Inequality about infinite cardinals

Question

The following is a question I'm interested in solving:

For an infinite cardinal $\kappa$, $\aleph_0 \leq 2^{2^\kappa}$

The idea is to solve it without using AC. As @Asaf Karagila pointed out in the comment section:

"Even if $\kappa$ is an infinite cardinal, you can't assume that $\kappa \geq \omega$ without using AC".

I am still trying to think this through: My reasoning being, that as $\kappa > n \forall n \in \omega$, then $\kappa \geq \sup\{n:n\in \omega\}=\omega$ as $\omega$ is a limit ordinal.

Where does my reasoning fail? And also, a detailed solution would be very appreciated.

Answer 1

Let's look at something else which we can prove by induction: For every $n$ there is a decreasing sequence of natural numbers of length $n$.

Given a sequence of length $n$, shift it all by $1$, and obtain a sequence of length $n+1$. Great!

Does that mean that there is an infinite decreasing sequence of natural numbers? No. Because for different $n$'s, the decreasing sequence is different, and we cannot stitch them together.

The induction only proved that we can find arbitrarily large finite sequences. Not an infinite sequence.

Similarly here, the induction proof only lets you find arbitrarily large finite sets. It does not let you combine them into a countably infinite set, nor it even lets you find a sequence of sets of arbitrary sizes.

Answer 2

While we can prove that $\kappa>n$ for all $n\in\omega$ by induction, that induction doesn't carry us any further forward. It takes a weak version of $\mathsf{AC}$ to say that every infinite set has a countably-infinite subset.

Remember that $\kappa>n$ means that there is an injection from $n$ into $\kappa.$ In fact, there are infinitely-many such injections for any such $n,$ if $\kappa$ is an infinite cardinal. One might think we could use injections $n\hookrightarrow\kappa$ to somehow build an injection $\omega\hookrightarrow\kappa,$ but in order to do so, we must make countably-infinitely-many choices. Can't do that without some Choice, since all we know about $\kappa$ is that it is infinite.