A proof for the equality $\aleph_0 card(X) = card(X)$ with $X$ an infinite set?

Question

Good evening,

I want to show that all bases of a vector space have the same cardinality, and it needs the following equality : Let $\aleph_0$ be the cardinality of $\mathbb{N}$ and $X$ an infinite set, then $$\aleph_0 \operatorname{card}(X) = \operatorname{card}(X).$$

Does anyone know where there is a proof for this equality?

Thanks in advance,

Duc Anh

Answer 1

We use the fact that $|X\times X|=|X|$, applying Zorn lemma to $S:=\{(B,f), B\subset X, f\colon B\times B\to B,f\mbox{ bijective}\}$ with the partial order $(B_1,f_1)\leq (B_2,f_2)$ if $B_1\subset B_2$ and $f_{2\mid B_1}=f_1$. Then we conclude by Cantor-Bernstein theorem.

Answer 2

Here's a more direct approach; I think something like this must be hidden in the intermediate step of Davide's proof anyway:

First, we need to assume the Axiom of Choice. So $X$ can be well-ordered, and without loss of generality we can assume that $X$ is an initial ordinal. In particular, then, $X$ is a limit ordinal.

One can easily prove that every ordinal can be uniquely written as $\alpha+n$ where $\alpha$ is zero or a limit ordinal and $n$ is finite.

Fix a bijection $f:\mathbb N\times\mathbb N\to \mathbb N$. Then $$ \langle n,\alpha+m\rangle \mapsto \alpha+f(n,m) $$ is a bijection $\mathbb N\times X \to X$.