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Talk:Continuum hypothesis/Archive 1

1 who believes CH or ~CH ?
- 1.1 Did Cantor believe CH?
2 Investigating the continuum hypothesis
3 Any set to demonstrate CH?
4 Independence of CH from large cardinal axioms
- 4.1 Forms of CH without AC
5 Original Sources
6 Hilbert's first has nothing to do with ZFC per se

who believes CH or ~CH ?

I don't entirely agree with the following passage:

Generally speaking, mathematicians who favor a "rich" and "large" universe of sets are against CH, while those favoring a "neat" and "controllable" universe favor CH.

See the Maddy reference I added, page 500, the sections entitled Not-CH is restrictive (in favor) and Modern forcing (in favor).

I'll try to come up with some wording that's not too awkward, illustrating the historical view reported in the existing Wiki article, while also pointing out that between models having all the same reals, it's the ones with more sets of reals that are more likely to satisfy CH. --trovatore

My attempt is now in place Trovatore 05:04, 27 Jun 2005 (UTC)

Did Cantor believe CH?

I would add to Trovatore's point that it is questionable (as stated on the page) that Cantor himself believed CH. Did he not alternate back and forth during his manic (and depressive) phases, sometimes believing CH, sometimes believing ~CH? See the BBC TV documentary "Dangerous Knowledge" willbown. 16 June 2008. —Preceding comment was added at 22:37, 16 June 2008 (UTC)

I'd never heard that he ever believed ¬CH, only that he sometimes despaired of ever being able to prove CH. Interesting if true, though. What did the documentary say exactly? Has anyone seen this claim in any source available online or at the library? --Trovatore (talk) 22:53, 16 June 2008 (UTC)

I've got the programme on tape, and I'll look it up and report back. Great programme, BTW. --Michael C. Price ^talk 18:08, 24 August 2008 (UTC)

The programme hints that Cantor wanted to believe in CH -- but it is not explicit. As for Cantor's alternating between CH and ¬CH, that was in reference to him thinking that he found a proof of CH at times and ¬CH at other times. Proof and belief are not quite the same.--Michael C. Price ^talk 08:25, 25 August 2008 (UTC)

Investigating the continuum hypothesis

I've removed this section:

If a set S were found that disproved the continuum hypothesis, it would be impossible to make a one-to-one correspondence between S and the set of integers, because there would always be elements of set S that were "left over". Similarly, it would be impossible to make a one-to-one correspondence between S and the set of real numbers, because there would always be real numbers that were "left over".

It's a good thing to add some intuition, but I don't know that this passage helped much. CH isn't about whether such a set S can be found in any ordinary sense, but just about whether one exists. --Trovatore 21:12, 22 October 2005 (UTC)

Let me add this thought:

Since CH cannot be disproven, a set that denies it cannot be found. This doesn't mean it does not exist, but we're close :-) Honnza (talk) 06:01, 17 May 2008 (UTC)

CH cannot be disproved from ZFC. No, I'm afraid that's not particularly close to settling the issue. --Trovatore (talk) 08:40, 17 May 2008 (UTC)

Any set to demonstrate CH?

The undecidability of CH begs the question: If there is a cardinality between $\aleph_0$ and $2^{\aleph_0}$ , then what sets might there be that have this cardinality?

In other words, is there any known set that is larger than $\aleph_0$ , but such that you need to set the truth value of CH to determine whether or not it's equivalent to continuum?

Moreover, is there any evidence pointing to the existence or non-existence of such a set, or whether it would be possible to find it if it does exist? -- Smjg (talk) 16:35, 28 February 2008 (UTC)

Well, you have to keep in mind here the distinction between a set and a definition of a set. Think of sets as collections of things just lumped together at random, not necessarily in accordance with any rule. Then it might happen, just by accident, that there's some rule such that everything in the set satisfies the rule, and everything outside the set does not -- in that case we say that the set is "definable", but you still can't identify the set with its definition; that way lies all sorts of trouble (ask Frege).

So rephrasing your question -- is there a definition for a set of reals such that ZFC neither proves nor disproves that the set of all reals satisfying the definition, has cardinality $2^{\aleph_0}$ (but does prove that the set is uncountable)? Sure, but possibly not a terriby interesting definition. Something like "the first $\aleph_2$ ordinal-definable reals in the natural wellorder on OD, or all the reals if there are not $\aleph_2$ OD reals". --Trovatore (talk) 18:07, 28 February 2008 (UTC)

Also, $L_{\omega_1} \cap \mathcal{P}(\omega)$ is the set of all subsets of the natural numbers (sometimes identified with the reals) which are constructed (see Constructible universe) before the first uncountable stage. It has cardinality $\aleph_1$ , I believe. JRSpriggs (talk) 08:34, 29 February 2008 (UTC)

No, not unless $\aleph_1^L=\aleph_1$ (which it doesn't, of course). --Trovatore (talk) 18:42, 29 February 2008 (UTC)

To Trovatore: Thanks for the correction. JRSpriggs (talk) 12:03, 1 March 2008 (UTC)

The easiest example of a set that has cardinality $\aleph_1$ is the set of all countable ordinals (where everything is relativized to a fixed model of ZFC). Each countable ordinal can be viewed as a linear order on the natural numbers and thus an element of

2 ω

. So if you use AC to choose for each countable ordinal a single real encoding that ordinal, you will obtain a set that has cardinality $\aleph_1$ , but will not have the cardinality of the continuum unless CH holds. — Carl (CBM · talk) 15:03, 1 March 2008 (UTC)

Quoting Carl (an inch above) "Each countable ordinal can be viewed as a linear order on the natural numbers..." Is the converse statement "Each linear order on the natural numbers can be viewed as a countable ordinal" necesarilly true? If yes, is there then a one-to-one correspondence between linear orders on the natural numbers and countable ordinals? Or is that last statement perhaps equivalent to the CH? (When thinking of the Cantor-Bernstein theorem, it might be that way. (I'm just speculating now of course.)) The number of orderings of a finite number n is n!. Can one by a naive extension of the factorial function say that CH can be "rephrased" as the statement

ω

! =

ω 1

? YohanN7 (talk) 17:40, 18 May 2008 (UTC)

Taking the last question first: Yes, that's a reasonable way of putting it, though not quite by the reasoning you used; the reasoning has some flaws but it's close enough to see that the factorial of $\aleph_0$ (not a term that's used much but it's pretty clear what it means) is $2^{\aleph_0}$ .

Getting more into the details, more specifically, not every linear order of the natural numbers corresponds to a countable ordinal, but only the wellorderings. To a wellordering of the naturals you associate a unique countable ordinal, namely the length of the wellordering. And going the other direction, any (infinite) countable ordinal can be represented as the length of some wellordering--but not a unique one; there will be lots of different wellorderings with the same length. So what this gets us is that there are at least as many wellorderings of the naturals as there are countable ordinals (strictly speaking, infinite countable ordinals, but the finite ones are easily accounted for). It doesn't tell us there aren't more. --Trovatore (talk) 18:20, 18 May 2008 (UTC)

Interesting! To quote yourself Trovatore, "It's a good thing to add some intuition...", but you are always very (too?;) careful to make intuitive arguments close to precise mathematical statements. I believe that the above reply does shed additional light on what the CH actually is and goes beyond intuition. (It actually sounds like a statement of ZF(C) in my ears, making it mathematics.) I also think that som revised form of it may qualify for the main article as an example of what CH and its negation would imply. Most people who knows what a one-to-one correspondence will know what a wellordering is.YohanN7 (talk) 22:34, 20 May 2008 (UTC)

Independence of CH from large cardinal axioms

So far, CH appears to be independent of all known large cardinal axioms in the context of ZFC.

But in the context of ZF, j:V into V refutes choice (according to the article on Reinhardt cardinals anyway), and GCH implies it (according to this article). So then j:V into V would refute GCH. Right? --Unzerlegbarkeit (talk) 15:58, 27 May 2008 (UTC)

It may be that J:V into V refutes ZF.Kope (talk) 16:29, 27 May 2008 (UTC)

Well, sure. It may be that any large cardinal axiom refutes ZF. But so far no such refutation is known. Nor is any proof, refutation, or independence of CH or GCH from any large cardinal axiom known, other than this one. Is that correct? --Unzerlegbarkeit (talk) 02:36, 28 May 2008 (UTC)

The Generalized Continuum Hypothesis is much stronger than the Continuum Hypothesis. So many propositions may be consistent with CH, but not with GCH. Reinhardt's cardinal cannot exist in ZFC so the sentence in question does not apply to it. JRSpriggs (talk) 06:37, 28 May 2008 (UTC)

Sure. And I did say "in the context of ZF" and "GCH". I guess my implicit assertion is that whatever makes the sentence in question worth mentioning makes this worth mentioning too, assuming I've stated the situation correctly. --Unzerlegbarkeit (talk) 10:24, 28 May 2008 (UTC)

(To Unzerlegbarkeit ) Not quite. A favorable solution would be to show that the consistency of some (axiom of choice) large cardinal implies the consistency of j:V into V, minus AC. Incidentally, GCH has more than one formulations, which are equivalent in the presence of AC, but not necessarily so in the absence of it. One of them implies AC. Therefore, I think, it is more correct to say that one form of GCH implies AC. Kope (talk) 06:45, 28 May 2008 (UTC)

To Kope: See Talk:Axiom of choice#GCH implies AC?? for a proof that GCH implies AC. This establishes that there is only one version of GCH, even in ZF (without choice being assumed otherwise). JRSpriggs (talk) 06:59, 28 May 2008 (UTC)

But CH doesn't, so might it have inequivalent formulations in the absence of AC? Is this also worth mentioning? --Unzerlegbarkeit (talk) 10:24, 28 May 2008 (UTC)

Yep. The following are not equivalent in the absence of AC:

$2^{\aleph_0}=\aleph_1$ , $2^{\aleph_0} \leq \aleph_1$ , $2^{\aleph_0}<\aleph_2$
There is no cardinal m such that $\aleph_0 < m < 2^{\aleph_0}$
$2^{\aleph_0} \not > \aleph_1$
$2^{\aleph_0} \not \geq \aleph_2$

The first two are considered somewhat standard formulations of CH; I only included the latter two so that it's clear they are all the same under the axiom of choice. — Arthur Rubin (talk) 17:22, 28 May 2008 (UTC)

Hmmm. In ZFU (ZF with Urelements), AH (aka Aleph Hypothesis) $\left (2^{\aleph_\alpha}=\aleph_{\alpha+1} \right )$ does not imply GCH (if X is an infinite set, there is no Y such that $X < Y < \mathcal{P}(X)$ (with "<" in the sense of cardinality)). (Source: my mother's book Set Theory for the Mathematician, I believe). I'm sure that fact that PW (the power set of a well-ordered set can be well-ordered, a trivial consequence of AH) does not imply AC in ZFU, although it does in ZF, is in both editions of my parents' book Equivalents of the Axiom of Choice. — Arthur Rubin (talk) 17:40, 28 May 2008 (UTC)

However, see Beth number#Generalization — if the ur-elements form a set which is equinumerous with a pure set (a set whose transitive closure contains no ur-elements), then I believe that ZFU+AH would imply AC and thus GCH. JRSpriggs (talk) 14:08, 29 May 2008 (UTC)

Yes, I believe that's correct. The proof escapes me at the moment.... — Arthur Rubin (talk) 14:34, 29 May 2008 (UTC)

Consider the class of pure sets. It satifies ZF+AH. Since bijections between pure sets are themselves pure sets, it has the same aleph numbers as the original model (which included urelements). So the class of pure sets satifies AC. Thus the pure set which is equinumerous with the set of urelements can be well ordered. Thus the set of urelements can be well ordered. Then do induction on the rank α to show that V_α can be well ordered. So the universe satisfies AC. So it satisfies GCH. OK? JRSpriggs (talk) 15:31, 29 May 2008 (UTC)

Forms of CH without AC

The form $\neg \exist m \; \aleph_0 < m < 2^{\aleph_0}$ makes the continuum hypothesis meaningful even if the continuum isn't well-ordered. The article states:

The continuum hypothesis is closely related to many statements in analysis, point set topology and measure theory. As a result of its independence, many substantial conjectures in those fields have subsequently been shown to be independent as well.

Do we have any examples which would make sense without AC, and would they depend on the form of CH? Also, on the other side of the coin, my understanding is that there could be no purely arithmetic consequences, roughly because ZF + V=L proves CH anyway (and even GCH), and relativising everything to L makes no difference to (first-order) arithmetic. If correct, I believe this is worth stating. Also, the article states:

Assuming the axiom of choice, there is a smallest cardinal number $\aleph_1$ greater than $\aleph_0$ , and the continuum hypothesis is in turn equivalent to the equality $2^{\aleph_0} = \aleph_1.$

The "assuming the axiom of choice" bit should really attach to the "in turn equivalent to", because even from the axioms ZF without AC, aleph_1 exists and is an immediate successor of aleph_0, right? --Unzerlegbarkeit (talk) 02:39, 1 June 2008 (UTC)

Without AC, there could be other cardinals which are larger than $\aleph_0$ but incomparable with $\aleph_1$ , for example, the cardinality of a set which is the union of natural numbers with a Dedekind finite infinite set. In this case, neither the new cardinal nor $\aleph_1$ can be said to be the smallest cardinal greater than $\aleph_0$ . JRSpriggs (talk) 06:38, 1 June 2008 (UTC)

Original Sources

I notice there aren't any references to Cantor's original writings. I'll try to dig some up, but has anyone seen any English translated letters, etc? Libertyblues (talk) 00:03, 3 August 2008 (UTC)

Hilbert's first has nothing to do with ZFC per se

This statement is revealing a transfinite-platonic bias. without a formalization like ZFC, it is not clear that it is possible to give a meaning to the statement that the continuum even has an ordinal size. Before Cantor insisted it was true, an infinite collection like the set of real numbers was not considered to have a definite size, let alone a definite ordinal size, and indeed, after Cohen, it is again a widespread point of view that the continuum does not have a definite size as an ordinal.

The idea that the set of all real numbers has definite properties in a platonic realm can be classified as a type of "transfinite platonism", which is just what platonism usually means nowadays. But transfinite Platonism can be logically separated from "computational platonism" (I made that up, but it needs a name)--- the position that all computer programs either halt or do not halt. The second position is what people mean by platonism in practice--- that the results of computations with symbols have a meaning, and questions about their outcome have a truth value. it was the position of Paul Cohen, shared by most mathematicians, that all questions about the integers/computer-programs are decided by a strong enough axiom system, which is a way of saying that they have a truth value in an axiom-system independent way.

But you can believe this, and still be a formalist regarding transfinite set theory. The position might be called "formalist", but "formalist" conflates two notions: "formalist regarding uncountable ordinals and sets the size of the continuum" and "formalist regarding countable infinity". A "formalist regarding countable infinity" would, for example, consider a nonstandard models of Peano arithmetic to be just as "true" a model of the integers as the standard model. Such a formalist would believe that some statements about diophantine equations like Fermat's Last Theorem, are undecidable in an absolute sense. This is the type of straw-man formalist that people argue against.

A "formalist regarding uncountable ordinals" on the other hand, would say that the truth value of the 3N+1 conjecture is well defined, but statements like the continuum hypothesis have no truth value except relative to an axiom system. This is the type of formalism which the forcing models foist upon you. If you accept this philosophy, which many people do, then you would not regard the continuum hypothesis as independent of ZFC.

This is the philosophical position which Cohen alludes to in his book, hopefully stated clearly enough there so that this exposition will be recognized as a clarification. I have been trying to figure out how to say it clearly for a while.Likebox (talk) 17:22, 24 August 2008 (UTC)

Hilbert wanted a proof of the continuum hypothesis (he doesn't seem to have explicitly countenanced a refutation). A proof from what axioms, I'm afraid he didn't really say. But whatever they were, they certainly weren't the axioms of ZFC, because his speech was in 1900, and ZFC wasn't formulated until a couple of decades later.

As for the situation today, it is certainly true that there are folks who are realists about the naturals but formalists beyond that, and that they mostly will not consider CH to have a well-defined truth value outside of its provability or refutability in some formal theory. There is still, however, no clear reason that that formal theory ought to be ZFC. --Trovatore (talk) 01:06, 25 August 2008 (UTC)

I see your point, thanks for clarifying. But ZFC is not arbitrary--- it's sort of natural--- you have the function axiom, and everything else just builds up big ordinals in sequence. It's like Peano arithmetic, anything else is going to be roughly similar. Hilbert had formalized logic in mind, and probably knew approximately what a formalized set theory would look like, and I think he meant ZFC. But I defer to someone who actually read more Hilbert.Likebox (talk) 02:51, 25 August 2008 (UTC)

In one sense, you're right, ZFC is not arbitrary — all its axioms are well-motivated (in retrospect) by the picture of the cumulative hierarchy. What's arbitrary is cutting off precisely at the strength of ZFC, no less and no more.

Now, the most natural and well-accepted enhancements to ZFC, the large cardinal axioms, don't settle CH either, or at least the ones we know so far don't settle it. It doesn't really follow that we won't ever find ones that do (though they would have to be of a somewhat different character from the currently known ones).

On the other hand, there are extensions to ZFC that someone or other has considered natural at some point that do settle CH. For example the axiom of constructibility settles it positively, whereas the proper forcing axiom settles it negatively. The big difference is, I'm afraid, a Platonistic one: there's not as compelling a reason to believe that either of these is true (and indeed there are extremely good reasons to believe that the axiom of constructibility is false).

The most interesting current line of inquiry into the question is one started by Woodin, who has a subtle and difficult argument that, if accepted, would imply that CH is false. --Trovatore (talk) 03:49, 25 August 2008 (UTC)

You like repeating the party line! But of course, you know that I think it's rubbishy. And Woodin's approach is neither subtle or difficult. It's just determinacy.Likebox (talk) 06:30, 25 August 2008 (UTC)

If you truly understand Woodin's approach, I'd be happy if you'd teach it to me; I know only some vague generalities about it. But it's certainly not "just determinacy". In fact I don't see that it has that much to do with determinacy, directly. What he claims is that he has something analogous to determinacy, one level up; that just as projective determinacy "settles all natural questions" (whatever that means exactly—note that the quotes are not intended to attribute the phrase to Woodin) about $H(\aleph_1)$ , his approach similarly "should" settle all natural questions about $H(\aleph_2)$ (one of which is CH).Now, just by the way, "just" determinacy can get extremely difficult and subtle. But Ω-logic is still quite another thing. --Trovatore (talk) 06:44, 25 August 2008 (UTC)

I thought he was just repeating determinacy, I guess I'm totally wrong. I didn't read his paper very closely at all.Likebox (talk) 10:31, 25 August 2008 (UTC)

I know (from a very superficial look at) 'The Higher Infinite' that there were some (at the time of their discovery) unexpected connections between large cardinal axioms and various versions of the axiom of determinacy, for what that's worth. Zero sharp (talk) 14:49, 25 August 2008 (UTC)

(deindent) I am pretty ignorant about determinacy--- I only have only the most superficial heresay knowledge, and no clear understanding. But I have a somewhat negative opinion anyway, probably unjustified.

Well, then just a friendly suggestion: refrain from making categorical statements that Woodin's approach is 'just determinacy' and from accusing people of 'just repeating the party line'. It just makes you look ignorant. —Preceding unsigned comment added by 67.118.103.210 (talk) 23:32, 26 August 2008 (UTC)

I am ignorant. So is everybody else. That's life.Likebox (talk) 01:53, 27 August 2008 (UTC)

When I think about the continuum hypothesis, it is linked in my mind to the notion of measure and probability--- the reason that many people have intuitions about the cardinality of the real numbers is that it is consistent to talk about randomly choosing real numbers in a way that is different from randomly choosing integers or elements of a well ordered set. You can't pick an integer uniformly at random--- the concept doesn't make sense. You need a probability distribution, and certain integers are more likely than others. The same goes for any countable ordinal--- you need to weight different positions with different probability.

But for the interval [0,1], the notion of picking a number at random seems to be perfectly well defined, because you can roll dice to find each digit in sucession. This converges to a real number, and any property of that number that has probability zero is false. If a random number can be thought of as an element of the mathematical universe, with the property of belonging or not belonging to any previously specified set, it means that this previously specified set has a well defined Lebesgue measure. So not only is the continuum hypothesis false with random reals, it is meaningless, because the real numbers can't be well ordered. Any ordinal, no matter how large, can be imbedded in the reals by inductively mapping each successive element into [0,1] at random, and the probability of picking the same real twice is always zero (the last statement is only self consistently true, it's true in a countable model).

Making this precise is the job of forcing, which, given the political situation in mathematics, phrases everything in terms of the picking process, although at the end it talks about "random reals". But a random real number is an obviously consistent idea, even though it is incompatible with choice.

So when I look at post-forcing axioms like determinacy, I am always looking for the probability interpretation, and in this case I couldn't see it. The determinacy axiom, as I heard it, was exciting because it had a completely different character--- it asserted something about infinite sets that somebody thought was intuitively consistent (I don't know why, but they turned out to be right, so they must have had a good idea). But the justification for this for those with a formal view of higher infinity is that a large cardinal axiom proved that it is equiconsistent with set theory. So determinacy, as far as theorems about integers are concerned, is a moderate large cardinal axiom, and is not as interesting as a completely new axiom (although Cohen's "article of faith" says there are not going to be any consistent new axioms that are not equiconsistent by virtue of a large enough cardinal). So a superficial glance at Woodin's article was disappointing for me, because it didn't give a perspective which was probabalistic. But that's a very self-centered point of view, so I'll give it another shot.Likebox (talk) 17:51, 26 August 2008 (UTC)