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I am attempting to solve this following problem, and am having difficulty understanding. The question is as follows:

Let $A$ be a set. Let $\phi: \mathcal{P}(A) \to \mathcal{P}(A)$ be defined by $\phi(X)=A-X$ for all $x \in \mathcal{P}(A)$. Prove that $\phi$ is bijective.

I understand that I must show that $\phi$ is injective and surjective, which obviously requires two parts. However, I do not understand how this is done. Can somebody please help me? Merci.

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Sorry, the function is $\phi:\mathcal{P}(A)→\mathcal{P}(A)$, not $\phi :\mathcal{P}(A)→\mathcal{P}(B)$. Original edited to reflect this. – user106251 Nov 7 '13 at 21:10


  • In this particular example the function is an inverse of itself, i.e. $$(\phi \circ \phi)(X) = X$$ because $X \subseteq A$, that is, $$A - (A - X) = A \cap (A \cap X^c)^c = A \cap (A^c \cup X) = (A \cap A^c) \cup (A \cap X) = A \cap X = X.$$
  • If the function is an inverse of itself, it has to be a bijection.

I hope this helps $\ddot\smile$

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