an elementary cardinal inequality

Question

if A,B,C,D are cardinal numbers and A$\le$B , C$\le$D prove that : $A^C \le B^D$
i know that there exists a 1-1 function from A to B and C to D, and i have to declare a function which takes every function from C to A to a function from D to B ,
can anyone help?

Answer 1

You will need that $C\neq\varnothing$ or that $D=\varnothing$ for this to hold generally. The reason being that if $C=\varnothing$, then the empty function is a function from $\varnothing$ to $A$, making $A^C=1$. If also $D\neq\varnothing$ and $B=\varnothing$, then there exists no function from $B$ to $D$, thus $B^D=0$.

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Consider the case where $B=\varnothing$ separately from $B\neq\varnothing$.

If $B\neq\varnothing$, fix some $x\in B$. When you take a function $f:C\to A$, and you have injections $\pi:A\to B$ and $\sigma:C\to D$, use these injections to define a function $f':\sigma(C)\to \pi(A)$, which is a partial function from $D$ to $B$. To make the function total, extend the domain of $f'$ by sending any $d\in D\setminus \sigma(C)$ to your fixed element $x\in B$.

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Secondly, if $B=\varnothing$, then also $A=\varnothing$. We have two cases: if $D=\varnothing$, then the empty function is a function from $D\to B$. Since $D=\varnothing$ implies $C=\varnothing$, we see that $A^C=B^D=1$.

On the other hand, if $D\neq\varnothing$, then $B^D=0$, since there exists no function with nonempty domain and empty range. Here we use that $C\neq \varnothing$ as well, so that also $A^C=0$.