Talk:Continuum hypothesis/Archive 1

Practical consequences?

Are there any "practical consequences" of presuming CH (or GCH) to be true or false? For example, are there any theorems which can be proved by presuming CH to be true or by presuming it to be false, but for which the proof is much simpler if we presume CH to be true, (or vice versa). -- SJK

In set theory and analysis, there are many statements which could be proven if GCH is assumed, and could be disproven if GCH fails. I don't have any examples right now. I don't know if there are theorems whose proofs get simpler by assuming GCH. --AxelBoldt

There are. I don't remember any examples, but I distinctly remember moaning "if only we could just use GCH..." doing a proof on an assignment. --

I don't remember ever needing the continuum hypothesis. A list of examples is sorely needed in the article, otherwise the paragraph about "substantial results" should be deleted. Also, since the GCH implies the axiom of choice, it is much more likely that just the axiom of choice would suffice. -- Miguel

Under CH, there exists bounded functions from the unit square into R, which are measurable in each coordinate, but so that the functions is not integrable. It is consistent with ZFC that no such function exists. (So CH is necessary.) Although if you permute any of the conditions slightly, the situation can be resolved in ZFC.

There are plenty of things known to be independent of ZFC but which don't line up so nicely with CH. (Kaplansky's problem and the Whitehead problem are two good examples.) --

---

Gödel's incompleteness theorems only say that if proof is identified with first-order logical derivation, then any consistent axiomatization will be incomplete. But his proof of the first theorem has two parts: the first proves that his wff U is unprovable; the second gives a proof of U (or rather its interpretation in N). The statement, "This statement is not first-order derivable from the given axioms" is surely provable, though not first-order derivable.

Likewise, CH has not been shown unprovable, but only underivable from ZFC.

Chris Freiling's "Axioms of Symmetry: Throwing Darts at the Real Number Line" (Journal of Symbolic Logic Vol. 51, Iss. 1, pp. 190-200) presents a (rather philosophical) argument against CH. --Archibald Fitzchesterfield

Yes, that's a good paper, I'll add it to the list of references. --AxelBoldt

---

I find this article ot be hard to understand. I had to go through several other articles to even understand what it was about. Maybe someone should add a short informal summary that explains what the continuum hypothesis means that is also easier to understand. Right now I'd guess that if you understand the article you already know what the continuum hypothesis is. The article then loses a big part of its use. -XeoX

better?

---

Is Chris Freiling's "statement about probabilities" Freiling's Axiom of Symmetry? If so, somebody should add a link.

techniccalyy there aren fractal dimensions that are applicableto set theory and cardinallity that raise questions as to whether Cantors statment

'There is no set whose size is strictly between that of the integers and that of the real numbers.'

is correct

---

Removed statement that it would be impossible to prove that ZF contains a contradiction.

Roadrunner 21:59, 20 Apr 2004 (UTC)

---

I've been told that logicians now have reason to think that the "continuum hypothesis is false." Apparently the situation is that people basically want to take the axiom of projective determinancy, and it's now been shown that the axiom of projective determinacy impiles the negation of the contiuum hypothesis. I'll have to leave editing the article to someone who actually knows what's going on.

I found a relevant link: http://math.berkeley.edu/~woodin/talks/Lectures.html

---

Absolutely, Woodin's work is relevant, and should be referred to somewhere in the article. Unfortunately it's rather hard to summarize in a way that would make much sense to more than, at a generous estimate, 1000 people in the world.

Projective determinacy does not in fact imply the negation of CH (ZFC+PD+CH is consistent, assuming ZFC+PD is itself consistent). Rather, Woodin offers PD as an example of a proposition that has a kind of indestructibility via forcing. Woodin proposes an allegedly similar sort of stability under forcing for the theory of (I think) $H(\omega _{2})$, and argues that this stability implies ~CH. Or something like that. I'm pretty fuzzy on the details myself. --Trovatore 00:43, 26 Jun 2005 (UTC)

So I didn't try to explain it, just added the ref and pointed the reader to it. Trovatore 05:04, 27 Jun 2005 (UTC)

Change hyperlink?

Shouldn't "Zermelo-Fränkel set theory axiom system" be a hyperlink to Zermelo-Fraenkel set theory instead of Axiomatic set theory?

Simplifying the language a bit

Quoting from the Manual of style:

Do not assume that your reader is familiar with the acronym or abbreviation you are using. The standard writing style is to spell out the acronym or abbreviation on the first reference (wikilinked if appropriate) and then show the acronym or abbreviation after it.

I say this because it took me about 10 seconds to realise that in ZF + GCH ⊦AC, AC stood for Axiom of Choice. I daresay there are a number of people who want to understand this article but wouldn't even know what ⊦ means (especially if they don't know to check an article about formal logic). Perhaps the first reference to Zemillo-Frankael should include (ZF) after it, and likewise include a brief statement somewhere that ZF + Axiom of Choice (AC) = ZFC. Confusing Manifestation 02:26, 24 February 2006 (UTC)

Quote by Martin

I removed the quote by Martin, since I dislike attaching someone's name to comments that have since been disavowed. I don't know if an official transcript of the Atlanta discussion was ever published. (I was there but only remember generalities, and the bandage on Cohen's head.) The closest thing to a transcript I could find on the web was here on the FOM list, but that summary was probably not reviewed by the speakers themselves. It does state that Martin retracted the quote from his 1976 paper. CMummert · talk 20:47, 5 February 2007 (UTC)

I think the bit about the axiom of determinacy is misleading; I'd take it out. It's true that AD implies a weak version of CH (namely that there's no cardinality strictly intermediate between $\aleph _{0}$ and $2^{\aleph _{0}}$, but it does not imply $2^{\aleph _{0}}=\aleph _{1}$, and in my view it's the latter that should be thought of as CH in non-AC or not-necessarily-AC contexts. (For one thing it's simpler in descriptive-set-theoretic terms, $\Sigma _{1}^{2}$ versus $\Pi _{2}^{2}$.) --Trovatore 20:54, 5 February 2007 (UTC)

Feel free to edit that section any way you like. My goal was mainly to get the "CH has no truth value at all" opinion into the article. The quote by Martin seemed like an easy way, since I already had that book out from the library. I'm sure there are other places where the entire section can be worked on. CMummert · talk 21:07, 5 February 2007 (UTC)

(Bumping, because I think it's relevent, re discussion in Talk:Axiom of choice).

I think AH(α)

$2^{\aleph _{\alpha }}=\aleph _{\alpha +1}$

and CH(ℵ_α)

$(\not \exists \mathrm {m} )(\aleph _{\alpha }<\mathrm {m} <2^{\aleph _{\alpha }})$

are both plausible interpretations of CH, in the absence of AC. They are not equivalent, even for α=0. — Arthur Rubin | (talk) 22:36, 21 May 2007 (UTC)

CH and Zorn's lemma

What is the easiest/quickest way to understand how CH is independent of AC? Rather, what is the easiest, most direct reference that still gives a rigorous exposition? I'm staring at something that involves Zorn's lemma, it seems CH-like, and I am confused about what I'm looking at. linas 04:18, 8 April 2007 (UTC)

It depends on what you mean by independent. To see why either CH or not CH can hold when AC holds, you need to read about L and about forcing. Unfortunately, I don't know of any undergrad-level expositions. You can start with the WP articles or with this expository essay. Kunen's book is a standard formal treatment that includes the details. CMummert · talk 13:06, 8 April 2007 (UTC)

Otheruses template

I added {{otheruses4}} in case somebody came here looking for this, but the parameters I used were kind of clunky and I don't think they read well. Feel free to change them. --superioridad ^(discusión) 23:17, 11 May 2007 (UTC)

Wow, learn something new every day. I wasn't aware of this meaning of "continuum hypothesis" but it seems to be well attested. Can't think of any tweaks to the message that would be obvious improvements, at the moment. --Trovatore 08:00, 12 May 2007 (UTC)

I hadn't heard of it until today, either. I went to the fluid mechanics page from a recent Wired article on bad writing in Wikipedia science articles, saw that section, and figured that this article could do with the dab link. --superioridad ^(discusión) 09:58, 12 May 2007 (UTC)

Contradiction ?

...and he showed that the set of integers is strictly smaller than the set of real numbers...

This is equivalent to:

$2^{\aleph _{0}}=\aleph _{1}$

I feel like there is something wrong somewhere. If it is smaller, it is not equal, right ? --86.215.106.35 23:35, 14 August 2007 (UTC)

Possibly the text could be more clear. He showed the set of integers is strictly smaller than the set of real numbers. What the continuum hypothesis states is different from that; it states that there is no set S such that yada yada yada. This (the latter "this") is equivalent to $2^{\aleph _{0}}=\aleph _{1}$. Do you understand it now? If so, would you like to take a crack at rewording in such a way that it wouldn't have confused you the first time? That would be a valuable service. --Trovatore 23:43, 14 August 2007 (UTC)

I made an attempt at expanding it. I think the middle part may be a little long, the part on equivalent formulations, but I left it. I was thinking of moving it to a lower section. — Carl (CBM · talk) 01:26, 15 August 2007 (UTC)

Powerset cardinality

Do we need the CH to prove that

$A<B\to 2^{A}<2^{B}$

for infinite cardinals? --Michael C. Price ^talk 12:01, 31 August 2007 (UTC)

GCH is adequate for that. Shelah's general results on cardinal exponentiation has shown, among other things, that it is consistent with ZF that:

$2^{\aleph _{0}}=\aleph _{1}$, $2^{\aleph _{1}}=\aleph _{3}$, $2^{\aleph _{2}}=\aleph _{3}$,

so that CH is not adequate. On the other hand, it is consistent that:

$2^{\aleph _{0}}=\aleph _{2}$, $2^{\aleph _{\alpha +1}}=\aleph _{\alpha +3}$, $2^{\aleph _{\lambda }}\leq \aleph _{\lambda +2}$ for limit ordinals $\lambda \$

so the above statement does not imply GCH. — Arthur Rubin | (talk) 12:54, 31 August 2007 (UTC)

I did mean GCH, not CH. Thanks for the results -- most interesting. (That was ZF+AC, I assume?) Thanks also for the pointer to Saharon Shelah. Do you have any arXiv links for his relevant papers? --Michael C. Price ^talk 15:17, 31 August 2007 (UTC)

The quoted result is Easton's theorem, not Shelah's. Kope 15:38, 31 August 2007 (UTC)

Exactly. That theorem states that, for regular cardinals, the only restrictions on the continuum function are weak monotonicity (α ≤ β → 2^α ≤ 2^β) and Konig's theorem (set theory) (α < cf 2^α). Any possible arrangement of regular cardinals and their powersets that meets these requirements is consistent with ZFC. This is a forcing proof covered both in Jech's book and Kunen's book on set theory. — Carl (CBM · talk) 16:58, 31 August 2007 (UTC)

Thanks everybody. Looks like a pretty good reason for assuming GCH. --Michael C. Price ^talk 17:20, 31 August 2007 (UTC)

How ya figure? --Trovatore 00:06, 1 September 2007 (UTC)

Since we need the GCH to demonstrate

$A<B\to 2^{A}<2^{B}$

which seems intuitively obvious. Ergo to reject the GCH is to accept "unreasonable" propositions, such as $2^{\aleph _{1}}=2^{\aleph _{2}}$.--Michael C. Price ^talk 01:19, 1 September 2007 (UTC)

Well, the strong monotonicity does seem like a natural enough thing to expect, by analogy with the finite case, but I think if you break it down it's a bit too complicated to claim any direct intuitions about. The proper forcing axiom implies that $2^{\aleph _{0}}=2^{\aleph _{1}}=\aleph _{2}$. I think PFA might be accepted as true by those set theorists who find Woodin's arguments convincing (the ones about Ω-logic) but I'm a little hazy on that point. --Trovatore 02:30, 1 September 2007 (UTC)

Intuition is a subjective thing; I'm just saying how it looks to me. But there again, what better guide is there to the acceptance of axioms? --Michael C. Price ^talk 14:39, 1 September 2007 (UTC)

I can see that strong monotonicity has an appeal from a naive point of view, but I do not know any source that suggests strong monotonicity as an intuitively obvious axiom.

Strong monotonicity is infinitely weaker than GCH. If you convince me that strong monotonicity should "obviously" hold, you are still far away from convincing me that GCH (or any other cardinal arithmetic implying strong monotonicity, such as $2^{\aleph _{\alpha }}=\aleph _{2^{\alpha }}$ for all successor ordinals α, plus SCH) is obvious.

--Aleph4 17:38, 1 September 2007 (UTC)

Fair enough, I see your point. Leaving aside the issue of axiomatic plausibility, does anyone object to a statement about strong monotonicity appearing in the "Implications of GCH for cardinal exponentiation" section? Is it fair to say that the addition of GCH to ZFC converts $A\leq B\to 2^{A}\leq 2^{B}$ to the stronger $A<B\to 2^{A}<2^{B}$? --Michael C. Price ^talk 07:59, 2 September 2007 (UTC)

That seems like a reasonable point to make. I don't much like the wording "converts", though. From the extra assumption, you can prove the stronger theorem -- that's the way I'd word it. --Trovatore 05:44, 3 September 2007 (UTC)

Done as suggested. I also said that the weak monotonic theorem was a consequence of Easton's theorem, since that seems the consensus here. --Michael C. Price ^talk 08:31, 3 September 2007 (UTC)

I wonder if we can, in the absence of GCH, strengthen the statement $A\leq B\to 2^{A}\leq 2^{B}$ to the stronger $A<B\to 2^{A}\leq 2^{B}$, since A = B seems to induce a bijection between their powersets, yielding $2^{A}=2^{B}$? --Michael C. Price ^talk 07:02, 16 September 2007 (UTC)

So the statement $A<B\to 2^{A}\leq 2^{B}$ is certainly a theorem of ZFC without needing GCH (and indeed you could drop AC and probably Replacement as well). But I don't see how it's a "stronger" statement. When you strengthen the hypothesis and get the same conclusion, you weaken the theorem, not strengthen it. --Trovatore 07:08, 16 September 2007 (UTC)

Thanks for correction. What about the article? If we accept all statements not excluded by Easton's & Konig's theorems are compatible with ZFC then the current article implies that $A=B\to 2^{A}<2^{B}$ is possible for some A & B. --Michael C. Price ^talk 07:34, 16 September 2007 (UTC)

I don't know if I follow. You can get from $A=B$ to $2^{A}=2^{B}$ by logic alone (it's one of the properties of identity; "law of substitution" or something like that). You don't need any set theory at all. I guess to refute the proposition above you do need to know that a cardinal is not less than itself. --Trovatore 07:49, 16 September 2007 (UTC)

I agree that $A=B$ implies $2^{A}=2^{B}$. But my point is statements obvious to you (or even me) are not obvious to the general reader. Perhaps a few examples (such as Arthur's) culled from the talk page would help the article.--Michael C. Price ^talk 08:00, 16 September 2007 (UTC)

Wait a minute, there's a subtlety here. You agree with the implication, but do you understand that it follows by logic alone, with no need for any set theory? What you said earlier was A = B seems to induce a bijection between their powersets, which is true, but uses set theory. --Trovatore 18:03, 16 September 2007 (UTC)

Well, if you read the = symbol as "has the same cardinality" rather than intensional equality,... But it's not really a big deal. — Carl (CBM · talk) 18:17, 16 September 2007 (UTC)

Yes, I did understand that the implication followed by logic alone, but I wasn't clear if that was relevant, since it also follows from set theory. Carl's point also seems relevant. --Michael C. Price ^talk 18:39, 16 September 2007 (UTC)

It's a bad idea to use the = sign to mean anything but literal identity, which is what I suspected Michael might have been doing.

I still don't understand what Michael wants to accomplish by stating the result in a (formally) weaker form; I don't see how that addresses his concerns at all. --Trovatore 18:52, 16 September 2007 (UTC)

(deindent) Well, my concern is that mathematically illiterate idiots like myself will look at $A\leq B\to 2^{A}\leq 2^{B}$ and conclude that $A=B\to 2^{A}<2^{B}$ for some A and B for ZFC without GCH, since the latter implication is consistent with the earlier. --Michael C. Price ^talk 19:04, 16 September 2007 (UTC)

To MichaelCPrice: It is enough that we are trying to provide information that is useful and true. To ask us to be responsible for dealing with the infinite variety of misconceptions which people may have is asking us to do the impossible. JRSpriggs 01:16, 17 September 2007 (UTC)

This is underscored by the number of times I have seen talk page comments where someone says "Why does the article claim every foo is a bar?" when the article says something like "note that not every foo is a bar." — Carl (CBM · talk) 01:48, 17 September 2007 (UTC)

Okay, but that is not the case here, where the article lacks any such caveat. Weakening an implication can make it more informative. Regarding JRSpriggs's "useful and true" comment, I am saying that I don't believe that the stronger implication is as useful as the weaker one. (Both implications are true, of course.) I have explained why a couple of time; no one has responded to the actual point of my criticism -- I assume because it is accepted as true.

If we are concerned about Trovatore's point about weakening the implication then I suggest replacing:

$A\leq B\to 2^{A}\leq 2^{B}$

by the pair of implications:

$A<B\to 2^{A}\leq 2^{B}$ and

$A=B\to 2^{A}=2^{B}$.

Would that make everybody happy? Now there is no loss of information; instead the pair of implications is actually more informative. --Michael C. Price ^talk 09:27, 17 September 2007 (UTC)

Aleph4 says: Comparing the two versions

1. $A<B\to 2^{A}\leq 2^{B}$
2. $A\leq B\to 2^{A}\leq 2^{B}$

I definitely prefer version #2, for several reasons (which are admittedly not really disjoint/orthogonal):

• The conclusion has the same form as the hypothesis, which is not only aesthetically more appealing, but also more practical. You can easily stack such inference, and get, for example $A\leq B\to 2^{2^{A}}\leq 2^{2^{B}}$ right away.
• $A\leq B$ is a simple statement: "there is an injection". On the other hand, $A<B$ is a more complicated compound statement: "there is an injection one way, but not the other way".
  (English language, as well as the symbols that we use, seems to indicate that "less than" is the basic concept, and "less than or equal" is the derived one. But in set theory the opposite is true.
• The second version is (slightly) stronger, at least apparently. (Of course they are really equivalent over a very weak base theory.)

--Aleph4 16:45, 17 September 2007 (UTC)

Hmmmm, well we can also get $A<B\to 2^{2^{A}}\leq 2^{2^{B}}$ pretty damn quickly.--Michael C. Price ^talk 19:01, 17 September 2007 (UTC)

Cohen's Proof

I tried to make the discussion accessible to anybody, and to follow Cohen as closely as possible. Unfortunately, I am not the worlds biggest expert on this stuff, and I am still confused on some points. I forgot the details of Shelah's really simple proof that CH is consistent using forcing, and I am still a bit confused about when ccc prevents cardinal collapse. I will try to come back to this, but any informal discussion, I think, is good, because this stuff is usually presented with a lot of logic jargon. Plus, since Cohen passed away, I think his point of view should be preserved. Maybe I didn't do it justice. I don't know. Likebox 02:58, 27 September 2007 (UTC)

There's some good stuff there, but there's some material that I don't think makes an awful lot of sense. The business about geometry at is the most obvious offender -- it does not appear to me that any of the stated results have anything to do with whether there's a wellordering of the reals such that every proper initial segment is countable. They seem to have more to do with coding the reals into the language of set theory, but that doesn't depend on CH. And the thing about $\aleph _{28}$ also seems completely irrelevant; you can consider the 28th iterated powerset of the naturals without needing to know its cardinality as an aleph, and it's not clear why any of the referenced "wonders" would depend on its cardinality as an aleph.

I'm tempted to revert the whole thing for now and suggest that it be discussed more carefully on the talk page; generally I prefer that discussion take place prior to such wholesale revisons of such an important article. I may yet do that; I'm taking it under advisement. --Trovatore 03:58, 27 September 2007 (UTC)

Point well taken, I was "being bold", as they advise. As far as the geometry business is concerned, it is designed to convey the intuition that set theory is bigger than geometry if you accept CH. Because then all of geometry is done with piddling little aleph 1, while super-geometries could be constructed which are as strong as aleph whatever. If you revert, I'll just bring up the points on the talk page- no harm done. Thank you for your consideration. Likebox 05:43, 27 September 2007 (UTC)

OK, so my point here is that "set theory is bigger than geometry" even if CH fails. In fact I don't see that CH has much to do with it -- why would the point be different if $2^{\aleph _{0}}$ were, say, $\aleph _{237}$ or $\aleph _{\epsilon _{0}+\omega ^{2}+3}$? You still have bigger sets around, and by an arbitrary amount. Still, I do think there's a lot worth saving from your additions. I'll try to go over it carefully when I have time. --Trovatore 06:39, 27 September 2007 (UTC)

Hi. I will try to explain better, because I don't think I am saying something controvertial. Every aleph of new cardinality, in completely modern intuition, is a set which is itself a model for all the sets of previous cardinality and so proves the consistency of set theory restricted to those earlier sets. This is a godel step, and allows you to prove new theorems about the integers. The number of "godel" steps is the aleph size, more or less. So where the continuum lies tells you how many godel steps you get by using the continuum. If the continuum is at aleph 1, you get exactly one godel step past countable set theory. That's not very much. If the continuum is at $\aleph _{\epsilon _{0}+\omega ^{2}+3}$ then you get $\epsilon _{0}$ or so godel steps by going to geometry. The point is that when the continuum can embed bigger cardinals, it can prove more theorems about the integers.

Cohen's method shows that whatever logical model you have for sets, you can embed the whole universe inside the continuum, because the continuum can "forcing-wise" embed an arbitrarily large set. So the continuum is behaving more like a proper class than a set. It is so enormous that it can prove any godelization of anything. Another way of saying this is that in a universe with random or generic points, geometry is more powerful than any set theoretic construction. It is more powerful than any axiom system. It is as strong as the strongest large cardinal. I think this point of view, which is expressed by Cohen in his book, should be represented.

The fact that you can intuitively get "larger" sets by going to the powerset of the continuum is a total red herring. Say you are working in a model where the continuum is aleph 2, and the powerset of c is aleph-epsilon. You then get aleph-epsilon strong theorems when you pass to the powerset of the continuum. Big deal. You could have just switched to a forcing model where the continuum was already aleph-epsilon strong and done your theorem there. As far as theorems about integers are concerned, if you don't believe there is any restriction on the size of the continuum, it doesn't pay to go past the continuum. So set theory is not necessarily stronger than geometry. Likebox 18:14, 27 September 2007 (UTC)

Frankly I think you have misunderstood some of this a bit. I won't have time to address it in detail during the working day. But just very quickly, what you seem to be talking about is not so much the alephs as the levels of the von Neumann hierarchy. The reals appear at level $\omega +1$ of that hierarchy, quite irrespectively of whether CH is true. --Trovatore 19:48, 27 September 2007 (UTC)

Hi again. I don't mean to rush you or put pressure, I was just trying to be as clear as possible, becuase I don't know for sure if I am fairly representing Cohen's views. I read about this Von Neumann hierarchy, its not what I'm talking about. I am talking about the ordinal number of the continuum, exactly the question of the continuum hypothesis. The question "what is the power of geometry?" means "what theorems about integers can set theory prove?" The answer depends on how big an ordinal geometry can fit in. If geometry is a small cardinal, then geometry is useless for proving theorems that require large cardinals. If geometry can be an arbitrarily big cardinal, then geometry is more powerful than any set theory. The question hinges on whether you admit "random" points as part of geometry or not. If you admit "random" points, then geometry can embed the structure of an arbitrarily large cardinal by forcing it in. If you don't admit random points, and allow only reals that are constructed by iterated set theoretic constructions you just get L, and then you can't prove large cardinal theorems with geometry, because there is no injection of large cardinals into the plane.132.236.54.79 21:18, 27 September 2007 (UTC)

Finally you've said something precise enough to be wrong :-). ZFC+CH proves exactly the same first-order theorems about the integers that ZFC proves, or that ZFC+¬CH proves, or that ZFC+$2^{\aleph _{0}}=\aleph _{17}$ proves. I'll give more detail later, if you like. --Trovatore 21:22, 27 September 2007 (UTC)

I did not say that ZFC + CH proves less or more theorems. I did not imply it, and I never believed it. I agree that it is obvious that ZFC+CH proves the same theorems by equiconsistency. What I did say was that those same theorems will be proved by using different sets. They would use the same ordinals and cardinals, but the cardinals would correspond to different levels of powerset. What would be proven in one system using powerset(powerset(powerset(Z))) would be proven in a different system using only powerset(Z). Since powerset(Z) is geometry, if you can fit high enough ordinals into geometry you don't need powerset more than once.

If you want me to be formal about it, consider ZFC-P, ZFC without powerset. This has a model which is all countable sets. The theorem "Consis ZFC-P" is a provable statement about the integers in ZFC, since ZFC has an uncountable set, but unprovable in ZFC-P.

But you don't need the full powerset axiom to prove consis ZFC-P. You can prove that with the much weaker axiom "There exists an uncountable set". This axiom, axiom A1, is a large cardinal axiom for ZFC-Power. It asserts the existence of a mysterious large cardinal aleph-1 which is inaccessible by countable operations. Once you have axiom A1, you can prove "consis ZFC-P" by considering everything smaller than aleph-1 as a model. But now consider the following new axiom: "There exists a set which cannot be bijected with aleph-1", this is a second "large" cardinal axiom, axiom A2. ZFC-P+A1+A2 proves the theorem "consis ZFC-P+A1", a theorem in arithmetic which can't be proven in ZFC-P+A1.

Now let me add countable powerset CP, "There exists a powerset for every countable set". I can ask the question, what can I prove in this system? I can prove "Consis ZFC-P", but can I prove consis "ZFC-P+A1"? This exactly depends on whether the continuum hypothesis is true or not. If the continuum hypothesis is true in this universe, then certainly not. If the continuum hypothesis is false in this universe, then yes. In this case, the continuum hypothesis can clearly be assumed true, and the answer to this question is "ZFC+CP" does not prove "consis ZFC-P+A1", but it does prove "consis-ZFC-P". I can also add axiom CPP "There is a powerset both for every countable set and for every powerset of a countable set", then ZFC+CCP will prove "consis-ZFC+P+A1". So I need two iterations of power set to prove this theorem. If I add axiom CPPP "there is power for countable, powers of countable, and powers of powers of countable", then I can prove "consis-ZFC+P+A1+A2", so you need three iterations of powerset to prove that theorem.

But suppose you take ZFC+CPP which proves the existence of aleph-2, and force aleph-2 into the continuum. Now "ZFC+CPP+Forcing" proves the existence of aleph-3, and proves "consis-ZFC+P+A1+A2" which required CPPP before. So with forcing, you only need to use two iterations of powerset to prove something that you needed three iterations before. That's what I was trying to say. I hope this is no longer vague.

The whole point of forcing is that you only need one iteration of powerset to prove all the theorems, because you could force the powerset of Z as high as you like. The real numbers are arbitrarily powerful, so geometry can be bigger than set theory. —Preceding unsigned comment added by Likebox (talk • contribs) 00:39, 28 September 2007 (UTC)

Would not some of the new material (e.g. on "forcing") be suitable for a new article?--Michael C. Price ^talk 06:22, 27 September 2007 (UTC)

There's a long standing article on forcing (mathematics), though admittedly it assumes considerable background. A lot of the rest of the new material also duplicates other articles. Some of it might be used to make those articles more accessible. --Trovatore 06:39, 27 September 2007 (UTC)

Hi, I thought about that, and I wanted to write it on forcing, but I thought that here is more appropriate for these reasons:

1. Because this is only as much forcing as is needed to make the undecidability of the continuum hypothesis obvious.

2. Because I think Cohen's view the the continuum hypothesis is "obviously false" is an intuition that needs to be preserved and explained, and I don't think a discussion of posets clearly explains it. I think that a discussion of axiomatic embeddings of sets into the real numbers does explain it. In particular, it seems to me that Cohen thought of the real numbers as bigger than any set you can think of. So big, that it is useless to compare them with such piddling little trifles as ordinal numbers, no matter what their degree is.

3. It is painful for me to cross reference mathematics pages which are written with heavy use of jargon, because I know that such pages are very difficult for students to read, and for no good reason.132.236.54.79 21:18, 27 September 2007 (UTC)

Regarding point 3, I strongly encourage Likebox (and others) to improve the other articles, especially the forcing article, no matter how painful the process (and it is painful, I know!). For Trovatore to say that the forcing article "assumes considerable background" is a tacit admission that he agrees with Likebox "that such pages are very difficult for students to read". And it sounds like we all have the goal of making "those articles more accessible". I do not possess the background to add much new material, but am willing to try to continue to improve existing material.--Michael C. Price ^talk 21:12, 28 September 2007 (UTC)

Feferman

I've been looking at Solomon Feferman's papers at http://math.stanford.edu/~feferman/papers.html and trying to resist quoting them all over the place, since they're very interesting and I don't know of another such concentrated collection of work on these topics. Anyway he discusses CH in (among other places) Does mathematics need new axioms? (symposium proceedings from 2000 based on an earlier article) starting at page 4:

But the striking thing, despite all such progress, is that--contrary to Gödel's hopes--the Continuum Hypothesis is still completely undecided, in the sense that it is independent of all remotely plausible axioms of infinity, including all "large" large cardinal axioms which have been considered so far.7 In fact, it is consistent with all those axioms--if they are consistent--that the cardinal number of the continuum is anything it "ought" to be, i.e. anything which is not excluded by König's Theorem. ...

My own view--as is widely known--is that the Continuum Hypothesis is what I have called an "inherently vague" statement, and that the continuum itself, or equivalently the power set of the natural numbers, is not a definite mathematical object. Rather, it's a conception we have of the totality of "arbitrary" subsets of the set of natural numbers, a conception that is clear enough for us to ascribe many evident properties to that supposed object (such as the impredicative comprehension axiom scheme) but which cannot be sharpened in any way to determine or fix that object itself. On my view, it follows that the conception of the whole of the cumulative hierarchy, i.e. the transfinitely cumulatively iterated power set operation, is even more so inherently vague, and that one cannot in general speak of what is a fact of the matter under that conception. For example, I deny that it is a fact of the matter whether all projective sets are Lebesgue measurable or have the Baire property, and so on.

What then--on this view--explains the common feeling that set theory is such a coherent and robust subject, that our ordinary set-theoretical intuitions are a reliable guide through it (as in any well accepted part of mathematics), and that thousands of interesting and prima facie important results about sets which we have no reason to doubt have already been established? Well, I think that only shows that in set theory as throughout mathematics, a little bit goes a long way--in other words, that only the crudest features of our conception of the cumulative hierarchy are needed to build a coherent and elaborate body of results. Moreover, one can expect to make steady progress in expanding this body of results, but even so there will always lie beyond this a permanently grey area in which such problems as that of the continuum fall.

I believe some of this should be used in the article but I'm not sure exactly how. Could some editors familiar with what the larger mathematical community thinks of Feferman's work figure out how to place it? 75.62.4.229 (talk) 01:25, 24 November 2007 (UTC)

Judging from your quotation, this statement of his is vague. His assertions are not accepted as true of ZFC by the mathematical community. Indeed, he seems to be suggesting that ZFC be replaced with some other set of axioms which he does not specify. Also, we know from Gödel's incompleteness theorems that no effective and consistent set of axioms can be complete. So the problem he is complaining about is insoluble. JRSpriggs (talk) 02:39, 24 November 2007 (UTC)

I don't think he's saying CH is ill-posed as a problem in the formal system ZFC. Rather, he takes an anti-Platonistic stance towards ZFC itself: i.e., ZFC (unlike PA) is just a formal system and its theorems aren't really meaningful as mathematics outside of the pure study of set theory, and CH is neither true nor false, it's just more formal symbols. He has another article claiming that just about all scientifically applicable mathematics (including most of 20th century functional analysis) can be done in PA or maybe some slight extensions of PA. Anyway his article is pretty readable and maybe my description isn't doing it justice, so it would be great if you could look at it. 75.62.4.229 (talk) 04:28, 24 November 2007 (UTC)

I can't comment on Feferman's work in general, but based on the way that people in my field are using set theory I wouldn't agree with this passage, at least not without some reservation. I am working in classification theory, a part of model theory founded by Saharon Shelah. This is at the intersection of logic and algebra. We study objects like "the theory of all algebraically closed fields of characteristic 0" and assign invariants to them. Most of these invariants are maps Card → Card. Often the number of classes in such a classification is finite, and is smaller under the GCH than under certain strong negations of it. At least one theorem that has nothing to do with set theory was first proved by Shelah assuming a strong negation of GCH. Since this negation was known to be consistent with ZFC, and since the statement was "absolute", this was sufficient to show that the theorem holds just under ZFC.

Most people in this field try to avoid such arguments, but my impression is that nowadays that's a matter of taste rather than philosophy. I should note, however, that many experts consider these set-theoretic invariants a guide to the corresponding elementary notions rather than the real thing. It reminds me of the continued use of Betti numbers for some time after homology theory had made them obsolete. --Hans Adler (talk) 10:51, 24 November 2007 (UTC)

I read the PDF of his paper to which you linked. It is all very metamathematical. I only got one fact from it which I think is usable here. I summarized it in my edit which said "So far, CH appears to be independent of all known Large cardinal axioms in the context of ZFC.". JRSpriggs (talk) 03:54, 25 November 2007 (UTC)

Right, he's a logician whose interest area is arithmetization of metamathematics. I've requested his book "By the light of logic" from my local library so I'll see if I can find some clearer statements about CH in it. His view that CH is a vague problem is clearly a minority viewpoint among mathematicians; what I don't know is whether it's considered far-fringe. Certainly he's a respectable old-time logician (student of Tarski, professor at Stanford, editor in chief of Gödel's collected works, etc.) and as such, if we're trying to describe different viewpoints on CH, I think we should mention his, especially if we can find a concise formulation. Interestingly, I think he quotes Gödel somewhere as also saying that CH could not be settled by LCA. Are any of the logic editors here reading this familiar with his work? I'm a little surprised since I figured he'd be well-known. 75.62.4.229 (talk) 05:27, 25 November 2007 (UTC)

I was aware of him before this. He has written about ordinal numbers which is a special interest of mine. However, the mere fact that he and his work are notable generally is not sufficient to justify including whatever he says into this article. You must justify its relevance here. JRSpriggs (talk) 07:02, 25 November 2007 (UTC)

There's a section of the article listing several viewpoints on CH ("Arguments for and against CH"), and I'm proposing including his viewpoint (which is certainly a minority one) as a sentence or two in that section. So my main question is whether set theorists or logicians have given much attention to his view ("attention" does not necessarily mean "agreement"). If they have, that shows relevance. My secondary quesiton would be how to summarize his argument. 75.62.4.229 (talk) 09:45, 25 November 2007 (UTC)

I am no longer hooked into the community of set theorists well enough to know "whether set theorists or logicians have given much attention to his view". I suggest that you try to summarize his view insofar as it is relevant to that section. If anyone disagrees with your interpretation or its relevance, he is free to change it. JRSpriggs (talk) 00:09, 26 November 2007 (UTC)

Let me suggest this: Feferman is by no means the only logician who thinks CH is not a statement with a well-defined truth value. That may even be the majority opinion. Nor did I see anything in his remarks on CH in the linked paper that I think is particularly unique to him. What is notable about Feferman is that, of the highly-respected commentators, he is one of the most skeptical on strong ontological claims of set theory (or at least, of those highly-respected commentators, is one of the ones most noted for his skepticism, which I suppose is a bit different). But while that distinction might possibly make him a good person to quote as a representative of this view, it shouldn't really be presented as somehow uniquely his; there are a lot of people that think that. (Of course, they're all wrong, but that's not important right now :-).

There are possibly several different skeptical positions that could be mentioned. Right now the article mentions mainly the ZFC-formalist one, which I think is a fairly superficial viewpoint; Feferman's view is at least more substantial than that one. Another one that could possibly be mentioned is the so-called "many worlds" or "plenitudinous Platonist" viewpoint that refuses to discriminate among models of ZFC and figures that, since some satisfy it and some don't, then those must be equally valid possibilities. That's superficially appealing but when you really look into it you see that it's actually the silliest of all the options -- if you're a realist about models then you're actually very close to being forced to admit that some models are better than others and that in fact it's not hard to specify what kind of model has to be correct about CH's truth value. Nevertheless it is the position of some otherwise sensible folks, so it may deserve a mention (would be nice to find a criticism of it too, because it's so eminently criticisable). --Trovatore (talk) 01:41, 26 November 2007 (UTC)

Thanks, I think you've pegged the situation pretty well--I'll defer to you if you want to add something like the above to the article. I've requested a copy of John Dawson's biography of Gödel from my library (along with Feferman's book) and it's likely to have some good material from the realist perspective. I'll probably get both books next week and may try adding something once I've looked at them. 75.62.4.229 (talk) 07:29, 26 November 2007 (UTC)

As the first Hilbert problem...

The first problem consisted of two parts as originally presented in the actual talk by Hilbert. The second part, which was emphasized just as much CH, was to prove Cantors belief that every set can be well ordered. That second part has been answered (not proved, but axiomatized) in the disguise of The Axiom of Choice. YohanN7 (talk) 16:38, 27 November 2007 (UTC)

I just read the translation here again, and although Hilbert does mention the problem of well-ordering the real numbers, the standard understanding is that "Hilbert's first problem" is the continuum hypothesis. See the refs by Foreman and Martin in this article, for example. — Carl (CBM · talk) 17:03, 27 November 2007 (UTC)

I know, that is the standard understanding of Hilberts first problem - and I don´t want to try to alter that. I just wanted to point out that the original formulation did involve two, very deep indeed, questions, both of (CH and to a much lesser extent AC) which are still controversial. YohanN7 (talk) 18:29, 27 November 2007 (UTC)