How Can ZFC/PA do much of Math - it Can't Even Prove PA is Consistent (EASY PROOF) [Math]

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From: Transfer Principle on 6 Jul 2010 21:29

On Jul 6, 3:29 pm, "K_h" <KHol...(a)SX729.com> wrote:
> "Transfer Principle" <lwal...(a)lausd.net> wrote in message
> > And according to Srinivasan, the theory NBG-Infinity, with NAFL as
> > the underlying logic, proves that N=On and so N is a proper class,
> How so? Why wouldn't NBG-infinity be agnostic on this issue?

Normally, in FOL, it would be. Apparently, there is something
peculiar about NAFL such that one can prove that N isn't a
set in NAFL. Srinivasan, the inventor of NAFL, can explain
further how NAFL proves that N is a proper class.

> Let's review some basics about proper classes. Logically, a proper class has a
> "cardinal" or "ordinal" size of absolute infinity. (I use the equivocal quotes
> there because it's technically an abuse of those terms.) So the idea is that a
> proper class has the largest size possible and so there cannot be another class
> with an "ordinal" or "cardinal" size bigger than that. But you CAN get a bigger
> "ordinal" object than N. For instance, consider the sum of (1/2)^n for all n in
> {0,1,2,3,...}. Each natural number n provides one term in the sum:
> 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
> When this is summed for all naturals we get 2. Can we now add more? Yes we can:
> we can add the number 3 to 2 to get 5 and then we can add the number 4 to 5 to
> get 9. So it makes sense to say that the number 3 is the w term in the sum and
> the number 4 is the w+1 term in the sum. Ordinals simply mean do this first, do
> that second, etc, in a given order -- e.g. addition. That's the motivation
> underlying the ordinals. If N were a proper class then its "ordinal" size would
> have to be absolute infinite and, in this example, the number 4 would correspond
> to an "ordinal" term greater than absolute infinite! That is plainly a
> contradiction and this simple fact is one reason why N really must be a set.

Interesting. Still, I'm curious as to how a _finitist_ like
Srinivasan would react to an _infinite_ series such as:

1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

Presumably, a finitist would be opposed to the mere
existence of infinite series, and so would be unmoved
by the use of infinite series to prove that N is a set.

(As an aside, note that there are sci.math posters,
including Tony Orlow and tommy1729, who don't
accept the existence of the ordinal omega+1 as
used in K_h's post, yet both do accept that N is a
a set and that P(N) is a larger set. Go figure.)

> > K_h states that the main reason to make a distinction between
> > sets and proper classes is Cantor's Theorem. Thus, in a theory
> > which refutes Cantor's Theorem, such as NFU, we expect there
> > to be no distinction between sets and proper classes. Indeed,
> > we see that this is exactly the case -- NFU proves that even V
> > is a set.
> Why do you think NFU refutes Cantor's theorem instead of being agnostic on it? I
> am just curious -- whatever you believe is your own business.

Let's find out more about NFU:

http://math.boisestate.edu/~holmes/holmes/nf.html

"Russell's paradox: "x is not an element of x" is not a stratified
predicate. But the universe, V = { x | x = x }, does exist in NF and
its
subsystems known to be consistent (see below).

"Cantor's paradox of the largest cardinal: Cardinal numbers are
defined
in NF as equivalence classes of sets of the same size. The form of
Cantor's theorem which can be proven in Russell's type theory asserts
that the cardinality of the set of one-element subsets of A is less
than
the cardinality of the power set of A. Note that the usual form
(|A| < |P(A)|) doesn't even make sense in type theory. It makes sense
in NF, but it isn't true in all cases: for example, it wouldn't do to
have
|V| < |P(V)|, and indeed this is not the case, though the set 1 of all
one-element subsets of V is smaller than V (the obvious bijection
x |-> {x} has an unstratified definition!).

"Sets with the property that the set P1(A) of one-element subsets of A
is the same size as A are called "Cantorian" sets; sets with the
stronger
property that the restriction to A of the singleton "map" x |-> {x}
exists are
called "strongly Cantorian" sets."

Let's see what NFU has to say about the set N of natural numbers:

"The set of natural numbers is provably Cantorian; the assumption that
it is
strongly Cantorian, known as Rosser's Axiom of Counting, is known to
strengthen NF (if it is consistent -- one of the few independence
results not
obtained by permutation methods) and known to be consistent with NFU.
Cantorian and strongly Cantorian sets clearly satisfy Cantor's theorem
in
its usual unstratified form (|A| = |P1(A)| < |P(A)|)."

So N is Cantorian, and therefore card(N) < card(P(N)), just as it is
in
standard theory.

Even though I was referring to posters other than Srinivasan when I
started discussing NFU, the following line caught my eye:

"Ali Enayat and Solovay have some new results relating the strength of
NFU + "the universe is finite" and some of its extensions to systems
of arithmetic."

NFU + "the universe is finite"? Unfortunately, the link doesn't
discuss
what any of these results are exactly.

From: Marshall on 7 Jul 2010 22:12

On Jul 7, 7:05 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote:
> Aatu Koskensilta wrote:
> > Nam Nguyen <namducngu...(a)shaw.ca> writes:
>
> >> MoeBlee wrote:
>
> >>> And, in that regard, I have said all along that there is no
> >>> finitistic proof of the consistency of PA.
> >> So you've agreed that there's no formal proof for PA's consistency and that
> >> if you go by formal proof only then you don't have knowledge of PA's
> >> consistency.
>
> > There are many formal proofs of the consistency of PA. None of them are
> > finitistic.
>
> So it doesn't seem you used the phrase "formal proof" in the standard
> way that textbook (e.g. Shoenfield's) would use. In that standard
> usage, a formal proof is a (finite) syntactical proof of a FOL formal
> system theorem.
>
> Given that standard definition of "formal proof", would you agree with
> my statement above that:
>
> >> there's no formal proof for PA's consistency and that if you go by
> >> formal proof only then you don't have knowledge of PA's consistency..
>
> ?
>
> --
> ----------------------------------------------------
> There is no remainder in the mathematics of infinity.
>
> NYOGEN SENZAKI
> ----------------------------------------------------

Give it up, loon. PA is provably consistent. Deal with it.

Marshall

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From: Transfer Principle on 8 Jul 2010 18:42

On Jul 7, 8:58 pm, Marshall <marshall.spi...(a)gmail.com> wrote:
> On Jul 7, 7:22 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote:
> > That's all the _technical_ arguments an intellectual clown like you, Marshall,
> > could ever say!
> PA is provably consistent. Learn why, or shut [...] up.

Yet that hasn't stopped the mathematician Ed Nelson from
searching for a proof that PA is inconsistent.

> Give it up, loon. PA is provably consistent. Deal with it.

If those who even entertain the possibility that PA is
inconsistent are "loons," then I guess that makes Nelson
and Nguyen (and myself, since I keep bringing up Nelson)
a bunch of "loons."

It would be poetic justice for Nelson to complete his proof
and someone to tell Spight:

"PA is provably _in_consistent. Learn why, or shut up!"

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