winter" wrote:
> >
> >
> > At no point will the union contain any string of infinite length,
> > because there is no single set that contains a string of infinite
> > length. So also in the limit the union will not contain any string
> > of infinite length. -- dik t. winter
>
> I get your point: no finite binary string suffices,
> since it ends in 0* for any finite length n.
No, no finite binary string ends in 0* (whatever that
is). Consider the finite binary stting 1010010111.
It doesn's have any zeros at the end at all.
> For infinite subsets, resp. reals in (0,1) - this representation
> by special permutations of the NB_table rows (disjoint cycles of
> adjacent rows) suffices, doesn't it?
Suffices for what exactly?
Nico Benschop
>
> Nico Benschop <n.ben...@chello.nl> writes:
> > "David C. Ullrich" wrote:
> >> Look. Say S_n is the set of all strings of 0's and 1's of
> >> length n. Then S_n has 2^n elements - this grows exponentially.
> >> But the union of the S_n is countable. [...] -- David C. Ullrich
> >
> > Indeed for finite n. Now let n --> w, yielding strings of
> > countably infinite length having the above given arithmetic
> > semantics as reals on (0,1).
>
> At no point will the union contain any string of infinite length,
> because there is no single set that contains a string of infinite
> length. So also in the limit the union will not contain any string
Right you are, I have to get used to the difference.
Even 1/3 in binary code has an infinite representation:
1/3 = .010101010101... thus .(01)* in repetition notation
2/3 = .101010101010... thus .(10)* with '...' : limit
+ -------------------------
1 = .111111111111... thus .(1)*
I get your point: no finite binary string suffices,
since it ends in 0* for any finite length n.
Natural Binary 'NB' (Cantor) table up to n=6 (including 0, for 2/3):
\ \ ^^^^^^^^^^^
0. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.
1. 1 0 0 0 0 0 0 1 1 0 0 0 0 0 2.
2. 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1.
3. 1 1 1 0 0 0 0 1 1 1 1 0 0 0 4.
4. 1 1 1 1 0 0 0 1 1 1 0 0 0 0 3.
5. 1 1 1 1 1 0 0 1 1 1 1 1 1 0 6.
6. 1 1 1 1 1 1 0 1 1 1 1 1 0 0 5.
\ \
D= 0 0 0 0 0 0 0 D= 0 1 0 1 0 1 0 <-> 1/3 <-> even n \in N.
In terms of Cantor Diagonals (D) - notice: uncomplemented -,
in binary code representing subsets of naturals set N,
with place 1 just right of the binary point, etc.:
1/3 <-> all even n \in N <-> D of all C2 (swap) perm's (2i-1,2i) i>0
2/3 <-> all odd n \in N <-> D of all C2 (swap) perm's (2i-2,2i-1) i>0
1 <-> all n \in N <-> -D : Complemented Diagonal of Nat_Bin table.
For infinite subsets, resp. reals in (0,1) - this representation
by special permutations of the NB_table rows (disjoint cycles of
adjacent rows) suffices, doesn't it?
Dik T. Winter
> "Dik T. Winter" wrote:
> > At no point will the union contain any string of infinite length,
> > because there is no single set that contains a string of infinite
> > length. So also in the limit the union will not contain any string
> > of infinite length. -- dik t. winter
>
> Right you are, I have to get used to the difference.
So, here there is exponential grow but the result is countable. And
you said that exponential grow implied uncountable, hence that was
wrong, showing that was the purpose of David Ullrich's example.
...
> Natural Binary 'NB' (Cantor) table up to n=6 (including 0, for 2/3):
Now you have an unlimited number of swaps. In this case, yes, you get
1's at all odd places. But that is of course something completely
different. I have a feeling of deja vu from a long time ago. The map
you have here is neither injective nor surjective.
> For infinite subsets, resp. reals in (0,1) - this representation
> by special permutations of the NB_table rows (disjoint cycles of
> adjacent rows) suffices, doesn't it?
Suffices for what? You mean:
> This maps each element (= subset of N) of Boolean lattice 2^N
> (coded by countably long binary strings) 1-1 into group N!
> of all permutations of N, generated by two permutations
? You give a map from sequences of swaps of the positive integers
to the reals. I do not see where the lattice comes in. And the
map you have given is not 1-1.
To show that it is not injective:
Consider a 4x4 table. There are 24 permutations but only 16
different possible diagonals.
To show that it is not surjective:
A real that terminates in its binary expansion is equal to some
real that goes on with a continuous series of 1's. That is, the
single swap (12) gives the same real as the infinite series of
swaps (23)(34)(45)(56)... (But the latter is of course not a
permutation of the integers.
Nico Benschop
>
> Nico Benschop writes:
> > "Dik T. Winter" wrote:
> > > At no point will the union contain any string of infinite
> > > length, because there is no single set that contains a string
> > > of infinite length. So also in the limit the union will not
> > > contain any string of infinite length. -- dik t. winter
> >
> > Right you are, I have to get used to the difference.
>
> So, here there is exponential grow but the result is countable. And
> you said that exponential grow implied uncountable, hence that was
> wrong, showing that was the purpose of David Ullrich's example.
> ...
> > Natural Binary 'NB' (Cantor) table up to n=6
> > (including 0, for 2/3):
>
> Now you have an unlimited number of swaps. In this case, yes, you get
> 1's at all odd places. But that is of course something completely
> different. I have a feeling of deja vu from a long time ago.
That's right, I'm occasionally working on it, see
http://home.iae.nl/users/benschop/cantor.htm and */ism.htm
> The map you have here is neither injective nor surjective.
>
> > For infinite subsets, resp. reals in (0,1) - this representation
> > by special permutations of the NB_table rows (disjoint cycles of
> > adjacent rows) suffices, doesn't it?
>
> Suffices for what? You mean:
>
> > This maps each element (= subset of N) of Boolean lattice 2^N
> > (coded by countably long binary strings) 1-1 into group N!
> > of all permutations of N, ...
>
> You give a map from sequences of swaps of the positive integers
> to the reals.
No, as mentioned: I map each subset of N (in other words an element
of lattice 2^N) - which is coded uniquely by a binary string as
diagonal in the Natural Binary table - into group N! of
permutations of N (here: a permutation of the rows of that table)
This is a 1-1 mapping into N! due to the construction of the perm'n,
corresponding uniquely to the ordered set of adjacent rows that
are cyclically permuted to obtain the desired diagonal = binary coded
subset of N. The mapping is from 2^n to n! elements, with 2^n < n!
for finite n>4 (good point, thanks).
PS:
That the full group N! (resp. Z!) be 'generated by two permutations'
is irrelevant here (so I drop that claim;-) since I map only to a
special subset of permutations, namely permuting cyclically disjoint
subsets of adjacent rows of the table.
These cycles are disjoint, so they can (I think) be obtained by
commuting sub sequences of three 'basic' permutations in A={a,b,c}
mentioned earlier: shift_up n+1, shift_down n-1, and 'swap' (0,1)
- where naturals N are replaced by integers Z (forming a group).
> I do not see where the lattice comes in.
> And the map you have given is not 1-1.
>
> To show that it is not injective:
> Consider a 4x4 table. There are 24 permutations but only 16
> different possible diagonals.
> To show that it is not surjective:
> A real that terminates in its binary expansion is equal to some
> real that goes on with a continuous series of 1's. That is, the
> single swap (12) gives the same real as the infinite series of
> swaps (23)(34)(45)(56)... (But the latter is of course not a
> permutation of the integers. -- dik t. winter
I'm now not concerned with reals in (0,1) - which have the problem of
multiple representations as you mention - but with 2^N (resp. 2^Z)
to be represented by certain type of permutations of N (resp. Z).
It's just strings in B* over B={0,1} <--> Diagonals <--> Row permut's.
-- NB
Dik T. Winter
> "Dik T. Winter" wrote:
> > > This maps each element (= subset of N) of Boolean lattice 2^N
> > > (coded by countably long binary strings) 1-1 into group N!
> > > of all permutations of N, ...
> >
> > You give a map from sequences of swaps of the positive integers
> > to the reals.
>
> No, as mentioned: I map each subset of N (in other words an element
> of lattice 2^N) - which is coded uniquely by a binary string as
> diagonal in the Natural Binary table - into group N! of
> permutations of N (here: a permutation of the rows of that table)
This is just verbiosity. You go from the subsets of N to sequences of
swaps to the reals. That is what I see. Note that *not* every
sequence of swaps is a permutation of N. And some of the sequence
of swaps indeed are not a permutation.
> This is a 1-1 mapping into N! due to the construction of the perm'n,
> corresponding uniquely to the ordered set of adjacent rows that
> are cyclically permuted to obtain the desired diagonal = binary coded
> subset of N. The mapping is from 2^n to n! elements, with 2^n < n!
> for finite n>4 (good point, thanks).
Well, it can not be 1-1 for any finite n, because the number of elements
in the two sets are not equal. So it is (at most) an injection. And,
yes, for finite n it is an injection when n >= 4.
> I'm now not concerned with reals in (0,1) - which have the problem of
> multiple representations as you mention - but with 2^N (resp. 2^Z)
> to be represented by certain type of permutations of N (resp. Z).
Ok.
> It's just strings in B* over B={0,1} <--> Diagonals <--> Row permut's.
And this kind of sentences make your articles so difficult to read.
Quite a bit of non-standard terminology without definitions.
Glen
You math people seem to unaware that there is only a limited amount of
US currency to own anyway, so unless you played this game with the
federal givnment and they decided to mint more for you, you would not
get an amount approaching infinity.
Henry wrote:
> On Sat, 13 Apr 2002 11:05:37 GMT, Nico Benschop <n.ben...@chello.nl>
> wrote:
>
>>Anything that grows exponentially, is uncountable
>> after countably infinite time (isn't it?-)
>>
>
> That depends. 2^n increases to the same infinity as n when n tends to
> infinity. But the power set of a infinite set is a higher infinity.
> (The difference is between ordinal and cardinal infinities.) It all
> depends on what you might mean by the meaningless phrase "after
> countably infinite time".
>
>
Tim Mellor
|B^*|=|RO1| is correct because we can code (0,1) by (legnth omega)
binary strings which is setting up a bijection detween the two sets
(and this is the definition of |B^*|=|RO1|).
There was some discussion earlier about lim_{n -> omega}(2^n) this is
the equivalent to the size of the set of all finite binary strings,
which is ofcourse omega as we can identify it naturally with a subset
of the rationals.
vIRCiated(tm)
"Koe Nic" <koe...@usa.com> escreveu na mensagem
news:4da3663d.02041...@posting.google.com...
> Here is a little trick that I like to try on people at times. Perhaps
> some of you have done the same thing. I know I've had fun with it
> several times.
>
> I would be in a conversation with someone I know. Eventually I would
> make a comment about seeing if they would give me a penny today and
> then 2 cents tomorrow, 4 cents the next day, and then just continue to
> double it each day for a month. Most people will agree to it since it
> only seems to them that they will be giving me not much more than a
> few cents.
>
> There was this one friend I would see on a daily basis and he actually
> did start doing this. However, after it reached $2.56 he finally
> started thinking and decided he didn't want to continue.
>
> The typical person might not stop and think that if he would do this
> for a month he would end up making me a millionaire, since it starts
> with such a small amount, but he would.
>
> If you know a few millionaires, this might be a good trick to play on
> them, or just with some friends who might not be aware of how much
> this would actually cost them.
It will grow infinity...
1 day -> 2c
2 day -> day(1)*2
3 day -> day(2)*2
4 day -> day(3)*2
5 day -> day(4)*2
6 day -> day(5)*2
...
nth day -> day(nth-1)*2
If I am not wrong the limit of this is infinity.
> --
> Almost Anything You Want To Know
> http://www.weeuniverse.com
Virgil
The limit to this depends on how good a con artist you are.
Adam Russell
"vIRCiated(tm)" <ma...@zipmail.com.br> wrote in message
news:abjqfe$i0ir0$3...@ID-88878.news.dfncis.de...
What a strange thing to say.
Dick Alvarez
When you get through with them, they may not be millionaires!
Anybody who is capable of earning a million dollars (if that is the unit
of currency under discussion), probably is smart enough not to fall for that
trick. But some people are more skilled at inheriting their wealth than at
taking care of it.
Dick Alvarez
alvarez at olagrande dot net