If \(A\) and \(B\) are finite, show that the set of all functions \(f: A \to B\) is finite.
Solution
Let \(|A|=n\) and \(|B|=m\). After choosing a bijection \[ A\cong\{1,\dots,n\}, \] a function \(f:A\to B\) is uniquely determined by the \(n\)-tuple \[ (f(1),\dots,f(n))\in B^n. \] Thus the set \(B^A\) of all functions \(A\to B\) is in bijection with \(B^n\). A finite cartesian product of finite sets is finite, so \(B^n\), and hence \(B^A\), is finite.
More precisely, when \(n,m>0\), \[ |B^A|=m^n. \] If \(A=\varnothing\), there is exactly one function \(A\to B\); if \(B=\varnothing\) and \(A\ne\varnothing\), there are none. These cases are finite as well.