From: herbzet on


|-|ercules wrote:
> "herbzet" wrote ...
> > |-|ercules wrote:
> >> herbzet wrote:
> >> > herbzet wrote:
> >> >> |-|ercules wrote:
> >> >>
> >> >> > I'll wait and see if someone else takes the bait.
> >> >> >
> >> >> > >> The proof of higher infinities than 1,2,3...oo infinity relies on
> >> >> > >> the fact that there is no box that contains all and only all the
> >> >> > >> label numbers of the boxes that don't contain their own label number.
> >> >> >
> >> >> > TRUE OR FALSE
> >> >>
> >> >> Um, false, so far as I know.
> >> >>
> >> >> We have
> >> >>
> >> >> 1) |N| < |P(N)|
> >> >> 2) |P(N)| <= |R|
> >> >> --------------
> >> >> .: |N| < |R|
> >> >>
> >> >> but neither of Cantor's proofs that |N| < |R| involves either of
> >> >> premises (1) or (2), as far as I can recall.
> >> >
> >> > Why do you ask?
> >> >
> >>
> >> Because the most widely used proof of uncountable infinity is the
> >> contradiction of a bijection from N to P(N), which is analagous to
> >> the missing box question.
> >
> > Perhaps so, but why do you ask?
>
> It's hard to explain to you when you answer FALSE and then PERHAPS to identical
> questions.

> Anyway, I'm going to sleep [...]

I think you need some sleep -- I didn't answer "perhaps" to any question,
much less a question identical to your poll question.

Just to clear up any ambiguity, my "perhaps" was directed to your
assertion that the most widely used proof of uncountable infinity
is the contradiction of a bijection from N to P(N) -- which I doubt --
not to your subordinate assertion that the proof is analagous to the
missing box question, an assertion whose truth I freely grant.

--
hz
From: herbzet on


herbzet wrote:

> Just to clear up any ambiguity, my "perhaps" was directed to your
> assertion that the most widely used proof of uncountable infinity
> is the contradiction of a bijection from N to P(N) -- which I doubt --
> [...]

That is, I doubt your assertion, not the proof.

English is funny, no?

--
hz
From: herbzet on


|-|ercules wrote:
> "herbzet" wrote ...
> > |-|ercules wrote:
> >> herbzet wrote:
> >> > herbzet wrote:
> >> >> |-|ercules wrote:
> >> >>
> >> >> > I'll wait and see if someone else takes the bait.
> >> >> >
> >> >> > >> The proof of higher infinities than 1,2,3...oo infinity relies on
> >> >> > >> the fact that there is no box that contains all and only all the
> >> >> > >> label numbers of the boxes that don't contain their own label number.
> >> >> >
> >> >> > TRUE OR FALSE
> >> >>
> >> >> Um, false, so far as I know.
> >> >>
> >> >> We have
> >> >>
> >> >> 1) |N| < |P(N)|
> >> >> 2) |P(N)| <= |R|
> >> >> --------------
> >> >> .: |N| < |R|
> >> >>
> >> >> but neither of Cantor's proofs that |N| < |R| involves either of
> >> >> premises (1) or (2), as far as I can recall.
> >> >
> >> > Why do you ask?
> >> >
> >>
> >> Because the most widely used proof of uncountable infinity is the
> >> contradiction of a bijection from N to P(N), which is analagous to
> >> the missing box question.
> >
> > Perhaps so, but why do you ask?
>
> It's hard to explain to you when you answer FALSE and then PERHAPS to identical questions.
>
> Anyway, I'm going to sleep and aren't posting any more,

May flights of angels SING thee to thy rest!

Hurry back, sweet Herkimer.

--
hz
From: Transfer Principle on
On Jun 4, 12:04 pm, MoeBlee <jazzm...(a)hotmail.com> wrote:
> On Jun 4, 1:37 pm, Transfer Principle <lwal...(a)lausd.net> wrote:
> > How about this: ZFC proves _neither_ or _both_ of
> > CH and its negation (Goedel and Cohen)?
> I see. So do you have any confidence that ZF is consistent?

Here is my opinion on this issue.

I am confident that ZF is consistent, because I believe
that if ZF were inconsistent, a proof of this would have
been found by now. The rapidity of how a proof of the
inconsistency of naive set theory was found does suggest
that a proof of ~Con(ZF) would've been found equally
quickly, if such a proof were possible.

If it does turn out that ZF is inconsistent, then there
would have been some underlying reason that the proof
wasn't discovered for over a century after the axioms
were first given. This is why I often refer to the
mathematician Ed Nelson, who's currently working on a
proof that PA (and hence ZF) is inconsistent. Nelson's
work depends on very large natural numbers -- naturals
so large that one has to come up with a new operation,
called tetration, to describe them. Tetration is an
operation that isn't considered by most mathematicians,
and so a proof which relies on tetration can easily be
overlooked for more than a century. The fact that the
consistency strength of PA (Gentzen) is epsilon_0, the
smallest ordinal which can't be constructed from omega
with finitely many additions, multiplications, and
exponentiations, but can be written with finitely many
_tetrations_ -- indeed, it is omega^^omega (Rucker) --
is suspicious. It indicates to me that the key to a
potential proof of ~Con(PA) (hence ~Con(ZF)) is the
operation of tetration.

But that being said, even though it's unlikely that
we'll ever see a proof that ZF is inconsistent (even
from Nelson), this doesn't mean that I must a priori
agree that Herc is wrong. Although in Cooper's polls,
I will vote for the choice which favors ZFC (and thus
Herc opposes), the other choice isn't "wrong." There
can be theories other than ZFC in which the other
choice in the poll is right. Similarly, I'll vote for
"0.999... = 1" in a poll asking for whether these two
values are equal, but I'll still consider theories
which prove that they are distinct.

To conclude, just because I believe that ZF is
consistent, it doesn't mean I believe that Cooper
must be wrong.
From: Transfer Principle on
On Jun 5, 6:14 am, herbzet <herb...(a)gmail.com> wrote:
> |-|ercules wrote:
> > Because the most widely used proof of uncountable infinity is the
> > contradiction of a bijection from N to P(N), which is analagous to
> > the missing box question.
> I'm not sure that this proof is really a "proof of uncountable infinity"
> anyway.  A finitist, for example, would reject the notion that the naturals
> constitute an infinite set in the first place, but I see no reason
> why she would reject the proof that for any set S, |S| < |P(S)|.

MoeBlee pointed this out too:

"Sure, but without the power set axiom, we can still prove that for
any
S, if S has a power set, then there is no surjection from S onto its
power set, which is the "essence" of Cantor's theorem."

Of course, whenever posters mention this, I immediately
point to the theory NFU. NFU proves the existence of
non-Cantorian sets, and a non-Cantorian set is precisely
a set S such that card(S) < card(P(S)).

The simplest example of such a set is the set V of all
possible sets -- a set whose existence is provable in
NFU (but not ZFC, of course). It's easy to find a
surjection from V to P(V) -- since P(V) = V, the
identity _bijection_ suffices.

Therefore, any poster who doesn't like Cantor's Theorem
ought to consider NFU instead of ZFC.