Request for Review of ZF Inconsistency Proof
Scott
I am in the process of learning the subtle details of the basics of
set theory, a little deeper than textbooks go into. In studying
Cantor's Theorem, I've run across what I believe is a proof that P(N)
is countable infinite. I would like to ask the group to review it and
point out any problems, fallacies, or other areas to focus on in my
studies. Thanks in advance.
Let us begin.
In this thread, Cantor's Theorem will be applied only to the set of
natural numbers -- the general case is left for the reader. Let S
denote the infinite set of denumerable binary strings with alphabet
{0,1}. An individual string s is denoted as s[0],...,s[i-1],s[i]. Let
n[j],n[j-1],...,n[0] denotate a natural number n in radix 2. Let P[x]
denote an element of P(N).
Proposition 2.1: There exists a function f from N into S.
Proof: Let f be f(n) = s[0],...,s[i-1],s[i] where s[x] = ((n/2^x) mod
2) using integer division. By construction, every s[x] is either 0 or
1 and so every natural number n has some corresponding string s.
Suppose there exists f(n)=s and f(n)=t where s != t. Then there exists
some y such that ((n/2^y) mod 2 ) != ((n/2^y) mod 2). Contradiction.
Therefore, f is an injective function from N to S.
Comments/Feedback so far?
Arturo Magidin
In other words, every natural number corresponds to a (finite)
sequence of 0s and 1s (or alternatively, an infinite sequence with a
"tail of zeros"), via reverse binary notation.
(Trying to anticipate you: if you will apply the diagonal process to
deduce that "natural numbers are not denumerable", have you checked to
see if the resulting sequence has a tail of zeros?)
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Jesse F. Hughes
> In this thread, Cantor's Theorem will be applied only to the set of
> natural numbers -- the general case is left for the reader. Let S
> denote the infinite set of denumerable binary strings with alphabet
> {0,1}. An individual string s is denoted as s[0],...,s[i-1],s[i]. Let
> n[j],n[j-1],...,n[0] denotate a natural number n in radix 2. Let P[x]
> denote an element of P(N).
>
> Proposition 2.1: There exists a function f from N into S.
>
> Proof: Let f be f(n) = s[0],...,s[i-1],s[i] where s[x] = ((n/2^x) mod
> 2) using integer division. By construction, every s[x] is either 0 or
> 1 and so every natural number n has some corresponding string s.
> Suppose there exists f(n)=s and f(n)=t where s != t. Then there exists
> some y such that ((n/2^y) mod 2 ) != ((n/2^y) mod 2). Contradiction.
> Therefore, f is an injective function from N to S.
Okay, there is an injection from N to S. That's not too hard to see.
But I doubt it will get you too far. What you'd really like is a
surjection from N to S.
--
Jesse F. Hughes
"So, either I've found the world's first perfect, non-quantum, random
number generator, or there's a way to make this [factorization] method
work." -- James S. Harris: It's Win-Win, Baby!
MoeBlee
> Hi All:
>
> I am in the process of learning the subtle details of the basics of
> set theory, a little deeper than textbooks go into.
Right, because to prove the inconsistency of set theory you have to
really burrow down much much deeper than the superficiality of
textbooks. When we get into the real subtle aspects that the textbooks
evade is when we see the inconsistency of set theory positively
glaring right at us. Most people, though, are not cut out for that
degree of intellectual honesty. So we really must be thankful for
people such as you who are our last hope for intellectual salvation.
> In studying
> Cantor's Theorem, I've run across what I believe is a proof that P(N)
> is countable infinite. I would like to ask the group to review it and
> point out any problems, fallacies, or other areas to focus on in my
> studies. Thanks in advance.
Okie dokie.
> Let us begin.
Yes, lettuce.
> In this thread, Cantor's Theorem will be applied only to the set of
> natural numbers -- the general case is left for the reader.
Oh, left for the reader. Like a real life textbook exercise. Gee, this
is challenging already.
> Let S
> denote the infinite set of denumerable binary strings with alphabet
> {0,1}. An individual string s is denoted as s[0],...,s[i-1],s[i].
That's not a denumerable string. That's a finite string. So, right
away, it's completely unclear what you have in mind. Do you mean for S
to be the set of denumerable binary strings or do you mean S to be the
set of finite binary strings? It is a bit more than a "subtle"
difference.
> Let
> n[j],n[j-1],...,n[0] denotate a natural number n in radix 2.
I know what you mean, so okay.
> Let P[x]
> denote an element of P(N).
I take it you mean:
P[x] iff x is a member of P(N).
> Proposition 2.1: There exists a function f from N into S.
Oh, proposition numbers already. This must be serious business indeed.
> Proof: Let f be f(n) = s[0],...,s[i-1],s[i] where s[x] = ((n/2^x) mod
> 2) using integer division. By construction, every s[x] is either 0 or
> 1 and so every natural number n has some corresponding string s.
> Suppose there exists f(n)=s and f(n)=t where s != t. Then there exists
> some y such that ((n/2^y) mod 2 ) != ((n/2^y) mod 2). Contradiction.
> Therefore, f is an injective function from N to S.
What the L? You need all that just to prove there's a function from N
into S?
You want a function from N into S?
Let x be any member of S.
{<n x> | n in N} is a function from n into S.
> Comments/Feedback so far?
Yes, are the exercises going to get much harder as we go along?
Because my brain is already worn out. I don't think I can cope with
all these subtleties.
MoeBlee
Jesse F. Hughes
Moe, with all due respect, you were more than a bit snippy. This guy
has not been a presumptuous twit, like most folks who claim to have a
proof that ZF is inconsistent. He seems fairly reasonable and is
asking for help evaluating his argument.
>> Proof: Let f be f(n) = s[0],...,s[i-1],s[i] where s[x] = ((n/2^x) mod
>> 2) using integer division. By construction, every s[x] is either 0 or
>> 1 and so every natural number n has some corresponding string s.
>> Suppose there exists f(n)=s and f(n)=t where s != t. Then there exists
>> some y such that ((n/2^y) mod 2 ) != ((n/2^y) mod 2). Contradiction.
>> Therefore, f is an injective function from N to S.
>
> What the L? You need all that just to prove there's a function from N
> into S?
>
> You want a function from N into S?
He *did* say "injective". Not that it is difficult to prove there's
an injection from N to S or that such an injection seems likely to
help him out, but his claim is a touch more difficult than you read
it.
> Let x be any member of S.
>
> {<n x> | n in N} is a function from n into S.
As I said above, he wants an injective function.
--
Jesse F. Hughes
"And I'm one of my own biggest skeptics as I had *YEARS* of wrong
ideas, and attempts that failed. Worse, for some of them it took
*MONTHS* before I figured out where I screwed up." -- James Harris
MoeBlee
> MoeBlee <jazzm...@hotmail.com> writes:
>
> Moe, with all due respect, you were more than a bit snippy. This guy
> has not been a presumptuous twit, like most folks who claim to have a
> proof that ZF is inconsistent. He seems fairly reasonable and is
> asking for help evaluating his argument.
His tone rubs me wrong: "Deeper than the textbooks". Okay, I may have
jumped ahead on this one, but I'm bettting crank and rounds of rounds
of the usual cycle of crank. You know, starts off with "I'm just
asking..." and soon enough reverts to the usual crank obstinancy.
Maybe I'm wrong though. We'll see...
> He *did* say "injective".
Only at the end of a proof of a propostion that says:
"Proposition 2.1 There exists a function f from N into S."
Okay, he made a typo of omission. That would be fair enough, and I
wouldn't pounce on such a thing if I didn't sense that the guy is a
crank from the start, as he STARTS by purporting he has attention to
detail: "I am in the process of learning the subtle details of the
basics of set theory, a little deeper than textbooks go into."
Meanwhile, he seems to be using 'denumerable' to mean 'finite'. From
what textbook did he "go deeper" to find that "subtlety"?
But like I said, perhaps I did jump early. We'll see...
MoeBlee
Peter Webb
"MoeBlee" <jazz...@hotmail.com> wrote in message
news:1181022336.5...@z28g2000prd.googlegroups.com...
I'm still waiting for somebody to ask him where 1/3 is found on his list of
all Reals [0,1].
What's the bet that it doesn't, even though we don't even yet know how his
list is constructed?
Aatu Koskensilta
> I am in the process of learning the subtle details of the basics of
> set theory, a little deeper than textbooks go into.
Which textbooks would these be? As mathematics goes there's nothing
particularly subtle to the details of the basics of set theory.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Jesse F. Hughes
> On Jun 4, 5:39 pm, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
>> MoeBlee <jazzm...@hotmail.com> writes:
>>
>> He *did* say "injective".
>
> Only at the end of a proof of a propostion that says:
>
> "Proposition 2.1 There exists a function f from N into S."
I missed the omission in the proposition statement, I guess.
> But like I said, perhaps I did jump early. We'll see...
Indeed.
--
"Quincy, would you rather do epistemology or conceptual analysis?"
"You know what? I'd rather fight on an aircraft carrier.... And Mama
and Baba (Papa) would fight on an aircraft carrier, too."
-- Quincy P. Hughes, age 3 1/2
Scott
> > Let S
> > denote the infinite set of denumerable binary strings with alphabet
> > {0,1}. An individual string s is denoted as s[0],...,s[i-1],s[i].
>
> That's not a denumerable string. That's a finite string. So, right
> away, it's completely unclear what you have in mind. Do you mean for S
> to be the set of denumerable binary strings or do you mean S to be the
> set of finite binary strings? It is a bit more than a "subtle"
> difference.
Thanks for pointing out this typing error. Let S denote the infinite
set of infinite denumerable binary strings with alphabet {0,1}.
> > Let P[x]
> > denote an element of P(N).
>
> I take it you mean:
>
> P[x] iff x is a member of P(N).
No. I original wrote P_x \in \mathfrak{P}(N) but thought it might be
too difficult to read. So I'm subsituting array notation to indicate
subscripts; ie, P[x] <-> P_x and P(N) <-> \mathfrak{P}(N).
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 4, 3:09 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>> > Let S
>> > denote the infinite set of denumerable binary strings with alphabet
>> > {0,1}. An individual string s is denoted as s[0],...,s[i-1],s[i].
>>
>> That's not a denumerable string. That's a finite string. So, right
>> away, it's completely unclear what you have in mind. Do you mean for S
>> to be the set of denumerable binary strings or do you mean S to be the
>> set of finite binary strings? It is a bit more than a "subtle"
>> difference.
>
>Thanks for pointing out this typing error. Let S denote the infinite
>set of infinite denumerable binary strings with alphabet {0,1}.
Then your function as defined is incorrect. The images are FINITE
strings, so they are not elements of S.
Of course, you can always take the finite string and add a "tail of
zeroes" to make it an infinite string. But you have to do it
explicitly, and you did not.
Scott
> MoeBlee <jazzm...@hotmail.com> writes:
> > On Jun 4, 5:39 pm, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
> >> MoeBlee <jazzm...@hotmail.com> writes:
>
> >> He *did* say "injective".
>
> > Only at the end of a proof of a propostion that says:
>
> > "Proposition 2.1 There exists a function f from N into S."
>
> I missed the omission in the proposition statement, I guess.
>
Is my terminology incorrect here? Is "f from N into S" the proper way
of stating "f is a injective function from N to S"? Or is it better to
use the latter?
Arturo Magidin
It is better to use the latter. In general, "f is a function from A
into B" may easily be interpret to mean "A is the domain, B is the
codomain", and nothing else.
Scott
> >Thanks for pointing out this typing error. Let S denote the infinite
> >set of infinite denumerable binary strings with alphabet {0,1}.
>
> Then your function as defined is incorrect. The images are FINITE
> strings, so they are not elements of S.
Why do you conclude the image contains finite strings? As defined
i=infinity and x=0..i so the image should contain infinite strings.
Jesse F. Hughes
It is better to use the latter.
--
Jesse F. Hughes
"Really, I'm not out to destroy Microsoft. That will just be a
completely unintentional side effect." -- Linus Torvalds
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 5, 9:47 am, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>> >Thanks for pointing out this typing error. Let S denote the infinite
>> >set of infinite denumerable binary strings with alphabet {0,1}.
>>
>> Then your function as defined is incorrect. The images are FINITE
>> strings, so they are not elements of S.
>
>Why do you conclude the image contains finite strings?
Because you ->said<- so.
Here is what you wrote:
> Let f be f(n) = s[0],...,s[i-1],s[i]
You'll notice the absence of an ellipsis at the end. Thus, you are
saying f(n) is a FINITE string.
If you meant it to continue, then it should have been written as:
Let f be f(n) = s[0],...,s[i-1],s[i],...
>As defined
>i=infinity and x=0..i so the image should contain infinite strings.
"As defined" you were unclear and imprecise. A bad start, to be sure.
As I said, easy to fix. But you'll want to be ->extremely<- precise,
->extremely<- clear, and cross ->all<- the t's and dot ->all<- the
i's, surely.
Scott
> > Let f be f(n) = s[0],...,s[i-1],s[i]
>
> You'll notice the absence of an ellipsis at the end. Thus, you are
> saying f(n) is a FINITE string.
So incorporating feedback:
Let S denote the infinite set of infinite denumerable binary strings
with alphabet {0,1}. An individual string s is denoted as
s[0],...,s[i-1],s[i],....
Proposition 2.1: There exists an injective function f from N to S.
Proof: Let f be f(n)=s[0],...,s[i-1],s[i],... where (as previously)...
Is this sufficiently cleaned up to convey the intent?
Scott
> (Trying to anticipate you: if you will apply the diagonal process to
> deduce that "natural numbers are not denumerable", have you checked to
> see if the resulting sequence has a tail of zeros?)
No, I won't be trying to prove "natural numbers are not denumerable".
Arturo Magidin
Yes; and trivial in any case. To every natural number there exists a
unique denumerable sequence of 0's and 1's, almost all terms (that is,
all but finitely many) equal to 0, namely, the sequence of digits in
the binary expression of the natural number.
Scott
Proposition 2.2: There exists an injective function f' from S to N.
Proof: Let f' be
f'(s) = Sum(x=0..infinity) of (2^x * s[x]).
By construction, every string s will generate some natural number n.
Suppose f'(s)=n and f'(s)=m where n != m. Then there exists some y
such that (2^y*s[y]) != (2^y*s[y]). Contradiction. Therefore, f' is an
injective function from S to N.
Comments/feedback?
MoeBlee
> Continuing:
>
> Proposition 2.2: There exists an injective function f' from S to N.
So I suppose you're clear now that S is the set of denumerable binary
strings.
> Proof: Let f' be
>
> f'(s) = Sum(x=0..infinity) of (2^x * s[x]).
>
> By construction, every string s will generate some natural number n.
No, you've not proven the existence of any such operation. Where in
set theory do you find a definition of infinite summation such as you
notate and whose range is a subset of N? Just throwing such notation
together doesn't prove the existence of such operations. In set theory
we prove the existence of the operations we use. You haven't proven
the existence of an operation that takes each denumerable binary
sequence and "sums" its entries to a natural number.
> Suppose f'(s)=n and f'(s)=m where n != m. Then there exists some y
> such that (2^y*s[y]) != (2^y*s[y]). Contradiction. Therefore, f' is an
> injective function from S to N.
>
> Comments/feedback?
Yes, why don't you study a good book on mathematical logic and one on
set theory so that you'll understand that just throwing a bunch of
notation together is not a proof of the existence of an operation.
>From this incident and others (including your earlier posts in this
thread and your posts in other threads) it is clear you don't have a
handle on the basics of the subject. What drives you to skip learning
the basics but to instead throw around a bunch of silly undefined
notation to "prove" something about a subject you haven't even studied
properly?
MoeBlee
Jesse F. Hughes
Let s be the string 11111... . What is f'(s)?
--
God made the bees
And the bees make honey.
The miller's man does all the work,
But the miller makes the money. --- Mother Goose
MoeBlee
wrote:
> On 2007-06-04, in sci.logic, Scott wrote:
>
> > I am in the process of learning the subtle details of the basics of
> > set theory, a little deeper than textbooks go into.
>
> Which textbooks would these be? As mathematics goes there's nothing
> particularly subtle to the details of the basics of set theory.
Might be off the same shelf from which he thought he had read a book
that mentions that Godel was considered a crank when he was a
university student.
MoeBlee
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>Continuing:
>
>Proposition 2.2: There exists an injective function f' from S to N.
>
>Proof: Let f' be
>
>f'(s) = Sum(x=0..infinity) of (2^x * s[x]).
The right hand side only makes sense if s[x]=0 for almost all
x. Otherwise, you have a "sum" with infinitely many nonzero terms, all
of them positive integers, and such a "sum" does not have any meaning
(in the real numbers, it diverges; in the natural numbers, it has no
meaning).
>By construction, every string s will generate some natural number n.
No. Only the strings s that are almost null (all but finitely many
terms nonzero) gives a natural number.
If s is the sequence s[x]=1 for all x, what natural number is
f'(s) = Sum(x=0 to infinity) 2^x
= 1 + 2 + 4 + 8 + 16 + ... + 2^n + ...
supposed to be?
If s is the sequence s[x] = 0 if x is even, s[x]=1 if x is odd, then
what natural number is
f'(s) = 2 + 8 + 32 + ... + 2^{2k+1} + ...
supposed to be?
What you have described is not a function from S to N. It is a
function from a SUBSET of S to N.
>Suppose f'(s)=n and f'(s)=m where n != m. Then there exists some y
>such that (2^y*s[y]) != (2^y*s[y]). Contradiction.
To what?
> Therefore, f' is an
>injective function from S to N.
Ehr, no. Even assuming that f' were a function, what you have here
does not prove that f' is injective; it would only show that f' is
well defined. To prove it is injective, you would need to show that if
f'(s)=n and f'(t)=n, then s=t; or else that if s is not equal to t,
then f'(s) is not equal to f'(t). You did neither.
>Comments/feedback?
(i) You did not define a function from S to N; your f' is only defined
on the subset S' = {s in S | s[x]=0 for almost all x}.
(ii) You did not prove that the function f':S'-->N is injective.
Note: (ii) is easy. (i) is the real problem.
Virgil
If, as seems to be intended from prior posts, s may have infinitely many
non-zero terms, then there is no such injection as described, as there
are some s for which the sum, by having infinitely many positive
integer terms is not defined.
On the other hand, if all s /are/ limited to only finitely many non-zero
terms, most infinite binary sequences have been omitted.
Scott
> In article <1181065310.361517.147...@q75g2000hsh.googlegroups.com>,
> >By construction, every string s will generate some natural number n.
>
> No. Only the strings s that are almost null (all but finitely many
> terms nonzero) gives a natural number.
>
> If s is the sequence s[x]=1 for all x, what natural number is
>
> f'(s) = Sum(x=0 to infinity) 2^x
> = 1 + 2 + 4 + 8 + 16 + ... + 2^n + ...
>
> supposed to be?
I guess this is the issue that I do not have a handle on yet. Let B be
a subset of S such that one and only one bit is set; eg, 0100... and
00010... are members but 0000... and 0101... are not. Then there is a
one-to-one correspondence between an n (n in N) and an a (b in B). So
in this case, an infinite string maps to a natural number. Why then
does this work but the equation above does not?
MoeBlee
I have no idea why you would think that the exisistence of a bijection
between N and a certain proper subset of S suggests that there exists
an operation of infinite summation of natural numbers such that the
range of the operation is a subset of N.
Do you not at all understand the point that in set theory we PROVE the
existence of certain operations before we just throw a bunch of
notation on the page as IF such an operation existed having certain
properties?
MoeBlee
Scott
> Let s be the string 11111... . What is f'(s)?
The rather large natural number ...11111. If natural numbers (widths)
are finite, then this would imply there is some boundary between
finite and infinite, some point at which an infinite string stops
mapping to a natural number. If such a boundary existed, then there
would be some largest number that is the boundary. But there is no
largest natural number so such a boundary does not exist.
I guess the crux of what I don't quite understand, and perhaps you can
shed some light on, is how an infinite number of things is considered
finite; eg, an infinite number of digits in a natural number is
considered finite.
MoeBlee
> On Jun 5, 11:15 am, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
>
> > Let s be the string 11111... . What is f'(s)?
>
> The rather large natural number ...11111. If natural numbers (widths)
> are finite, then this would imply there is some boundary between
> finite and infinite, some point at which an infinite string stops
> mapping to a natural number. If such a boundary existed, then there
> would be some largest number that is the boundary. But there is no
> largest natural number so such a boundary does not exist.
All of that is pure prose that is not in the mathematics of set
theory. We don't prove mathematical theorems by composing such prose.
Why don't you just learn the formal language of set theory and how to
prove things in it?
> I guess the crux of what I don't quite understand, and perhaps you can
> shed some light on, is how an infinite number of things is considered
> finite; eg, an infinite number of digits in a natural number is
> considered finite.
There is no such basis representation of a natural number unless the
representation is one that has all 0's or has all 0's before reaching
a finite substring that is not all 0's.
If you just learn the basics of this subject - set theoretical
definitions, and then the actual basis representation theorem for
natural numbers, then you'll see that your questions evaporate as they
were premised on a lack of understanding the basics of the subject.
MoeBlee
Jesse F. Hughes
> On Jun 5, 11:15 am, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
>
>> Let s be the string 11111... . What is f'(s)?
>
> The rather large natural number ...11111. If natural numbers (widths)
> are finite, then this would imply there is some boundary between
> finite and infinite, some point at which an infinite string stops
> mapping to a natural number. If such a boundary existed, then there
> would be some largest number that is the boundary. But there is no
> largest natural number so such a boundary does not exist.
This is, of course, nonsense. Every natural number is finite.
> I guess the crux of what I don't quite understand, and perhaps you can
> shed some light on, is how an infinite number of things is considered
> finite; eg, an infinite number of digits in a natural number is
> considered finite.
No natural number has an infinite number of non-zero digits. ....11111
is not a natural number at all.
--
Jesse F. Hughes
"A factor is simply something that multiplies against another factor
to produce a 'product'." -- James Harris offers a definition.
Aatu Koskensilta
> I guess the crux of what I don't quite understand, and perhaps you can
> shed some light on, is how an infinite number of things is considered
> finite; eg, an infinite number of digits in a natural number is
> considered finite.
Eat more chili and all will be clear. Your hand will be free of any
extraneous digits.
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 5, 11:15 am, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
>
>> Let s be the string 11111... . What is f'(s)?
>
>The rather large natural number ...11111.
There is no such natural number.
Every natural number has FINITE binary expansion. Prove it by
induction.
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 5, 11:45 am, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>> In article <1181065310.361517.147...@q75g2000hsh.googlegroups.com>,
>> >By construction, every string s will generate some natural number n.
>>
>> No. Only the strings s that are almost null (all but finitely many
>> terms nonzero) gives a natural number.
>>
>> If s is the sequence s[x]=1 for all x, what natural number is
>>
>> f'(s) = Sum(x=0 to infinity) 2^x
>> = 1 + 2 + 4 + 8 + 16 + ... + 2^n + ...
>>
>> supposed to be?
>
>I guess this is the issue that I do not have a handle on yet.
No kidding.
>Let B be
>a subset of S such that one and only one bit is set; eg, 0100... and
>00010... are members but 0000... and 0101... are not.
So?
> Then there is a
>one-to-one correspondence between an n (n in N) and an a (b in B).
So?
> So
>in this case, an infinite string maps to a natural number. Why then
>does this work but the equation above does not?
There is nothing which in principle prevents you from associating a
natural number and an "infinite string" in ->SOME<- way.
However, not EVERY way of attempting to do the association will make
sense.
You have attempted to define the association via a summation. That
summation only makes sense if the number of nonzero terms is
FINITE. Otherwise, your expression is meaningless and does not
correspond to a natural number.
If you wish to try to define some OTHER association between sequences
of 0s and 1s and natural numbers, you are free to try to do so, but
the one you proposed is not well defined; it only works for a very
small subset of the sequences.
The expression
Sum_{x to infinity} 2^x*s[x]
only makes sense if s[x]=0 for almost all values of x. Since not every
s has this property, your proposed "definition" fails to define a
function which is meaningful at every s in S. That is all.
Virgil
Because one of them claims to be able to add up infinitely many positive
integers, which does not work, and the other does not.
Virgil
> On Jun 5, 11:15 am, "Jesse F. Hughes" <j...@phiwumbda.org> wrote:
>
> > Let s be the string 11111... . What is f'(s)?
>
> The rather large natural number ...11111.
Every natural number of my acquaintance has a left-most 1 in its binary
representation. That above does not, so is not a natural number, at
least by my standards.
For another thing, every natural must allow adding 1 to it to form
another natural. What natural do you get when you try to add 1 to your
alleged "rather large natural number ...11111"?
> If natural numbers (widths)
> are finite, then this would imply there is some boundary between
> finite and infinite, some point at which an infinite string stops
> mapping to a natural number. If such a boundary existed, then there
> would be some largest number that is the boundary. But there is no
> largest natural number so such a boundary does not exist.
This is a sort of 'pons asinorum' issue. That there can be an infinite
set of objects of unlimited size without any of them needing to be
themselves infinite.
>
> I guess the crux of what I don't quite understand, and perhaps you can
> shed some light on, is how an infinite number of things is considered
> finite; eg, an infinite number of digits in a natural number is
> considered finite.
Which natural are you declaring needs more than a finite number of
non-zero digits?
1 only needs a finite number of digits
If natural number 'n' only needs a finite number of digits, then 'n+1'
is a natural which also only needs finitely many.
But this covers ALL natural numbers, so that no natural need more than
finitely many digits.
The above style argument is called finite induction, and is central to
understanding the properties of the naturals.
Scott
> All of that is pure prose that is not in the mathematics of set
> theory. We don't prove mathematical theorems by composing such prose.
> Why don't you just learn the formal language of set theory and how to
> prove things in it?
Yes, I did notice all your arguments are in prose and not the formal
language of set theory.
Scott
> 1 only needs a finite number of digits
> If natural number 'n' only needs a finite number of digits, then 'n+1'
> is a natural which also only needs finitely many.
> But this covers ALL natural numbers, so that no natural need more than
> finitely many digits.
>
> The above style argument is called finite induction, and is central to
> understanding the properties of the naturals.
Thanks. This is helpful.
MoeBlee
What a puerile comment that is. First, most of my arguments in this
context have not been to prove any particular theorems of set theory
but rather to explain to you where you've gotten yourself confused
about the subject. Second, when I do make an argument for a proof in
set theory, of course I use informal prose, but I do so in a manner
that anyone familiar with the formal language would can see that that
argument can be formalized. I don't pretend to support mathematical
statements by free floating undefined prose such as you just posted.
Why don't you just learn some basic mathematical logic and the precise
definitions and theorems of set theory? Then see if you can carry out
your arguments in actual set theory rather than with undefined prose
and botched and undefined notations?
MoeBlee
Aatu Koskensilta
> language of set theory.
Why would anyone be interested in tediosuly detailed formulations in the
language of set theory? Sheer perversity?
MoeBlee
wrote:
> On 2007-06-05, in sci.logic, Scott wrote:
>
> > Yes, I did notice all your arguments are in prose and not the formal
> > language of set theory.
>
> Why would anyone be interested in tediosuly detailed formulations in the
> language of set theory? Sheer perversity?
I wrote, "All of that is pure prose that is not in the mathematics of
set
theory. We don't prove mathematical theorems by composing such prose.
Why don't you just learn the formal language of set theory and how to
prove things in it?"
I would have thought that he would have taken that in the reasonable
sense of learning how to perform proofs in a formal theory as that
would contribute to an appreciation of the difference between a bunch
of undefined mathematical sounding prose and an argument informally
given but in principle formalizable (especially since his agenda is to
show a contradiction in ZFC). But he took what I said as material for
a strawman as if I could avoid hypocrisy only by myself giving
arguments strictly in a formal language. Such responses as that are
indicative of someone who does not wish to actually understand the
subject at hand (even if with the ultimate goal of making an attempt
to prove inconsistency).
MoeBlee
Aatu Koskensilta
> I would have thought that he would have taken that in the reasonable
> sense of learning how to perform proofs in a formal theory as that
> would contribute to an appreciation of the difference between a bunch
> of undefined mathematical sounding prose and an argument informally
> given but in principle formalizable (especially since his agenda is to
> show a contradiction in ZFC). But he took what I said as material for
> a strawman as if I could avoid hypocrisy only by myself giving
> arguments strictly in a formal language. Such responses as that are
> indicative of someone who does not wish to actually understand the
> subject at hand (even if with the ultimate goal of making an attempt
> to prove inconsistency).
Yes, quite evil of him. Perhaps -- not that it is any business of mine --
that might suggest it's entirely pointless to explain, explicate, show
stuff to the idiotic individuals?
MoeBlee
wrote:
> On 2007-06-05, in sci.logic, MoeBlee wrote:
>
> > I would have thought that he would have taken that in the reasonable
> > sense of learning how to perform proofs in a formal theory as that
> > would contribute to an appreciation of the difference between a bunch
> > of undefined mathematical sounding prose and an argument informally
> > given but in principle formalizable (especially since his agenda is to
> > show a contradiction in ZFC). But he took what I said as material for
> > a strawman as if I could avoid hypocrisy only by myself giving
> > arguments strictly in a formal language. Such responses as that are
> > indicative of someone who does not wish to actually understand the
> > subject at hand (even if with the ultimate goal of making an attempt
> > to prove inconsistency).
>
> Yes, quite evil of him.
Oh come now [that's one I'm borrowing from you know who], I didn't say
it's evil.
> Perhaps -- not that it is any business of mine --
> that might suggest it's entirely pointless to explain, explicate, show
> stuff to the idiotic individuals?
In the case of the most entrenched cranks, such posts to them are
pointless if the expectation is that of getting through to them. But
one may have one's own reasons for such seemingly unrewarding posting
to cranks, which - as with any such endeavor, habit, or vice - are not
necessarily so very rational in terms of expectations of particular
outcomes.
MoeBlee
MoeBlee
> On Jun 5, 5:02 pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> > Perhaps -- not that it is any business of mine --
> > that might suggest it's entirely pointless to explain, explicate, show
> > stuff to the idiotic individuals?
>
> In the case of the most entrenched cranks, such posts to them are
> pointless if the expectation is that of getting through to them. But
> one may have one's own reasons for such seemingly unrewarding posting
> to cranks, which - as with any such endeavor, habit, or vice - are not
> necessarily so very rational in terms of expectations of particular
> outcomes.
P.S. I didn't say, even suggest, or even assent that he's an idiot.
MoeBlee
Aatu Koskensilta
> Oh come now [that's one I'm borrowing from you know who], I didn't say
> it's evil.
Doing as Scott did is, if not evil, at least somewhat silly. I'm afraid he
has revealed himself as a second class mathematical crank. As you reminded
me, he's had some interesting things to say about Gödel, incompleteness,
inconsistency, in the past, too.
> In the case of the most entrenched cranks, such posts to them are
> pointless if the expectation is that of getting through to them. But
> one may have one's own reasons for such seemingly unrewarding posting
> to cranks, which - as with any such endeavor, habit, or vice - are not
> necessarily so very rational in terms of expectations of particular
> outcomes.
True. Hanging out on Usenet is in itself a somewhat queer hobby.
Aatu Koskensilta
> that mentions that Godel was considered a crank when he was a
> university student.
Possibly. That reminds me; several people, over the years, have talked about
"Gödel's paradox", here in our cozy sci.logic. This has usually been met
with derision -- quite unjustifiedly, I now find; the renowned proof-theorist
Jean-Yves Girard mentions Gödel's paradox in his positively hilarious _Blind
Spot_, available on-line. Next time someone comes along and starts going on
about Gödel's paradox we, in light of this, need to take on our most solemn
expression, think of Girard, and endure the trial.
MoeBlee
wrote:
> On 2007-06-05, in sci.logic, MoeBlee wrote:
>
> > Might be off the same shelf from which he thought he had read a book
> > that mentions that Godel was considered a crank when he was a
> > university student.
>
> Possibly.
Too bad though that when I asked Scott about it, he couldn't produce a
specific reference to such a tantalizing biographical morsel.
> Jean-Yves Girard mentions Gödel's paradox in his positively hilarious _Blind
> Spot_, available on-line.
That sounded like so much fun that I aimed my browser right to the
page, only to find it's offline. So first it's Scott teasing with
hints that Godel himself once suffered a reputation as a crank but no
actual book or article to read, then it's you dropping word of some
real funny stuff on the Internet about mathematical logic but only at
a site that can't be viewed. This is what I mean when I way that
people are EVIL!
MoeBlee
Aatu Koskensilta
> page, only to find it's offline. So first it's Scott teasing with
> hints that Godel himself once suffered a reputation as a crank but no
> actual book or article to read, then it's you dropping word of some
> real funny stuff on the Internet about mathematical logic but only at
> a site that can't be viewed. This is what I mean when I way that
> people are EVIL!
It's evil of Girard to have taken the wonderful _The Blind Spot_ off-line!
The book's full of all sort of great stuff and rambling tirades in grand
style. It also seems it's due to be published (in French) sometime soon, or
so I read in a random blog I stumbled upon.
As a (small) consolation, the book _Proofs and Types_ is still available
on-line, at
http://www.cs.man.ac.uk/~pt/stable/Proofs+Types.html
Alas, it's not even remotely as entertaining as _The Blind Spot_.
MoeBlee
wrote:
> As a (small) consolation, the book _Proofs and Types_ is still available
> on-line, at
>
> http://www.cs.man.ac.uk/~pt/stable/Proofs+Types.html
Very nice! Like a greedy little squirrel finding a plump acorn, I
snatched that one right up and tucked it away in my goodies folder.
Not so small a consolation. Thanks.
> Alas, it's not even remotely as entertaining as _The Blind Spot_.
It'll probably be back online sometime soon. I'll look forward to it.
MoeBlee
David C. Ullrich
>Continuing:
>
>Proposition 2.2: There exists an injective function f' from S to N.
>
>Proof: Let f' be
>
>f'(s) = Sum(x=0..infinity) of (2^x * s[x]).
This is simply one version of precisely the standard
error that people were expecting: This formula makes
sense for finite strings s, or for strings s such
that s[x] = 0 for all but finitely many x. The formula
does not make sense, or at least it does not define
a natural number, for an arbitrary s in S.
>By construction, every string s will generate some natural number n.
>Suppose f'(s)=n and f'(s)=m where n != m. Then there exists some y
>such that (2^y*s[y]) != (2^y*s[y]). Contradiction. Therefore, f' is an
>injective function from S to N.
>
>Comments/feedback?
************************
David C. Ullrich
David C. Ullrich
Huh? You say you don't understand this - what makes you
think it's so? An infinite string of digits does _not_
define a natural number.
(Announcing a proof that ZF is inconsistent at a point where
there are still details you don't understand is a little
silly, by the way.)
************************
David C. Ullrich
Aatu Koskensilta
> (Announcing a proof that ZF is inconsistent at a point where
> there are still details you don't understand is a little
> silly, by the way.)
It's standard practice, however.
Scott
> In article <1181077048.979617.320...@g4g2000hsf.googlegroups.com>,
Scott
while trying to cut and paste.)
On Jun 5, 2:48 pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
> In article <1181077048.979617.320...@g4g2000hsf.googlegroups.com>,
> The expression
>
> Sum_{x to infinity} 2^x*s[x]
>
> only makes sense if s[x]=0 for almost all values of x. Since not every
> s has this property, your proposed "definition" fails to define a
> function which is meaningful at every s in S. That is all.
Just to active listen back to you: If A subset S such that
s[0],...,s[i],0... are the only elements of A, then the expression
above is defined and makes sense? And such would be a mapping from A
to N?
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>(Please excuse the previous post as I accidently hit the send button
>while trying to cut and paste.)
>
>On Jun 5, 2:48 pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>> In article <1181077048.979617.320...@g4g2000hsf.googlegroups.com>,
>> The expression
>>
>> Sum_{x to infinity} 2^x*s[x]
>>
>> only makes sense if s[x]=0 for almost all values of x. Since not every
>> s has this property, your proposed "definition" fails to define a
>> function which is meaningful at every s in S. That is all.
>
>Just to active listen back to you: If A subset S such that
>s[0],...,s[i],0... are the only elements of A
then you are confused and speaking nonsense (again).
S was defined to be the collection of all infinite sequences of 0s and
1s. Thus, if A is a subset of S, then its elements must be
SEQUENCES. But in your notation, s[i] represents the i-th term of the
sequence s, so "s[0]", "s[1]", ..., "s[i]", and "0" are each of them
either "0" or "1", not sequences.
So it is impossible for them to be elements of A at all.
You REALLY need to learn to express yourself clearly and
ACCURATELY. Just like you need to learn to read the replies you get
carefully. In my antepenultimate reply you seemed to think the
objection raised had something to do with the fact that the elements
of s were infinite, when there was absolutely nothing in what was
written that could be interpreted that way except through careless and
confused reading.
>then the expression
>above is defined and makes sense? And such would be a mapping from A
>to N?
A sequence s of numbers is said to be "almost null" or "almost always
equal to 0" if and only if s[i]=0 for all but finitely many values of
i. This is equivalent to the existence of a natural number N such that
s[i]=0 for all i>=N.
If we let A be the subset of S
A = { s in S | s is almost null }
then the expression
f'(s) = sum_{i=0 to oo} 2^i*s[i]
defines a natural number if and only if s lies in A.
So your expression defines a function from A to N, but is meaningless
for any element of S\A.
Scott
> For another thing, every natural must allow adding 1 to it to form
> another natural. What natural do you get when you try to add 1 to your
> alleged "rather large natural number ...11111"?
>
How does the reals handle this situation? For example: Suppose sum =
pi + pi. Then for every digit in sum we have sum[x] = pi[x] + pi[x] +
carry(x+1) where x=0 refers to the digit immediately to the right of
the decimal point. But since pi has inifinite digits, there is no
carry(infinity+1) defined for the last right digit as there is no last
right digit. Therefore, in this case, addition is undefined.
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 5, 3:04 pm, Virgil <vir...@comcast.net> wrote:
>> In article <1181077821.087256.311...@q66g2000hsg.googlegroups.com>,
>> For another thing, every natural must allow adding 1 to it to form
>> another natural. What natural do you get when you try to add 1 to your
>> alleged "rather large natural number ...11111"?
>>
>
>How does the reals handle this situation? For example: Suppose sum =
>pi + pi. Then for every digit in sum we have sum[x] = pi[x] + pi[x] +
>carry(x+1)
What makes you think that this is the definition of sum of real
numbers?
Scott
> In article <1181161222.638465.242...@x35g2000prf.googlegroups.com>,
> >How does the reals handle this situation? For example: Suppose sum =
> >pi + pi. Then for every digit in sum we have sum[x] = pi[x] + pi[x] +
> >carry(x+1)
>
> What makes you think that this is the definition of sum of real
> numbers?
How then would you add them? Especially using our familiar decimal
notation?
Scott
finite string, I ran across the following in a college course online
notes:
"There are also kinds of objects for which the principle of induction
fails. One example is the set of subsets of natural numbers. One can
prove that the empty set is finite, and if every proper subset of a
set A is finite, then A is finite. Bit it would be incorrect to
conclude that every set of natural numbers is finite."
1. Is this statement correct?
2. Under what conditions is it correct?
3. If correct, then does not the proof of n+1 possibiliy fall under
this case; ie, it is incorrect to conclude every string is finite?
MoeBlee
You're extrapolating from the case of an algorithm for addition in
which the numbers are represented by finite sequences to the general
case of real numbers some of which are not represented by a finite
sequence. That is not set theory. If you want to understand rigorous
set theory and real analysis, then you need to do it by understanding
what are proofs from axioms and what are proper definitions in a
language as opposed to getting that all mixed up with all of your
floating notions that are pre-axiomatic.
There are different apporaches to defining the set of reals - usually
either the Dedekind cut approach or the equivalence class of Cauchy
sequences approach, and for each of those there are rigorous
definitions of the operation of addition and your ill-conceived notion
of extrapolating from addition for finite sequences does not apply.
There is also an approach to defining the reals by taking each real to
itself be a certain denumerable sequence. I haven't yet studied that
approach, but I'm confident that if you studied it then you'd find
that the definition of addition in this approach is not broken by your
ill-conceived extrapolation from addition of finite rows.
What set theory textbook are you working in?
MoeBlee
MoeBlee
Addition of reals is defined in both the Dedekind cut method and
equivalence class of Cauchy sequences method. Most set theory
textbooks have one or the other. What set theory textbook are you
working in? As to the decimal expansion approach, in Enderton's set
theory text, he mentions that the decimal expansin approach is found
in Claude Burrill's 'Foundations Of Real Numbers'. I haven't had a
chance to see that book yet, but I'd expect that addition of reals is
defined there for the decimal expansion approach.
Anyway, meanwhile, either the Dedekind cut approach or the equivalence
class of Cauchy sequences approach are shown in ordinary set theory
textbooks. For especially detailed proofs for the Dedekind cut
approach see Mendelson's 'Number Systems And The Foundations Of
Analysis'. And for the Cauchy sequences approach, if yout textbook
doesn't have it, I can, if memory will guide me, point you to a self-
contained document on the web that goes through the whole construction
proving a complete ordered field.
Also, a Dedekind cut variation having some attractive features is
found in Levy's 'Basic Set Theory'.
MoeBlee
Virgil
> On Jun 5, 3:04 pm, Virgil <vir...@comcast.net> wrote:
> > In article <1181077821.087256.311...@q66g2000hsg.googlegroups.com>,
> > For another thing, every natural must allow adding 1 to it to form
> > another natural. What natural do you get when you try to add 1 to your
> > alleged "rather large natural number ...11111"?
> >
>
> How does the reals handle this situation?
When one is working only with naturals, asking how the reals handle
addition is quite irrelevant, and deliberately avoids the issue.
The correct response would be to admit that "...1111" does not and
cannot represent a natural number.
Similar arguments will show that any such infinite-to-the-left string
which contains infinitely many 1's does not represent a natural number.
Virgil
Scott <ToaT...@gmail.com> wrote:
Carefully.
Virgil
Scott <ToaT...@gmail.com> wrote:
> In searching for the proof that n+1 can always be represented as a
> finite string, I ran across the following in a college course online
> notes:
>
> "There are also kinds of objects for which the principle of induction
> fails. One example is the set of subsets of natural numbers. One can
> prove that the empty set is finite, and if every proper subset of a
> set A is finite, then A is finite. Bit it would be incorrect to
> conclude that every set of natural numbers is finite."
>
> 1. Is this statement correct?
Yes.
> 2. Under what conditions is it correct?
All.
>
> 3. If correct, then does not the proof of n+1 possibiliy fall under
> this case; ie, it is incorrect to conclude every string is finite?
It is correct to conclude that every string representing a natural
number in the usual binary integer notation is finite, or, if allowed to
be infinite, has all but a finite number of its digits equal to zero.
You seem unable to grasp that the set of naturals is infinite without
any of its members being infinite.
This seems to be a pons asinorum ini dealing with infinite sets.
MoeBlee
> In searching for the proof that n+1 can always be represented as a
> finite string, I ran across the following in a college course online
> notes:
>
> "There are also kinds of objects for which the principle of induction
> fails. One example is the set of subsets of natural numbers. One can
> prove that the empty set is finite, and if every proper subset of a
> set A is finite, then A is finite. Bit it would be incorrect to
> conclude that every set of natural numbers is finite."
>
> 1. Is this statement correct?
It's not at all the way I would present these notions, but the
statement is correct in the general sense of what it is driving at. It
is correct that it is incorrect to conclude that A is finite on the
basis that the empty set is finite and that if every proper subset of
A is finite then A is finite.
But I don't see that that has anything much to do with induction
anyway.
Perhaps the most general notion of induction is that of a set S being
inductive with regard to a set of relations on S. This involves
certain conditions regarding that set of relations vis-a-vis S. I find
that Enderton's 'A Mathematical Intoduction To Logic' has one of the
best entrances to the subject. There is a particular exercise in that
book that puts the matter of induction into beautiful perspective.
> 2. Under what conditions is it correct?
It seems to me that the only conditions are pretty much from the
definitions and some basic set theory not even including the axiom of
infinity. If I'm not mistaken, "every proper subset of S is finite,
then S is finite" doesn't need any choice principle to prove:
If S is nonempty, then let xeS. Then S\{x} is a proper subset of S.
And if S\{x} is finite, then (S\{x})u{x} = S is finite (based on a
theorem that we prove that adding a member to a finite set results in
a finite set).
> 3. If correct, then does not the proof of n+1 possibiliy fall under
> this case; ie, it is incorrect to conclude every string is finite?
No, because we prove that the successor operation on N satisfies the
conditions for inductivity with regard to N. We PROVE that certain
sets are inductive with respect to certain relations on that set, and
only then, after giving such a proof are we allowed to apply inductive
arguments with regard to the set and said relations.
MoeBlee
Arturo Magidin
You are confusing an operation with an algorithm that is use to
evaluate the operation in certain instances.
What you describe is the ALGORITHM for adding real numbers ->with
terminating decimal notations<-. It is not, however, the definition of
"sum" for real numbers. It does not even work to add 1/3 with 2/3.
(Of course, for RATIONAL numbers we have an algorithm that ->always<-
works: given two rational numbers r and s, let a, b, c, and d be any
integers such that r = a/b and s=c/d, and then the algorithm says that
r+s is equal to (ad+bc)/(bd); the addition an multiplication is of
integers, and ->that<- can be evaluated by the usual add-and-carry
method).
How you define the addition of real numbers (note, how it is DEFINED,
not how it may be EVALUATED) depends on how you define the real
numbers. But for example, if you define real numbers as equivalence
classes of Cauchy sequences, then the easiest way to define the sum of
two real numbers is as follows:
Let r and s be two real numbers. Let {r_i} be any Cauchy sequence
of rational numbers that represents r (that is, converges to r); let
{s_i} be any Cauchy sequence of rational numbers that represents s
(that is, converges to s). Then r+s is defined to be the
equivalence class of the Cauchy sequence {r_i+s_i}.
Note that since r_i and s_i are rationals, their sum can be defined
and evaluated as above.
"Add-and-carry" is not the DEFINITION of sum. It is a ->method<- (more
precisely, an ->algorithm<-, called, unsurprisingly, the "sum
algorithm") for EVALUATING the sum in ->certain<- circumstances.
It is really mind boggling that you would even presume to believe that
you might have a proof of inconsistency of set theory, when you prove
yourself time and again incapable of even the most basic discernment
and careful speech in mathematcs, and you prove yourself again and
again incapable of simple careful reading of mathematical
statements. It is the exact same shortcomings you evidenced some time
ago when you pretended to yourself you had objections to the proof of
Cantor's Theorem.
It is a waste of YOUR time, and a waste of everyone else's time.
Scott
> It is a waste of YOUR time, and a waste of everyone else's time.
Not really, as I enjoy learning the subtlies of set theory. There is
no better way then taking theories and performing some excercises to
discover the why of things.
As to grasping language, statements, etc... I must confess that I was
not born with complete knowledge like you were, but must stumble
through life learning to improve myself. Hopefully, my posts are
moving in a positive direction.
Anyways, I want to thank everyone for taking the time to point out the
flaws. I was hoping to get a little farther in that the summation
would generate a finite string, but apparently not.
I think Virgil hit it on the nose: I am having a difficult time
understanding how N is infinite with only finite subsets. A theorem
stated elsewhere states:
"Every infinite set can be placed in a one-to-one correspondence with
at least one of its proper subsets."
This is the crux of what I'm trying to undersand as it apparently does
not apply to the set of natural numbers. If you wish to address this
issue, please do so. Otherwise, good luck in life and I hope to talk
with you in another post.
Virgil
Scott <ToaT...@gmail.com> wrote:
Then you do not understand very much about the set of natural numbers.
The mapping from N = {1,2,3,...} to N\{1} = {2,3,4,...} which maps each
n in N to n+1 in N\{1} is a one-to-one correspondence between N and its
proper subset, N\{1}.
MoeBlee
> On Jun 6, 3:34 pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>
> > It is a waste of YOUR time, and a waste of everyone else's time.
>
> Not really, as I enjoy learning the subtlies of set theory. There is
> no better way then taking theories and performing some excercises to
> discover the why of things.
To do that coherently you need to study the axioms, definitions, and
proofs.
> As to grasping language, statements, etc... I must confess that I was
> not born with complete knowledge like you were, but must stumble
> through life learning to improve myself. Hopefully, my posts are
> moving in a positive direction.
Who said anything about innate knowledge? People learn from books and
lectures just as would be expected, not from innate knowlege. And you
move in a positive direction faster just by reading one among the
various books that the people you're talking to have read.
> Anyways, I want to thank everyone for taking the time to point out the
> flaws. I was hoping to get a little farther in that the summation
> would generate a finite string, but apparently not.
>
> I think Virgil hit it on the nose: I am having a difficult time
> understanding how N is infinite with only finite subsets.
What? N has infinite subsets (it's a subset of itself just for
starters) and infinite proper subsets (in fact uncountably many
infinite proper subsets). What you might have meant to say is that for
any given basis representation system, each member of N is represented
by a finite sequence.
> A theorem
> stated elsewhere states:
>
> "Every infinite set can be placed in a one-to-one correspondence with
> at least one of its proper subsets."
That's a definition of 'Dedekind infinite':
x is Dedekind infinite <-> Ey(y prpoer subset of x & x equinumerous
with y)
and if se define 'infinite' as follows
x is infinite <-> ~En(n is a natural number & x equinumerous with x)
then we prove
x is Dedekind infinite -> x is infinite
and with the axiom of denumerable choice we prove
x is infinite -> x is Dedekind infinite.
> This is the crux of what I'm trying to undersand as it apparently does
> not apply to the set of natural numbers.
Yes is does! (And even without any choice principle.)
Please, what set theory book are you studying? Really, you're saying
all these confused things, so it would help to point you to the right
place in whatever book you're studying.
> If you wish to address this
> issue, please do so. Otherwise, good luck in life and I hope to talk
> with you in another post.
What's to address? N is Dedekind infinite (even without invoking any
choice principles) and with any given basis representation system,
each member of N is represented by a finite sequence and no member of
N is represented by an infinite sequence that doesn't have all 0's at
the front (as the sequences are ordinarily displayed growing from
right to left, in contrast with ordinary left to right notation for
mathematical formulas in general).
MoeBlee
David C. Ullrich
>On Jun 6, 3:34 pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>> It is a waste of YOUR time, and a waste of everyone else's time.
>
>Not really, as I enjoy learning the subtlies of set theory. There is
>no better way then taking theories and performing some excercises to
>discover the why of things.
You deleted the statement of _what_ he said was a waste of time.
Yes, to attain a deep understanding of something it's often good
to try to work things out for yourself. He didn't say anything
about that. In fact claiming that you've shown something
like "ZF is inconsistent" when you don't even know what the
things you're talking about _mean_ is a waste of time.
For example:
This subthread started with your silly comments about how pi + pi
was undefined. It was clear that you didn't know the _definition_
of the sum of two real numbers. Saying things about addition of
reals _before_ learning what addition of reals _is_ is in fact
a waste of time.
>As to grasping language, statements, etc... I must confess that I was
>not born with complete knowledge like you were, but must stumble
>through life learning to improve myself.
That's not what you're doing, when you just _guess_ at what things
(like x + y for real x, y) mean instead of reading the definition.
>Hopefully, my posts are
>moving in a positive direction.
>
>Anyways, I want to thank everyone for taking the time to point out the
>flaws. I was hoping to get a little farther in that the summation
>would generate a finite string, but apparently not.
>
>I think Virgil hit it on the nose: I am having a difficult time
>understanding how N is infinite with only finite subsets.
Huh? Nobody said that every subset of N was finite! Every _element_
of N is finite.
This is another example. If you're going to accomplish something
in math you simply need to be much much more careful.
> A theorem
>stated elsewhere states:
>
>"Every infinite set can be placed in a one-to-one correspondence with
>at least one of its proper subsets."
>
>This is the crux of what I'm trying to undersand as it apparently does
>not apply to the set of natural numbers.
Of course that applies to N - nobody's said anything that would
indicate it didn't.
>If you wish to address this
>issue, please do so. Otherwise, good luck in life and I hope to talk
>with you in another post.
You really do need to read and write much more carefully. When
someone says something about "element" and you think he said
something about "subset" you're being so sloppy that you're
really not going to get anywhere.
That's true, regardless of whether you believe it. If you
prefer not to believe it, and hence prefer to continue to
make a fool of yourself in public, that's your right.
************************
David C. Ullrich
David C. Ullrich
>In searching for the proof that n+1 can always be represented as a
>finite string, I ran across the following in a college course online
>notes:
>
>"There are also kinds of objects for which the principle of induction
>fails. One example is the set of subsets of natural numbers. One can
>prove that the empty set is finite, and if every proper subset of a
>set A is finite, then A is finite. Bit it would be incorrect to
>conclude that every set of natural numbers is finite."
>
>1. Is this statement correct?
Yes.
>2. Under what conditions is it correct?
>
>3. If correct, then does not the proof of n+1 possibiliy fall under
>this case; ie, it is incorrect to conclude every string is finite?
No. The other two replies so far seem to be missing the point:
The statement you quote about subsets of N is simply _irrelevant_
to the statement you cite about every natural number n!
Because a natural number is an _element_ of N, not a _subset_ of N.
************************
David C. Ullrich
Daryl McCullough
>"Every infinite set can be placed in a one-to-one correspondence with
>at least one of its proper subsets."
>
>This is the crux of what I'm trying to undersand as it apparently does
>not apply to the set of natural numbers.
Sure it does. The set of natural numbers 0, 1, 2, ...
is infinite, and it can be placed into a one-to-one correspondence
with the proper subset 1, 2, 3, ...
0 <-> 1
1 <-> 2
2 <-> 3
...
>If you wish to address this issue, please do so.
There is no issue.
--
Daryl McCullough
Ithaca, NY
Arturo Magidin
>> It is a waste of YOUR time, and a waste of everyone else's time.
>
>Not really, as I enjoy learning the subtlies of set theory.
Do you also enjoy taking things out of context? Here you are pretty
close to lying about what I said by implication, by not putting my
quote in correct context.
>There is
>no better way then taking theories and performing some excercises to
>discover the why of things.
Nowhere did I say or even imply that this would be a bad thing. What I
->did<- say was a waste of your time and everyone else's, however,
was:
It is really mind boggling that you would even presume to believe
that you might have a proof of inconsistency of set theory, when
you prove yourself time and again incapable of even the most basic
discernment and careful speech in mathematcs, and you prove
yourself again and again incapable of simple careful reading of
mathematical statements.
It is the presumption of talking about "proofs of inconsistency of set
theory" given your level and lacks of knowledge which are a waste of
time. It is discussions about "how could you possibly define pi+pi
without carry" when you don't even seem to have the most basic grasps
about the natural numbers which are a waste of time.
>As to grasping language, statements, etc... I must confess that I was
>not born with complete knowledge like you were,
Ha, ha.
> but must stumble
>through life learning to improve myself.
And the best way is to read and ask questions, not to presume that you
have a proof of inconsistency of set theory, to claim that Cantor's
proof is incoherent, or to ask about just exactly why Goedel is
crank.
You waste time because you try to leap ahead and rush through without
taking the care of learning the BASICS. That is what is a waste of
time.
You waste time because you are careless. In this thread, in your
alleged proof of the second proposition, you made two mistakes. One
was subtle but common: you did not realize that your summation
expression was meaningless for sequences that had an infinite number
of nonzero terms.
But the second mistake was much more basic; I noticed it, and I
mentioned it, and as far as I can tell you did not even acknowledge
that very serious, very basic, very catastrophic mistake.
You were trying to prove that a function f' was one-to-one.
To do this, you proved that you cannot have f'(s)=n and f'(s)=m with
n!=m.
This is a serious error. That is not how one checks if a function is
one-to-one. Did you even notice that mistake, even after it was
pointed out? Or were you too busy trying to misunderstand the
objection on the summation as one about the infinity of the sequence,
instead of what was very clearly stated, a problem with the definition
of an infinite sum of positive integers?
> Hopefully, my posts are
>moving in a positive direction.
Your posts are wasting a lot more time than they are inching you in
the right direction. If you take some time to learn the BASICS instead
of all this silly jumping around and all the unfathomably silly claims
of "inconsistencies", then you would be moving in a good direction at
a much better pace.
>I think Virgil hit it on the nose: I am having a difficult time
>understanding how N is infinite with only finite subsets. A theorem
>stated elsewhere states:
>
>"Every infinite set can be placed in a one-to-one correspondence with
>at least one of its proper subsets."
>
>This is the crux of what I'm trying to undersand as it apparently does
>not apply to the set of natural numbers.
See? This is utterly silly.
Take the map that sends every natural number to twice that natural
number. The set of even natural numbers is a proper subset of the
natural numbers, and it was just placed in one-to-one correspondence
with the full set.
Arturo Magidin
>I think Virgil hit it on the nose: I am having a difficult time
>understanding how N is infinite with only finite subsets.
I disagree: the reason you are having difficulty is that, by your own
admission, you are trying to do things without a careful study of the
basics.
This is a perfect example.
Why on Earth would you believe that N has "only finite subsets"? What
could ->possibly<- lead you to believe such a silly statement? Is N
itself a finite set? Is N itself a subset of itself?
What, other than sheer carelessness and thoughtlessness, could
possibly lead you to believe that "N [has] only finite subsets"?
Scott
> S was defined to be the collection of all infinite sequences of 0s and
> 1s. Thus, if A is a subset of S, then its elements must be
> SEQUENCES. But in your notation, s[i] represents the i-th term of the
> sequence s, so "s[0]", "s[1]", ..., "s[i]", and "0" are each of them
> either "0" or "1", not sequences.
>
> So it is impossible for them to be elements of A at all.
Uh, A is a subset of S such that all elements have the form
s[0],...,s[i],0... which means all elements of A have a finite number
of s[i]=1 digits;eg, 01000..., 111000... are members, but 111111....
is not. How is this different from what you wrote below? Would it have
been clearer if somehow I could have written the bar over the 0
indicating infinite repetition? Or are you trying to make claim that
such a template is not equivalent to what you wrote below?
>
> You REALLY need to learn to express yourself clearly and
> ACCURATELY. Just like you need to learn to read the replies you get
> carefully. In my antepenultimate reply you seemed to think the
> objection raised had something to do with the fact that the elements
> of s were infinite, when there was absolutely nothing in what was
> written that could be interpreted that way except through careless and
> confused reading.
No I did not think that. I thought you were making reference to the
fact that there needed to be a finite number of s[i]=1 digits.
Scott
> In article <1181077821.087256.311...@q66g2000hsg.googlegroups.com>,
> For another thing, every natural must allow adding 1 to it to form
> another natural. What natural do you get when you try to add 1 to your
> alleged "rather large natural number ...11111"?
>
How does the reals handle this situation? For example: Suppose sum =
pi + pi. Then for every digit in sum we have sum[x] = pi[x] + pi[x] +
Scott
> For another thing, every natural must allow adding 1 to it to form
> another natural. What natural do you get when you try to add 1 to your
> alleged "rather large natural number ...11111"?
>
Would appreciate a critique of the following line of reasoning:
Suppose ...11111 is a natural number n. Then, by PA, there is a
sequence ...SSSSS0, where S is the successor function, for n. Since n
+1 is equivalent to 1+n, then ...SSSSSS0 is the sucessor sequence to
n. Therefore, there exists a natural number n+1 such that n < n+1.
Aatu Koskensilta
> How does the reals handle this situation? For example: Suppose sum =
> pi + pi. Then for every digit in sum we have sum[x] = pi[x] + pi[x] +
> carry(x+1) where x=0 refers to the digit immediately to the right of
> the decimal point. But since pi has inifinite digits, there is no
> carry(infinity+1) defined for the last right digit as there is no last
> right digit. Therefore, in this case, addition is undefined.
If you're interested in how addition is defined for reals, I suggest you
have a peek at any decent textbook on analysis including the construction of
the reals. It's a pain to define addition, multiplication and so on, in
terms of decimal expansions. That, too, can be done, however, and you'll
find the details in the literature.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Aatu Koskensilta
>
> Suppose ...11111 is a natural number n. Then, by PA, there is a
> sequence ...SSSSS0, where S is the successor function, for n. Since n
> +1 is equivalent to 1+n, then ...SSSSSS0 is the sucessor sequence to
> n. Therefore, there exists a natural number n+1 such that n < n+1.
Your reasoning is perfectly valid. What point you see to it is somewhat
obscure, though, since ...11111 is not a natural number.
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 6, 1:19 pm, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>
>> S was defined to be the collection of all infinite sequences of 0s and
>> 1s. Thus, if A is a subset of S, then its elements must be
>> SEQUENCES. But in your notation, s[i] represents the i-th term of the
>> sequence s, so "s[0]", "s[1]", ..., "s[i]", and "0" are each of them
>> either "0" or "1", not sequences.
>>
>> So it is impossible for them to be elements of A at all.
>
>Uh, A is a subset of S such that all elements have the form
>s[0],...,s[i],0... which means all elements of A have a finite number
>of s[i]=1 digits;
Which is what you should have said. Which is what I said, and you
proceded to mangle in your usual unimitable manner.
>eg, 01000..., 111000... are members, but 111111....
>is not. How is this different from what you wrote below?
It is not. However, it is VERY different from what YOU wrote
which you deleted. You talked about s[0], s[1], etc, being elements of
A.
> Would it have
>been clearer if somehow I could have written the bar over the 0
>indicating infinite repetition?
No. It would have been clearer if you had not written nonsense
originally, and if you did not try to hide that fact by removing
context to make it appear that you did not.
>Or are you trying to make claim that
>such a template is not equivalent to what you wrote below?
No, I was saying that what you WROTE originally was incoherent
silliness. What you write ->NOW<- is different from what you wrote
before. If you cannot tell the difference between what you wrote in
this message and wha tyou wrote before then this is more evidence that
you cannot understand mathematical sentences.
David C. Ullrich
People have already explained to you why this is nonsense! The
(standard) definition of the addition of two reals is _not_
given in terms of the decimal expansion.
You really should look up the actual definition, as has been
suggested by several people. It really looks silly, the way
you insist on posting these things, saying you're trying to
learn this and that, while at the same time you simply
refuse to actually _learn_ what you need to learn to discuss
these things intelligently.
(It's _possible_ to define the sum of two reals in terms of
the decimal expansions, but this is not the easiest way to
give the definition, and the definition is not what you
think it is...)
************************
David C. Ullrich
David C. Ullrich
This is nonsense. (I'm puzzled by Aatu's comment that the reasoning
is perfectly valid.)
It's nonsense because the very first line, "Suppose ...11111 is
a natural number n.", is _meaningless. It's meaningless until
you give us the _definition_ of ...111.
Possibly you'll reply that ...111 is by definition the sum
of 2^j, for j = 0 to infinity (or maybe you meant 10^j.)
That's not a valid definition until you give us the
_definition_ of that _sum_ - the sum of infinitely many
natural numbers is undefined. (When I say it's undefined
I mean there's no standard definition - you're free to
define it to be whatever you want, but you have to _say_
what the definition _is_.)
Ok, then possibly you mean that the sum of 2^j for j = 0 to
infinity is defined to be a certain limit, as in calculus.
If _that's_ what you mean then (i) it's awesomely clear that
this is not a natural number, and (ii) the _next_ line,
"Then, by PA, ..." is nonsense, because there's nothing
in PA about limits.
************************
David C. Ullrich
David C. Ullrich
<aatu.kos...@xortec.fi> wrote:
>On 2007-06-08, in sci.logic, Scott wrote:
>> Would appreciate a critique of the following line of reasoning:
>>
>> Suppose ...11111 is a natural number n. Then, by PA, there is a
>> sequence ...SSSSS0, where S is the successor function, for n. Since n
>> +1 is equivalent to 1+n, then ...SSSSSS0 is the sucessor sequence to
>> n. Therefore, there exists a natural number n+1 such that n < n+1.
>
>Your reasoning is perfectly valid.
??? How can this be valid, without definitions of things
like ...111 and ...SSS0?
> What point you see to it is somewhat
>obscure, though, since ...11111 is not a natural number.
************************
David C. Ullrich
Aatu Koskensilta
> On 8 Jun 2007 02:21:53 -0700, Scott <ToaT...@gmail.com> wrote:
>
>>Suppose ...11111 is a natural number n. Then, by PA, there is a
>>sequence ...SSSSS0, where S is the successor function, for n. Since n
>>+1 is equivalent to 1+n, then ...SSSSSS0 is the sucessor sequence to
>>n. Therefore, there exists a natural number n+1 such that n < n+1.
>
> This is nonsense. (I'm puzzled by Aatu's comment that the reasoning
> is perfectly valid.)
It's valid the same way an argument starting with "Suppose the set R of real
numbers is a natural numbers. Then there is a natural n such that n = S(R),
..." is. It's utterly pointless, of course. Here, of course, I'm assuming
...11111 stands for a mathematical object, the most natural choice being a
"reverse-sequence" of type omega*.
Aatu Koskensilta
> like ...111 and ...SSS0?
Ah, I didn't pay attention to the ...SSSSSS0. You're quite right, the
argument as it stands is not only pointless but meaningless as well.
Scott
> Please, what set theory book are you studying? Really, you're saying
> all these confused things, so it would help to point you to the right
> place in whatever book you're studying.
Set Theory and its Philosophy, by M. Potter.
By the way, I've heard mention of another book that is really good as
an introduction but I was unable to write it down (and can't find the
post). Something like "the magic of sets" or "the life of sets". Not
much to go on, but if you know it, I would appreciate the title/
author. Thanks.
Scott
> But the second mistake was much more basic; I noticed it, and I
> mentioned it, and as far as I can tell you did not even acknowledge
> that very serious, very basic, very catastrophic mistake.
>
> You were trying to prove that a function f' was one-to-one.
> To do this, you proved that you cannot have f'(s)=n and f'(s)=m with
> n!=m.
Sorry I can't win with you. Yes, I realized it was a mistake when you
pointed it out. However, in a previous thread you told me not to
bother fixing the second part of a proof if the first part is
incorrect. So I didn't.
Aatu Koskensilta
> By the way, I've heard mention of another book that is really good as
> an introduction but I was unable to write it down (and can't find the
> post). Something like "the magic of sets" or "the life of sets". Not
> much to go on, but if you know it, I would appreciate the title/
> author. Thanks.
Perhaps you're thinking of _The Joy of Sets_ by Keith Devlin, indeed a nice
read.
Arturo Magidin
Scott <ToaT...@gmail.com> wrote:
>On Jun 7, 8:27 am, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>Sorry I can't win with you.
You might want to start by not taking things out of context; you are
treading dangerously close to the purposely misleading.
(You might want to continue by trying to learn the material from the
books ->first<-, instead of, as you so laughably claim, trying to
understand "subtleties" not covered in the textbooks before you are
able to grasp the non-subtle parts of the theory, such as the
difference between saying:
"A is a subset of S such that s[0], s[1], ..., s[i], 0, are all elements
of A"
and
"A is a subset of S such that each element s of A is of the form
s[0], s[1], ..., s[i], 0,..."
or thinking that the problem with the first formulation has to do with
the lack of ellipsis after the 0. Quite simply, you do not seem to be
at a point in your education of set theory when you can even begin to
grasp subtleties, let alone investigate them. You are still too weak
on the fundamentals and the obvious stuff)
> Yes, I realized it was a mistake when you
>pointed it out. However, in a previous thread you told me not to
>bother fixing the second part of a proof if the first part is
>incorrect. So I didn't.
I said "acknowledge", not "fix". You gave absolutely no indication
that you even realized the mistake had occurred. Of course there was
no point in fixing the "second part" (it assumed a function was
well-defined, when it wasn't). On the other hand, realizing you made a
mistake there as well was important too.
MoeBlee
> On Jun 6, 5:07 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> Set Theory and its Philosophy, by M. Potter.
That book has some good recommendations, and it seems to me that it
can give some good insights (I plan to get it and read it eventually).
But it takes an approach that I glean is equivalent with the ordinary
axiomatizations but is different from them. I'm mentioning that not to
discourage you from studying the book (it has been highly recommended
to me) but only to say that I can't do much to follow along with you
on that particular book since I don't have it.
> By the way, I've heard mention of another book that is really good as
> an introduction but I was unable to write it down (and can't find the
> post). Something like "the magic of sets" or "the life of sets". Not
> much to go on, but if you know it, I would appreciate the title/
> author. Thanks.
As Aatu mentioned, it might be Keith Devlin's 'The Joy Of Sets'.
MoeBlee
Aatu Koskensilta
> But it takes an approach that I glean is equivalent with the ordinary
> axiomatizations but is different from them.
The only essential idiosyncrasies in Potter's _Set Theory and Its
Philosophy_ are that non-sets are allowed and replacement is not adopted as
an official axiom.
MoeBlee
wrote:
> On 2007-06-12, in sci.logic, MoeBlee wrote:
>
> The only essential idiosyncrasies in Potter's _Set Theory and Its
> Philosophy_ are that non-sets are allowed and replacement is not adopted as
> an official axiom.
What I meant is that he develops quite a bit about levels (or whatever
the exact word he uses) before he does more of the usual stuff.
Doesn't he even axiomatize through things about levels (or whatever
exact words he uses)? If I am recalling correctly about this, this is
what I LIKE about his book: he gives theorems about levels (from some
axioms that are not the usual but turn out to provide an equivalent
axiomatization as the usual?) (except as you mention, allowing
urelements?) to the usual treatments. That gives a "look at it both
ways" perspective that I like.
MoeBlee
Aatu Koskensilta
show your support by posting sense).
MoeBlee <jazz...@hotmail.com> writes:
> If I am recalling correctly about this, this is what I LIKE about
> his book: he gives theorems about levels (from some axioms that are
> not the usual but turn out to provide an equivalent axiomatization
> as the usual?) (except as you mention, allowing urelements?) to the
> usual treatments.
Potter does base his development on levels, yes, and obtains Zermelo
set theory with ur-elements. The justification for the various axioms
of set theory on basis of the idea of sets being "built" in stages is
carefully considered, though in a style that, for some reason or
other, I found not quite to my liking.
One of my favourite observations that people usually find either
glaringly obvious or fail to appreciate entirely can also be found in
Potter's book. There is an important difference between the
transfinite recursion scheme
If G is a definable class function, there is a unique definable class
function F such that, provably in ZFC, F(alpha) = G(F restricted to
alpha) for all ordinals alpha
and the reflection scheme
For every P in the language of set theory, there is, provably in ZFC
an alpha, s.t P <--> (V_alpha |= P)
In the first case, the argument we give does not in any way depend on
the definability of G and F -- the only reason we mention definability
is that we're working in the language of set theory and have no class
quantification at our disposal; but with the reflection scheme, we
have to carry out an induction on the complexity of P, that is, it
does matter that P is a statement about sets expressible in the
language of set theory.
In _Reflection Principles, Large Cardinals, Elementary Embeddings_[1]
Reinhardt writes, talking about obtaining (small) large cardinals by
reflection:
Utilizing informal criteria to judge when the passage from "for all
definable F" to "for all F" is acceptable, one can in this way obtain
all the Mahlo cardinals. (The informal step is in each case analogous
to the passage from "V satisfies replacement for definable functions"
to "V is inaccessible.")
This passage is somewhat puzzling, since no "informal criteria"
suggest themselves that would allow one to ever pass from "V satisfies
replacement for definable functions" to "V is inaccessible". Rather,
the only reason anyone would think "V satisfies replacement for
definable functions" is that they were already convinced replacement
holds for whatever functions there happen to be, and, accepting the
legitimacy of functions definable in the language of set theory,
concluding on that basis that, in particular, replacement holds for
definable functions. (Of course, it is possible, perhaps even likely,
this is just what Reinhardt has in mind with "informal criteria").
The distinctions at issue can be captured formally, for example by
introducing class constants that effectively act as variables: no
properties of these class constants are stipulated, and thus the only
properties that of them we can use in proofs are that they are
classes. In addition, we add a rule of inference allowing a class
constant to be replaced with an open formula. The transfinite
recursion theorem can then be formulated as: G is a function -->
F(alpha,y) <--> <alpha,y> in G({beta,x} | F(beta,x) & beta < alpha}),
where F is a formula containing G. We get the instances of the usual
schematic form by replacing G with a formula in the language of set
theory. The reflection schema, and other schematic results that really
do only hold for definable properties, functions, statements
expressible in the language of set theory, have no such
reformulations; these results have no uniform proof, that is, the
structure of a proof of an instance of the scheme depends essentially
on the formula.
Footnotes:
[1] It is in this article the now famous "ultimate large cardinal
axiom", sometimes known as Reinhardt's axiom, is introduced, that there
is a non-trivial elementary embedding of the universe in itself. Kunen
showed that this axiom is, in presence of choice, inconsistent -- a
result occasionally referred to as the "Kunen inconsistency".
Chris Menzel
> On Jun 12, 3:30 pm, Aatu Koskensilta <aatu.koskensi...@xortec.fi>
> wrote:
>> On 2007-06-12, in sci.logic, MoeBlee wrote:
>>
>> The only essential idiosyncrasies in Potter's _Set Theory and Its
>> Philosophy_ are that non-sets are allowed and replacement is not
>> adopted as an official axiom.
>
> What I meant is that he develops quite a bit about levels (or whatever
> the exact word he uses) before he does more of the usual stuff.
> Doesn't he even axiomatize through things about levels (or whatever
> exact words he uses)?
I believe Potter draws on Dana Scott's paper "Axiomatizing set theory"
in which he developed an alternative axiomatization of ZF that takes
levels as primitive.
Peter_Smith
> I believe Potter draws on Dana Scott's paper "Axiomatizing set theory"
> in which he developed an alternative axiomatization of ZF that takes
> levels as primitive.
He does indeed. The book is a very good read (and I don't say that
just because he is an admired colleague!)
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