Show that 2S = S for all infinite sets I am a little ashamed to ask such a simple question here, but how can I prove that for any infinite set, 2S (two copies of the same set) has the same cardinality as S? I can do this for the naturals and reals but do not know how to extend this to higher cardinalities.
 A: In order to prove this is true for all infinite sets you have to use the axiom of choice in one way or another (or at least a fragment of it). And indeed we cannot prove that in naive set theory.
If you wish to prove that for "known" infinite sets, e.g. $\Bbb R$, then this can be done without the axiom of choice indeed.
Using the axiom of choice, $S$ can be put in bijection with an ordinal $\delta$ - which we will only require to be a limit ordinal. Now we can easily define the map from $2\times\delta$ into $\delta$ by the following injection, for $\beta$ limit ordinal, $n\in\omega$ and $i\in\{0,1\}$ we define $$(i,\beta+n)\mapsto\beta+2n+i.$$ This is an injection, since given $(i,\beta+n)$ and $(i',\beta'+n')$ if $i\neq i'$ then clearly the results differ, if $i=i'$ and $n\neq n'$ the result must again differ, and similarly for $\beta\neq\beta'$. So two pairs are mapped to two different pairs.
The above map is in fact a bijection (given $\gamma<\delta$ we can "decode" the pair that was mapped to it), but if one finds the proof of surjectivity any less than immediate, then we can define an injection from $\delta$ into $2\times\delta$ simply by $\alpha\mapsto(0,\alpha)$. Using the Cantor-Bernstein theorem we have that $|\delta|=2|\delta|$, and since $\delta$ and $S$ have the same cardinality we finish.
