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

Question

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 $\left|A^B\right| \leq \left| A^C \right|$ — Ben Grossmann, Jul 02 '14 at 16:33
@Omnomnomnom that's basically the same as $\alpha ^{\beta}\le \alpha ^{\gamma}$. — shinzou, Jul 02 '14 at 16:45
Note that you have to assume $\alpha > 0$ because $0^0 > 0^1$. — Mark, Mar 07 '19 at 01:52

score 2 · Accepted Answer · edited Jan 24 '23 at 14:33

Given the injection $f:B\to C$ and some $a_0\in A$, each function $h: B \to A$ can be associated with a function $g : C \to A$ defined so that $g(y)=h(f^{-1}(y))$ whenever $y\in f[B]$, and otherwise $g(y)=a_0$. Since $f$ is an injection, the $g$'s will be distinct whenever the $h$'s are, so the map $h\mapsto g$ is an injection $A^B\to A^C$.

Notice the above proof assumes there's some $a_0\in A$, so it assumes $\alpha\neq 0$. This is important, because the case where $\alpha=\beta=0\neq \gamma$ will have $\alpha^\beta=1>0=\alpha^\gamma$, which contradicts the assertion we're trying to prove. That's the only exception however: in case $\alpha=0\neq \beta$ then $\alpha^\beta=0\leq \alpha^\gamma$, and in case $\alpha=\beta=\gamma=0$ then $\alpha^\beta=1=\alpha^\gamma$.

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

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