I'm interested to know if the following statement is true:
If $A,B \subseteq X$ are equinumerous (i.e. $|A|=|B|$, or there is a bijection $A \to B$), then $X \setminus A$ and $X \setminus B$ are equinumerous as well.
This seems true even for infinite sets, but I couldn't prove it in the infinite case. I tried constructing a bijection $f:X \setminus A \to X \setminus B$ explicitly using the bijection $ A \to B$, and came up with an incomplete piecewise function
$$f(x)=\begin{cases} x & x \in X\setminus(A \cup B) \\? & x \in B \end{cases}, $$ but this doesn't seem to work. My question is, is there a nice way of proving this at all?
Thank you!