What is the cardinality of countable union of infinite sets?

Let $X=\underset{n\in\mathbb{N}}{\bigcup} X_n$ and each $X_n$ has infinite cardinality $\alpha$. What is the cardinality of $X$ ? Is this $\aleph_0\alpha$ ?

Solution 1:

This answer uses the Axiom of Choice.

In fact, $|X|=\alpha$. To prove this, we start with the following preparation.

Lemma: $|X|\leq \aleph_0\cdot\alpha$.

Proof. We show this by constructing a corresponding injection. For each $x\in X$, choose some $n(x)\in\mathbb N$ for which $x\in X_{n(x)}$ (this choice can be done without the Axiom of Choice: just take the smallest). Further, for each $n\in\mathbb N$, choose a bijection $f_n:X_n\to\alpha$ (they exist by assumption; here, the Axiom of Choice is required). Now, it is easy to see that the map $X\ni x\mapsto (n(x),f_{n(x)}(x))\in\mathbb N\times\alpha$ is injective, hence the Lemma.$\quad\small\blacksquare$

Proposition: For any ordinals $\lambda\geq\mu$, it holds that $\aleph_\lambda\cdot\aleph_\mu=\aleph_\lambda$.

Proof. This is Corollary 7.2.2 in Introduction to Set Theory by K. Hrbacek and T. Jech.$\quad\small\blacksquare$

We are now ready to prove the initial

Claim: $|X|=\alpha$.

Proof. We wish to prove both inequalities, then apply the Schröder–Bernstein Theorem. Firstly notice that, $\alpha\leq|X|$ (as $X_0\subseteq X$). Conversely, it holds that $|X|\leq \aleph_0\cdot\alpha$ by the Lemma, and $\aleph_0\cdot\alpha\leq \alpha$ by the Proposition, since $\alpha$ is equal to some aleph (by the Axiom of Choice).$\quad\small\blacksquare$