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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!

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    @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.2011-10-19

2 Answers 2

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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.

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    @barf: Well, I wrote more than a few answers on this site defining ordinals, cardinals, cofinality, and so on.2011-10-20
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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.