Cardinal arithmetic: $\aleph_2^{\aleph_0}$ = max( $\aleph_2$ , $2^{\aleph_0}$). For those interested this is Exercise 4.5 from Ernest Schimmerling's "A couse in set theory".
What i want to prove is $\aleph_2^{\aleph_0}$ = max( $\aleph_2$ , $2^{\aleph_0}$).
The inequality max(...) $\le$ $\aleph_2^{\aleph_0}$ is pretty straightforward.
Now for the other inequality.
If $\aleph_2$ $\le$ $2^{\aleph_0}$ the result follows easily.
My problem is if $2^{\aleph_0}$  < $\aleph_2$. It follows that $2^{\aleph_0}$ = $\aleph_1$, ie, we're assuming the continuum hypothesis. Is it possible to prove, assuming this, that $2^{\aleph_1}$ = $\aleph_2$ ? The remaining inequality would follow easily.
Help and thanks!
 A: No, ZFC+CH does not prove that $2^{\aleph_1}=\aleph_2$. In fact, beyond the obvious there are no restrictions on how the continuum function on regular cardinals can behave in ZFC, e.g. $$2^{\aleph_0}=\aleph_1,\quad 2^{\aleph_1}=\aleph_{17}, \quad 2^{\aleph_2}=\aleph_{17},\quad 2^{\aleph_3}=\aleph_{\omega^2\cdot 14 + 159}$$ is totally possible.
EDIT: singular cardinals are weird, and there are extremely interesting rules for how the continuum function behaves on singular cardinals. E.g. it turns out that if $\aleph_\omega$ is a strong limit cardinal (that is, $2^{\aleph_n}<\aleph_\omega$ for all $n<\omega$), then we have $2^{\aleph_\omega}<\aleph_{\omega_4}$. This is a result of pcf theory, due to Shelah, and it is still fairly mysterious - Shelah is on record as asking "Why the hell is it four?".
Instead, note that any map from $\omega$ to $\omega_2$ (= any element of $\aleph_2^{\aleph_0}$ - it's a bit easier to think in terms of ordinals, here) has range bounded by some element of $\omega_2$ (why? think about cofinality). Anything $<\omega_2$ has cardinality $\omega_1$, and there are $\omega_2$-many things $<\omega_2$; so we have $$\aleph_2^{\aleph_0}=\aleph_2\cdot(\aleph_1^{\aleph_0})$$ (fill in the steps here). But assuming CH, $\aleph_1^{\aleph_0}=\aleph_1$ (why?).
