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!

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    $\begingroup$ 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 $n\ne m$. $\endgroup$ – anon Oct 19 '11 at 7:44
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    $\begingroup$ 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 $\mathfrak{c}=\aleph_2$. $\endgroup$ – André Nicolas Oct 19 '11 at 7:45
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    $\begingroup$ [large-cardinal] is a technical term in set theory, it does not mean "cardinals which are very large." $\endgroup$ – Asaf Karagila Oct 19 '11 at 8:18
  • $\begingroup$ 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) $\endgroup$ – Henrique Tyrrell Oct 19 '11 at 8:21
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    $\begingroup$ If any of us might have the winning lottery ticket, it doesn't mean that you might be me. $\endgroup$ – JDH Oct 19 '11 at 13:36

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.

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    $\begingroup$ @ 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.) $\endgroup$ – Dave L. Renfro Oct 19 '11 at 17:02
  • $\begingroup$ @Dave: You are correct of course. I will remedy that now. $\endgroup$ – Asaf Karagila Oct 19 '11 at 19:32
  • $\begingroup$ Thanks for answer, Asaf - insightful as ever. Thanks as well as all other commenters and answerers. $\endgroup$ – barf Oct 19 '11 at 22:40
  • $\begingroup$ @barf: No problem. Are you familiar with concepts such as cofinality? $\endgroup$ – Asaf Karagila Oct 19 '11 at 22:56
  • $\begingroup$ @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. $\endgroup$ – barf Oct 20 '11 at 7:32

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.


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