Cardinality of Cartesian product of Binary Sequences

Question

Let $S = \{ s\mid s\text{ is an infinite binary sequence}\}$ be the set of all infinite binary sequences.

By Cantor's Diagonal Argument, $S$ is uncountable.

But does $|S \times S| = |S|$ also hold?

Answer 1

Yes. Simply separate each sequence into two sequences. If $s$ is a sequence, then we can decompose it to $u$ and $t$ such that $u(n)=s(2n)$ and $t(n)=s(2n+1)$. Namely, taking the even coordinates as the sequence $u$ and the odd coordinates as the sequence $t$.

It is not hard to check that the map $s\mapsto(u,t)$ as defined above is indeed a bijection between $S$ and $S\times S$.