From: rupert4050@my-deja.com
Subject: Re: Continuum Hypothesis
Date: Thu, 16 Dec 1999 01:14:10 GMT
Newsgroups: sci.math
Keywords: Natural assumptions which imply negation of Continuum Hypothesis
In article <19991215155248.09489.00000890@ng-co1.aol.com>,
seraphsama@aol.com (Seraph-sama) wrote:
> I think I asked this a long time ago but I don't remember receiving any
> reponses.
>
> I'm trying to figure out whether the Continuum Hypothesis is "true" or not.
You know a lot of people think it hasn't got a truth-value. They think
our concept of 'set' is given just by the axioms we've got, and we're
free to take either CH or ~CH as a new axiom, or just leave it be,
whatever we feel like. But that's certainly not my view.
> This isn't really mathematics; it's more epistemology or metaphysics or
> whatever you decide to call it. Perhaps, "trying to find an axiom that is
> intuitively obvious that leads to a proof of the Continuum Hypothesis."
You should take a look at the sci.math FAQ, that has a good summary of
some of the important bits of literature and references to them.
>The truth of 2 + 2 = 4 or "there exists one line through every two points" are
> "visualizable" truths. The truth of "For every homomorphism f:G -> H, G/ker f =
> Im f," is not readily visualizable to the beginning group theory student, but
> it can be visualized with some effort. My question is, has anyone ever really
> *understood* (i.e., visualized) the truth/falsity of the Continuum Hypothesis?
Not really, no. At least, not in any way that everyone else has agreed
is convincing. There have been a lot of arguments both for and against.
Like I say, you should take a look at the FAQ.
Here is one interesting argument. Say you've got a set S which is a
subset of the unit square, every row of which is countable. That is
it's a set of points such that 0<=x<=1 and 0<=y<=1, and for each
x the set of y such that is in S is countable. That set is pretty
thin in the unit square, right? Now, consider its reflection in the
diagonal y=x. That's the set of points such that is in S.
That set also is pretty thin in the unit square, right? And the union
of two such thin sets shouldn't fill up the entire unit square, right?
Well, given the axiom of choice, the proposition that it's indeed
always impossible for the union of S with its reflection in the
diagonal to cover the whole unit square, is equivalent to not-CH. I'll
give the proof at the end of this post.
> I understand that Paul Cohen (?) invented a technique called "forcing" to
> create a model in which CH was false. Because of CH's independence, "forcing"
> may or may not be in a "weird" set theory, comparable to the denial of the
> (independent) parallel postulate to produce consistent, yet "weird" geometries.
> (Of course "weird" is subjective, but so long as we are acquainted intuitively
> with the objects of mathematics we use, it seems we should be able to pick out
> the right set theory.) I want to know if the Continuum Hypothesis problem is
> still open. (As in, finding an intuitive axiom that proves it.)
>
>
Yep, it is.
Okay, let Axiom A be the assertion that if S is a subset of the unit
square every row of which is countable, then the union of S and the
"inverse" of S (the set of all such that is in S) cannot be
the entire unit square. Axiom A is equivalent to not-CH, given the
axiom of choice.
Proof (this argument is adapted from sci.math FAQ):
Suppose CH. Then the real numbers in [0,1] can be put into one-to-one
correspondence with the countable ordinals. Let r_alpha be the real
corresponding to the ordinal alpha. Let S be
{:beta<=alpha}. Then the union of S with its inverse
covers the entire unit square.
Suppose not-CH, and S is a subset of the unit square every row of which
is countable. We prove that the union of S with its inverse can't cover
the entire unit square. Any such subset S can be thought of as a
function f from [0,1] to the countable subsets of [0,1]. Like f(x) is
the set of all y such that y is in f(x). Now S together with its
universe cover the whole unit square iff for all real numbers x and y,
both in [0,1], either x is in f(y) or y is in f(x). We'll find two real
numbers x and y in [0,1] such that neither x is in f(y) nor y is in f
(x), thereby completing the proof. Let Y be a subset of [0,1] of
cardinality aleph-one, it has to be a proper subset of [0,1] since not-
CH. Let X be the union of all the sets f(y) for y in Y, this is union
of aleph-one countable sets, so its cardinality is at most aleph-one.
So the union of X and Y has cardinality aleph-one, so there's a real
number in [0,1] but not in X or Y, pick one and call it x. f(x) is
countable, so pick a real number y in Y but not in f(x). y is not in f
(x), and since y is in Y but x is not in X, x is not in f(y).
Sent via Deja.com http://www.deja.com/
Before you buy.
==============================================================================
From: Mike Oliver
Subject: Re: Continuum Hypotheisis
Date: Tue, 21 Dec 1999 12:18:08 -0800
Newsgroups: sci.math
Jonathan Hoyle wrote:
> It fails to be true once PP is dropped from the list
> of axioms of geometry, and in fact becomes false if ~PP is added to your
> axiom list. Where we differ, Mike, I think is simply our approaches.
> You see some model which pre-exists out there and we try to approximate
> it by a set of axioms. I, on the other hand, see the model as being
> generated from the axioms you supply.
Look, axioms do not walk out of the ocean on an oyster shell surrounded
by putti, nor are they chosen arbitrarily in some Sartresque act of the
will. They are, generally, an attempt to formalize an image that
the axiomatizer has in his mind.
This does not necessarily imply that such an image corresponds to some
real object in a Platonic heaven. Maybe it does, maybe not; I decline
to express an opinion. What's certain, though, is that the axiom was
not chosen as a meaningless string of symbols in a formal language.
Look up the story of the Magnificrab in "Goedel, Escher, Bach"
to see what I mean.
> Asking if CH is "really true" or not is a lot like asking if the
> Parallel Postulate is "really true" or not. Sure one model [...]
Quibble here: It will be easier to understand each other if you
use standard language. CH is not part of a "model", it is part
of a "theory". A theory is a collection of sentences (syntactic
objects), whereas a model is a collection of objects about
which the sentences purport to tell us something (the model
provides the semantics; before there was only syntax).
> ZFC + CH I am sure has some fascinating results, and I am
> sure ZFC + ~CH has some other different, yet equally fascinating
> results. We have barely begun to examine the differences yet.
Very considerable work exists involving the consequences of axioms
that imply either CH or ~CH. For example Diamond implies CH,
and the Proper Forcing Axiom (PFA) implies ~CH, and a great many
of their consequences in infinitary combinitorics are known.
Unfortunately there is no clear intuitive picture motivating
either axiom (except that Diamond is true in L and other core
models).
> Yes fascinating question. And I find that this is the most persuasive
> line of discussion you have on me. After all, we know there exists a
> set T in ZFC+~CH with cardinality Aleph1 (with Aleph1 < c), yet what can
> A be in ZFC+CH? How can it seem to exist in one model and not the
> other? Yes, this puzzles me too. (Hopefully I am expressing myself
> correctly here.)
Since you've used "model" to mean "theory" in the past, I can't be
sure which you mean here. But let's suppose you really meant "model",
which makes a lot more sense in context.
Suppose we have two models M and N of (some fragment of) set theory,
and to avoid technical unpleasantnesses let's suppose they're transitive
epsilon models; that is, for x,y \in M, M believes that x is an element of y
just in case x really *is* an element of y, and moreover if x \in M
and y \in x then y \in M (same two conditions for N).
Now suppose there are real numbers in N that just are not in M at all.
Then we might have M |= CH, but N |= ~CH ; M underestimates the cardinality
of R because it just doesn't have enough reals in it.
Or we might have that M and N have the same real numbers, but N has
more *sets* of reals. In that case we could have M |= ~CH, but
N |= CH. In this case what's going on is that M fails to find
the 1-1 correspondence between the reals and the countable ordinals,
because that correspondence is coded by a set of reals that M
doesn't have.
But the question is, what happens once we've got *all* the reals and
*all* the sets of reals? Any two (transitive epsilon) models containing
all sets of reals must *agree* on the truth value of CH. What we
don't know is in which way they agree.
> However, it is clear that the answer changes depending on your model.
> For example, suppose we have a function predicate F, such that F(0) is
> true, F(1) is true, F(2), F(3), etc. Yet instead of AxF(x), we add
> ~AxF(x) as an axiom. Seeming contradiction...but only from the point of
> view of the natural numbers...in Non-standard analysis, it is perfectly
> consistent.
But the point is, once you know that you have a model that satisfies
F(n) for every n but does not satisfy (All n)(F(n)), you know that
that model has a mistaken idea of what the natural numbers are.
You like to point out that such models are useful for NSA, and
that's true, they are -- but to make use of them in NSA, you have
to *recognize* the distinction between the naturals of the model
and the genuine naturals. That's how you define, for example,
the standard part of a hyperreal.