Implications of continuum hypothesis and consistency of ZFC

Question

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!

Answer 1

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.

Answer 2

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.