Bijection between sets and complements

Question

Suppose $f: A\rightarrow B$ is a bijection. For $A,B \subseteq C$. Show that a bijective map $h: C\setminus A \rightarrow C\setminus B$ exists.

I'm not sure how to proceed, may I have a hint please?

Answer 1

Suppose $C$ to be a finite set:

To be a bijection $|A|=|B|$ and since $ A,B \subseteq C \Longrightarrow |C|-|A|=|C_1|=|C|-|B|=|C_2|$

$|C\setminus A| = |C_1| = |C_2| = |C\setminus B|$

So there is indeed the option to map $h: C\setminus A \longrightarrow C\setminus B$ , and the option $h^{-1}:C\setminus B \longrightarrow C\setminus A$

Answer 2

Based on the comments I think the question has been answered collectively by fleablood, CoffeeArabica, and me.

If $C$ is finite, the conjecture is true. To show that a bijection exists between $C \setminus A$ and $C \setminus B$, it suffices to show the sets have the same cardinality. Since $A$ and $B$ are bijective, $|A| = |B|$. Therefore $$ \left|C \setminus A\right| = |C| - |A| = |C| - |B| = \left|C \setminus B\right| $$

If $C$ is infinite, the conjecture is false. For a counterexample, let $A = 2\mathbb{Z}$ and $B=C=\mathbb{Z}$. Let $f \colon A \to B$ be the map $x \mapsto \frac{1}{2}x$, a bijection. Then $C \setminus A$ is the set of odd integers (which is infinite), and $C \setminus B$ is the empty set. Therefore a bijection between these complements is impossible.