In a model solution it is stated that the cardinality of a set which is the countable union of sets of cardinality $2^{\aleph\_0}$ is $\aleph\_0\times 2^{\aleph\_0}$, and, **using the Axiom of Choice**, $\aleph\_0\times 2^{\aleph\_0} = 2^{\aleph\_0}$.

But, isn't the Cantor/Shroeder-Bernstein Theorem enough? I mean, given this:

$2^{\aleph\_0}\le \aleph\_0\times 2^{\aleph\_0}\le 2^{\aleph\_0}\times 2^{\aleph\_0} = 2^{\aleph\_0+\aleph\_0} = 2^{\aleph\_0}$

**EDIT**: so, the set in question is the set of all real-valued sequences $s = (s\_0,s\_1,\dots)$ such that for all but finitely many $n$, $s\_n = 0$. And so I see that we have to get a countable union of sets of size $|\Bbb{R}^n|=2^{\aleph\_0}$ for each $n$. Is there an (hand-wavy should be fine) way to set up the bijection needed with AC?