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I've been a bit confused whilst doing some reading. I think the confusion arises because I am trying to read Wikipedia on topics without being able to work along. Anyway,

I have read that, of course, $\mathfrak{c} = \aleph_1$ and $\mathfrak{c} \neq \aleph_1$ are both consistent with $\mathsf{ZFC}.$ However, I have also read on wikipedia that $\mathfrak{c} = \aleph_n$ is consistent for all $n\in \mathbf{N}.$

Does this not yield the absurdity that

$$\aleph_n=\aleph_m \text{ for all } n,m \in \mathbf{N}$$

is also consistent with $\mathsf{ZFC}$?

I understand that, of course, if $\mathfrak{c}$ were equal to any fixed cardinal number, it could not be equal to another, because $\mathsf{ZFC}$ is consistent. However, this seems like almost circular reasoning given the undecidability of CH. I am interested in getting an explanation why there is not a problem here.

I appreciate some clarification!

Asaf Karagila
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barf
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    No. You only tack on one of the equations $\mathfrak{c}=\aleph_n$ to $\mathsf{ZFC}$ at a time, not more than one, so you can never say $\aleph_n=\aleph_m$ for any $n\ne m$. – anon Oct 19 '11 at 07:44
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    For any fixed $n \in \mathbb{N}$, if ZFC is consistent, so is ZFC plus $\mathfrak{c}=\aleph_n$. You are certainly right that we cannot have simultaneously $\mathfrak{c}=\aleph_1$ and $\mathfrak{c}=\aleph_2$. – André Nicolas Oct 19 '11 at 07:45
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    [large-cardinal] is a technical term in set theory, it does not mean "cardinals which are very large." – Asaf Karagila Oct 19 '11 at 08:18
  • Also, we don't know if ZFC is consistent. It's not provable in ZFC :/ ZFC is consistent with (ZFC is inconsistent) ZFC is also consistent with (ZFC is consistent) – Henrique Tyrrell Oct 19 '11 at 08:21
  • @Henrique: The assertion $\psi$ is consistent with $ZFC$ is the same as to say $Con(ZFC)\rightarrow Con(ZFC+\psi)$. – Asaf Karagila Oct 19 '11 at 08:25
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    If any of us might have the winning lottery ticket, it doesn't mean that you might be me. – JDH Oct 19 '11 at 13:36
  • @Asaf, you should say relatively consistent in that case. I take "$\psi$ is consistent with ZFC" to mean the same as Con(ZFC+$\psi$). – JDH Oct 19 '11 at 15:02
  • @JDH: While I very much agree that in a book or a paper one should use "relatively" or something similar to ensure no misunderstand about the consistency of ZFC rises, I do know from my experience that in the every day discussions it can be fine to be slightly less accurate and say "consistent with ZFC". However, this may very well be the translation to Hebrew which makes the lack of accuracy more apparent. – Asaf Karagila Oct 19 '11 at 15:40

2 Answers2

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You mixed the quantifiers. Assuming the consistency of $ZFC$, it is consistent that for every $n>0$ there is a model of $ZFC$ such that $\frak c=\aleph_n$.

In fact, the result is even better. Let us introduce a new term:

We say that an ordinal $\alpha$ has cofinality $\omega$ ($\omega=\aleph_0$, the set of natural numbers) if we can find an increasing sequence $\langle\alpha_n\mid n<\omega\rangle$ such that $\alpha_n<\alpha$ and $\sup_n\alpha_n=\alpha$.

For example, $\omega_1+\omega$ has cofinality $\omega$, simply by $\alpha_n=\omega_1+n$. However $\omega_1$ does not have cofinality $\omega$ since a countable union of countably ordinals is countable.

Theorem: Let $\alpha$ be a finite number or an ordinal whose cofinality is not $\omega$, it is consistent that $\frak c=\aleph_\alpha$.

(Such result is proved through forcing. I will not get into the proof.)

Using another theorem we also have that this is pretty much all there is to say about this problem.

Theorem: The continuum does not have cofinality $\omega$.

This is of course a minor corollary from a much more general case, however it gives us that if $\alpha$ is a finite number, or does not have cofinality $\omega$ then it is possible that $\frak c = \aleph_\alpha$.

So to your original question: For all $n$ it is consistent with $ZFC$ that $\frak c=\aleph_n$. It does not mean that it is consistent with $ZFC$ that for all $n$, $\frak c=\aleph_n$.

From the above theorems, we have that $\aleph_\omega$, the first cardinal which is bigger than all the $\aleph_n$ cannot be $\frak c$. So not every uncountable cardinal can have the cardinality of the continuum.

Asaf Karagila
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    @ Asaf Karagila: The following in your answer is not (presently) correct: "From the above theorem, we have that $\aleph_\omega$, the first cardinal which is not $\aleph_n$ cannot be $\frak c$." As stated, the theorem only tells us which alephs can be $\frak c$. The theorem (as stated) tells us nothing about which alephs can't be $\frak c$. (Of course, the assertion "$\aleph_\omega$ is the least uncountable cardinal that cannot be $\frak c$" is correct.) – Dave L. Renfro Oct 19 '11 at 17:02
  • @Dave: You are correct of course. I will remedy that now. – Asaf Karagila Oct 19 '11 at 19:32
  • Thanks for answer, Asaf - insightful as ever. Thanks as well as all other commenters and answerers. – barf Oct 19 '11 at 22:40
  • @barf: No problem. Are you familiar with concepts such as cofinality? – Asaf Karagila Oct 19 '11 at 22:56
  • @Asaf No, I'm afraid not. I am an undergrad and I'm now taking upper division courses (split with grad students) in real analysis, abstract algebra, etc. But I don't think higher set theory will be accessible to me, at least not from a lecture type class, for a few years. I'd really welcome better reading material than Wikipedia (sloppy), Mathworld (quite sparse), and the Arxiv (far too technical). Most all i know of sets is the basic algebra of sets needed for analysis, point-set topology, etc. – barf Oct 20 '11 at 07:32
  • @barf: Well, I wrote more than a few answers on this site defining ordinals, cardinals, cofinality, and so on. – Asaf Karagila Oct 20 '11 at 08:11
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The answer to your question is no, here's why. Even if it's true for each $n \in \mathbb N$ the theory $\textbf{ZFC}+\mathfrak c = \aleph_n$ is consistent, it's not true that for any pair $n,m \in \mathbb N$ the theory $\mathbf{ZFC}+\mathfrak c = \aleph_n+\mathfrak c = \aleph_m$ is consistent, indeed this is consistent if and only if $n=m$. The reason for that is that in ZFC you can prove that for each pair $n,m \in \mathbb N$ you have $\aleph_n=\aleph_m$ just when $n=m$, if add those two axiom above to ZFC with $n \ne m$ then you get an inconsistent theory.

Hope this could help you.

Giorgio Mossa
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