Given any (not necessarily finite) set $X$ as well as arbitrary $A,B\subseteq X$ is it true that $|A|=|B|\implies |X\setminus A|=|X\setminus B|$?

Question

Given any (not necessarily finite) set $X$ as well as arbitrary subsets $A\subseteq X$ and $B\subseteq X$ if we know there exists a bijection $f:A\to B$ does this imply there exists a bijection $g:X\setminus A\to X\setminus B$?

I'm some what embarrassed to ask as it seems trivially true, though I am not totally certain.

Answer 1

Consider $A = \mathbb{R}$ and $B=\mathbb{R}\setminus\{0\}$ and $X=\mathbb{R}$. Then $|A|=|B|$, but $|X\setminus A|=0 \neq 1 = |X \setminus B|$.

Answer 2

Take $\mathbb{R}$, $A = [0, 1]$ $B = \mathbb{R} \setminus\{0\}$. Then $|A| = |B|$, but $|X \setminus B| = 1$ and $|X \setminus A| = |\mathbb{R}|$.