Proving $|S \times \mathbb{N}|=|S|$ when $S$ is an infinite set I am trying to prove that, given a collection $\{S_i:i \in \mathbb{N}\}$ of infinite sets which all have the same cardinality, that $|S_1 \cup S_2 \cup \cdots|=|S_1|$. WLOG, I'm going to assume the $S_i$ are disjoint.
My thought was to let each $f_i:S_1 \to S_i$ be a bijection and to define $g: S_1 \times \mathbb{N} \to S_1 \cup S_2 \cup \cdots$ by $g(x,n)=f_n(x)$. I see that $g$ is a bijection. 
So, the problem is reduced to "merely" proving that $|S_1 \times \mathbb{N}|=|S_1|$. I say "merely" with tongue in cheek because it is by no means obvious to me how to do it.
If $S_1$ is countable, I know how to proceed by the standard diagonal argument used to show $\mathbb{Q}$ is countable. But if $S_1$ is uncountable, I don't know what to do, even though it seems intuitive to me that crossing $S_1$ with $\mathbb{N}$ won't make the set "bigger" since it doesn't make it bigger in the countable case.
 A: There may be far simpler ways of doing this, but here's an idea. You could use the following theorem (which uses Zorn's lemma),

Theorem. If $S$ is an infinite set, there exists a set $I$ and a collection $\{S_i\}_{i \in I}$ of disjoint, countably-infinite subsets of $S$ such that 
  $$
S = \bigsqcup_{i \in I}S_i.
$$

A proof can be found in Kaplansky's Set Theory and Metric Spaces, Theorem 12, p.40. The idea of the proof is to consider the set of collections of countable disjoint subsets of $S$ with a certain order, so that (via Zorn's lemma) the maximal element of this set behaves as desired.
Now, by hypothesis we know that $S_1 = \bigsqcup_{j \in J}S_j$ with each $S_j$ a countable subset. Thus,
$$
S_1 \times \mathbb{N} = \bigsqcup_{j \in J} S_j \times \mathbb{N}
$$
Since $S_j$ is countable, we have $S_j \times \mathbb{N} \simeq \mathbb{N}^2 \simeq \mathbb{N} \simeq S_j$ where $\simeq$ denotes the existence of a bijection. This shows that
$$
S_1 \times \mathbb{N} = \bigsqcup_{j \in J} S_j \times \mathbb{N} \simeq \bigsqcup_{j \in J} S_j \simeq S_1.
$$
