Continuum Hypothesis - Why does my "proof" fail?

Question

While thinking about the Continuum Hypothesis, I stumbled across a way of thinking about it that seems to intuitively make sense to me. But, being that I'm not a mathematician and Gödel/Cohen together have shown that the $\sf{CH}$ is independent of the $\sf{ZFC}$ axioms, I understand that this is $99.999999999999\%$ likely to be wrong. Anyway, here is my thinking, and if possible, I'd like to know where it is that I went awry:

Idea:

The assumption underlying my "proof" is that the number of natural numbers you need to "index" a set corresponds to its cardinality. Therefore, if you have a set $A$ with a cardinality $a$, and a set $B$ with a cardinality $b$, and you need $i$ indices to index $A$, and $j$ indices to index $B$, and you can show that there exists no number of indices between $i$ and $j$ that would generate a different cardinality than either $a$ or $b$, then there must be no cardinalities between $a$ or $b$.

So, let me show you what I mean by indices. The members of the set of natural numbers need only $1$ natural number to index them. Duh. $1$ gets labeled by $1$, $2$ by $2$, etc. So, a set with cardinality $\aleph_0$ only needs $1$ index. We also know that ordered $n$-tuples (with finite $n$) also are countable, requiring only $1$ index to label them. What is the next possible number of indices you could require? $\aleph_0$. The very next possible number of indices that would generate a different cardinality is $\aleph_0$. If you have $\aleph_0$ indices, each index being a natural number, the set of all such objects would basically be an ordered $\aleph_0$-Tuple. The cardinality of such a set is that of the continuum. Therefore, since there is no possible number of indices to label members of sets between finite natural numbers and $\aleph_0$, there are no cardinalities between $\aleph_0$ and the cardinality of the continuum.

Why wouldn't something like this work? It seems like either something outside my assumption is wrong, or my assumption is independent of $\sf{ZFC}$, and it's just one of many possible axioms that $\sf{ZFC}$ could be extended with...

Noah Schweber · Accepted Answer · 2016-09-03T14:15:07.803

27

Your underlying assumption that "everything can be indexed nicely" basically translates to the following:

For every infinite set $X$, there is some set $Y$ such that $X\equiv (\aleph_0)^Y$, that is, the set of "$Y$-tuples" of natural numbers.

This, however, is provably false in ZFC! It turns out that we can show in ZFC using Koenig's Theorem that $(\aleph_0)^Y$ can never have size exactly $\aleph_\omega$.

In general, when reasoning about infinities, you need to be very careful and precise. Informal arguments like what you've written above - which hinge on appeals to intuition (how do you justify that every set can be "indexed" nicely?) - are more likely to confuse you than to help, until you have learned enough set theory to know when to trust them.

Incidentally, it's worth pointing out that we know that ZFC (assuming it's consistent!) can't prove the Continuum Hypothesis: if you give me a model of ZFC, I can produce (via forcing) a model of ZFC in which the Continuum Hypothesis is false (and another in which it is true).

edited Sep 03 '16 at 14:15

answered Aug 29 '16 at 22:54

Noah Schweber

245,398

1
Thanks for the answer. I appreciate it :)
Point to me an example of a way in which you could explicitly construct a set of cardinality $\aleph_\omega$? Or could you show that it's even in principle possible to imagine a set of such objects? It's my hunch that while infinite ordinals are valid, it's really wonky to just declare that an arbitrary class of ordered things automatically must have an omega'th item, simply because there are Aleph Null of them. For example, we know there isn't an Aleph Null'th integer, unless we make one up. Which is what I think about $\aleph_\omega$.

SSD

Aug 29 '16 at 23:17

@SSD This is a consequence of the Axiom of Replacement. Specifically, if I have sets $A_i$ of cardinality $\aleph_i$ respectively, then $\bigcup_{i\in\mathbb{N}} A_i$ has cardinality $\aleph_\omega$ (I don't even need to assume that the $A_i$s are disjoint!). This keeps going - for each ordinal $\alpha$, there is a cardinal $\aleph_\alpha$. Incidentally, it is consistent with ZFC that $2^{\aleph_0}$ (that is, the cardinality of $\mathbb{R}$) is much bigger than $\aleph_\omega$, although it can never be $\aleph_\omega$ exactly. – Noah Schweber Aug 29 '16 at 23:22

2

@SSD Here's an explicit set of cardinality $\aleph_\omega$: the set of ordinals which have cardinality $\aleph_n$ for some $n<\omega$. – Noah Schweber Aug 29 '16 at 23:37

1

Thanks, this is a direct answer to my concern, and I think that you've hit the nail on the head as to what the flaw was in my reasoning. – SSD Aug 30 '16 at 00:36

Just one more quick question - in your original reply - Y can be ANY cardinal number? Such as $\aleph_1$, $\aleph_2$, etc.? – SSD Aug 30 '16 at 00:42

1

@SSD Yes. There is no set at all - of any cardinality - such that $(\aleph_0)^Y\equiv \aleph_\omega$. – Noah Schweber Aug 30 '16 at 01:02

No no no. First you say for Every set X. Clearly there are no finite set X such that X = ( aleph_0 ) ^ Y. ( or card X ). So the basic idea is already wrong. Secondly : aleph_omega is WAY bigger than the continuum so it is NOT related to CH !! In fact it is even too big for a connection to GCH. In other words , I do not accept this answer , not even as an argument !! This might be unpopular but still. – mick Sep 03 '16 at 11:31

@mick As to your comment re: $X$, yes, that was a bit sloppy of me. I was restricting attention to infinite sets $X$, since that's the context of this question; I've edited to make that clear. As to the rest of your comment, I'm sorry, but that's nonsense. It is consistent with ZFC that the continuum is much larger than $\aleph_\omega$. This was proved by Solovay; see also https://en.wikipedia.org/wiki/Easton%27s_theorem (Easton proved a much more general result). I believe you are confusing the $\aleph$ numbers with the $\beth$ numbers. – Noah Schweber Sep 03 '16 at 14:14

@mick Re: the possible value of $2^{\aleph_0}$, see http://mathoverflow.net/questions/218407/how-far-wrong-could-the-continuum-hypothesis-be or http://mathoverflow.net/questions/104519/around-continuum-hypothesis?rq=1 or http://mathoverflow.net/questions/26959/can-the-continuum-be-a-singular-cardinal or the last paragraph of https://en.wikipedia.org/wiki/Continuum_hypothesis#Independence_from_ZFC. Solovay's paper proving the result is hard to find online, but the title is "$2^{\aleph_0}$ can be anything it ought to be" and it was published in the volume "Theory of Models" (1965). – Noah Schweber Sep 03 '16 at 14:23

@NoahSchweber, "Incidentally, it's worth pointing out that we know that ZFC (assuming it's consistent!) can't prove the Continuum Hypothesis." Not quite. We know that ZFC+Con(ZFC) proves the statement "ZFC cannot prove CH." But what if ZFC is consistent, but arithmetically unsound? In this case, "ZFC cannot prove CH" might be false, despite that ZFC+Con(ZFC) proves it. There's an old, poorly written question of mine that talks about this kind of thing. – goblin GONE Sep 03 '16 at 14:23

@goblin No, that's false. Assuming ZFC is consistent, it has countable models; and any countable model (even ill-founded!) of ZFC has a forcing extension in which CH is false, so ZFC can't prove CH. Arithmetical soundness doesn't enter into it at all. – Noah Schweber Sep 03 '16 at 14:25

@NoahSchweber, all that proves is that ZFC proves "ZFC cannot prove CH." The model-theoretic argument is used to derive an arithmetical statement, via the completeness theorem for first-order logic. So arithmetical soundness does, in fact, enter into it. – goblin GONE Sep 03 '16 at 14:27

@goblin That's not really true. The background theory for forcing arguments (done syntactically) is much less than ZFC; PA alone is already more than enough. So the arithmetical soundness of ZFC is just not relevant here. If you want to debate the arithmetical soundness of fragments of PA, then that's relevant; but also a bit silly (to me anyways). This was treated at an old mathoverflow question. – Noah Schweber Sep 03 '16 at 14:29

@NoahSchweber, in the question I linked, I write: "It is tempting to argue that, if we prove the consistency of a first-order theory using a sufficiently weak foundations, then this is more than just evidence of its consistency, it is proof. We might therefore try to reclassify certain beliefs and start calling them theorems. There's at least 2 reasons to fight this temptation." I still agree with this sentiment. – goblin GONE Sep 03 '16 at 14:31

@goblin Fine, but it's very misleading to say "this needs the arithmetical soundness of ZFC!" when in fact all that's needed is arithmetical soundness of a truly tiny fragment of PA (the question I was looking for was this one, specifically Emil's answer). It's simply blatantly false that the arithmetical soundness of ZFC is at all relevant here. – Noah Schweber Sep 03 '16 at 14:33

Sorry, I didn't mean to suggest we move this to chat. I agree that the arithmetical soundness of ZFC isn't really relevant. But its also not true that we know that ZFC, if consistent, cannot prove the continuum hypothesis. What I think you should say is that we have very strong evidence that ZFC, if consistent, cannot prove the continuum hypothesis. And spare a line to mention the formal system needed to derive this. – goblin GONE Sep 03 '16 at 14:38

@goblin Sorry, but I disagree. Do you also ask that every answer on this site about basic number theory be prefaced by "Assuming $PA$ . . ."? That ZFC doesn't prove CH is less controversial (in terms of axioms used) than e.g. the finite version of Ramsey's theorem. Despite being about set theory, it's not any more mysterious, and I think it's silly to single it out. (If you really do want every answer to begin "assuming $PA$," then that's consistent - but I honestly don't see the point, and I don't have that kind of strong finitism.) – Noah Schweber Sep 03 '16 at 16:14

@mick Getting back to your comment, the $\aleph$ numbers and the continuum function's possible behavior can be a source of confusion for beginners in set theory. This comment thread is already too long, but feel free to email me if you have any questions about the sources I linked to! (You can find my email through my website, which is linked to on my profile page.) – Noah Schweber Sep 03 '16 at 16:17

Yes sorry I meant the Beth Numbers. But since the card R = beth_1 , AND it has been proven card R < aleph_omega my opinion is unchanged ! – mick Sep 03 '16 at 23:00

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@mick it has NOT been proven that $\vert \mathbb{R} \vert$ is less than $\aleph_\omega$. You are simply wrong. It is consistent with ZFC that the continuum is $\aleph_{\omega+1} $, for instance. This is a consequence of Solovay result, which you can Google (I gave you citations above). – Noah Schweber Sep 04 '16 at 00:27

Then I have been misinformed. Sorry. – mick Sep 08 '16 at 12:21

score 14 · Answer 2 · answered Aug 29 '16 at 22:56

14

You are right that for any uncountable set, if you try to index it with tuples of natural numbers, you will need at least $\aleph_0$ indices, so your set of all possible tuples you can create has cardinality $\aleph_0^{\aleph_0}=\mathfrak{c}$. But what if you don't actually need every possible tuple to index the elements of your set? That is, maybe if you try to label all the elements of your set with $\aleph_0$-tuples of natural numbers, you can do so, but no matter how you do it, you will have some $\aleph_0$-tuples left over at the end which you haven't used. This would mean that your set actually has cardinality smaller than $\mathfrak{c}$. And yet the set still might be uncountable: just because you can't label all the elements with finite tuples, doesn't necessarily mean that if you use infinite tuples, you will need to use all of the possible tuples.

answered Aug 29 '16 at 22:56

Eric Wofsey

330,363

Hey, sorry to grave dig this, but looking back... is it not the case that even if you didn't need all tuples, as long as you needed infinite tuples, and less than or equal to $\aleph_0$, since there is no infinity less than $\aleph_0$, it follows that your cardinality is still that of the continuum? – SSD Jul 17 '18 at 02:55
No. Why would that follow? The cardinality of the continuum is the cardinality you would get by using $\aleph_0$-tuples with no tuples left over. – Eric Wofsey Jul 17 '18 at 03:13
Because if you have finitely fewer tuples, then you've still got $\aleph_0$ tuples. And if you have infinitely fewer tuples, since there is no infinity less than $\aleph_0$, you must be left with finitely many tuples. – SSD Jul 17 '18 at 03:15
Perhaps a finite example would be more helpful. Let's say you have some set of things, and you want to label it with $n$-tuples of $0$s and $1$s. For instance, if your set has $4$ elements, you can use $n=2$, using the labels $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$. Now suppose that you need tuples of at least length $3$ to label all the elements of your set. Does that mean that your set has to have $2^3=8$ elements? No, it might have fewer--specifically, it could have only $5,6,$ or $7$. – Eric Wofsey Jul 17 '18 at 03:16
Right, but that infinite set of tuples needed must have no fewer than $\aleph_0$ members, since there is no infinity less than $\aleph_0$. And it can't have anymore either obviously. So by a sort of squeeze theorem logic, it must have $\aleph_0$ members. So this set of tuples must result in a set of cardinality continuum, since any set that requires $\aleph_0$ labels will have such a cardinality. – SSD Jul 17 '18 at 03:19
Your argument applied to my finite example would say that a set with $5$ elements must actually have $8$ elements, because we can't label it with $2$-tuples but we can label it with $3$-tuples. After all, there's no (cardinal) number between $2$ and $3$! – Eric Wofsey Jul 17 '18 at 03:21
Let us continue this discussion in chat. – SSD Jul 17 '18 at 03:22

Continuum Hypothesis - Why does my "proof" fail?

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