# Let $\alpha, \beta, \gamma$ be cardinals, $\beta \leq \gamma$, prove $\alpha ^{\beta}\le \alpha ^{\gamma}$

Let $|A|=\alpha, |B|=\beta, |C|= \gamma$ be cardinals and $\beta \leq \gamma$. Prove $\alpha ^{\beta}\le \alpha ^{\gamma}$.

So from the given we know that there's an injection $f:B\to C$ and some functions $h:B\to A, g: C\to A$. We want to prove there's an injection $l_1:A\to C$. It appears that $f$ doesn't help here.

Trying to take representatives from $A$ and show they're in $C$ and there's an injection doesn't work so maybe the function should be $l_2: h \to g$ but I don't know how to work with it.

• It suffices to show that $\left|A^B\right| \leq \left| A^C \right|$ – Omnomnomnom Jul 2 '14 at 16:33
• @Omnomnomnom that's basically the same as $\alpha ^{\beta}\le \alpha ^{\gamma}$. – shinzou Jul 2 '14 at 16:45
• Note that you have to assume $\alpha > 0$ because $0^0 > 0^1$. – Mark Mar 7 at 1:52

Given the injection $f$, for each function $h: B \to A$ you can associate a $g(y): C \to A$ by $g(f(x))=h(x)$ if $y \in f(B),$ otherwise $g(y)=$something in $A$ Since $f$ is an injection, the $g$'s will be distinct whenever the $h$'s are.