I need to prove that $\{x\}^A \approx A$, that is, that the set of all functions from a set with a single element (namely, x) to another set A is equipotent to the set A.
Now the reasoning behind why this is true is clear: say A had only one element $y_0$, then the only function from $\{x\}$ to A is the function $f(x) = y_0$. Any other function would either not be a function (because $x$ would equal two different $f(x)$) or not belong in $\{x\}^A$ (because it wouldn't land in A). If A has two elements $y_0$ and $y_1$, then the only distinct functions from $\{x\}$ to A are $f(x) = y_0$ and $f(x) = y_1$. Clearly both sets always have the same cardinality, as the constant functions are always in $\{x\}^A$ . My problem is I don't know how to write this as a formal proof.
Previously I proved two sets were equipotent by finding a bijective function between them. But since in this case one of the sets is a set of functions, I don't know how to define a function whose parameter is another arbitrary function. I also thought to do it by induction (suppose cardinality of A to be n and prove it works for n+1) but I don't feel its the right way.
I also thought to do it by transitivity of equipotence, but since A can be any set, I'm not sure how to formulate this. Finally, I thought about contradiction, supposing them not to be equipotent, but I don't see what the contradiction will be.
Any hint in the direction to start the proof would be appreciated. Thanks!
Usually, the set of all functions from a set $B$ to another set $C$ is denoted by $C^B$. Anyway, yes, the natural way to show this is finding a bijection between those sets, doesn't matter that the elements of one of them are functions.
For $a \in A$, define the function $f(a) : \{x\} \to A$ by $f(a)(x) = a$. So, $f$ is in fact a function from $A$ to $A^{\{x\}}$, and this is the desired bijection.