Properties of the pre-image of a function

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Statement

Let $X$ and $Y$ be sets and let $f:X\rightarrow Y$ be a function between them. Then:

1. For $\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{P}(Y)[f^{-1}(\bigcup_{\alpha\in I}A_\alpha)=\bigcup_{\alpha\in I}f^{-1}(A_\alpha)]$
2. For $\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{P}(Y)[f^{-1}(\bigcap_{\alpha\in I}A_\alpha)=\bigcap_{\alpha\in I}f^{-1}(A_\alpha)]$
3. For $A,B\in\mathcal{P}(Y)[f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$
4. For $A\in\mathcal{P}(Y)[f^{-1}(Y-A)=X-f^{-1}(A)]$ - corollary to 3

Proof

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These are easy and routine

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3

1. $f^{-1}(A-B)\subseteq f^{-1}(A)-f^{-1}(B)$ (we use the implies-subset relation to see this is equivalent to $\forall x\in f^{-1}(A-B)[x\in f^{-1}(A)-f^{-1}(B)]$
   • Let $x\in f^{-1}(A-B)$ be given, then $f(x)\in A$ and $f(x)\notin B$ (as if $f(x)\in B$ then $f(x)\notin A-B$ so $x\notin f^{-1}(A-B)$)
     • so $x\in f^{-1}(A)$ and $x\notin f^{-1}(B)$ (as if $x\in f^{-1}(B)$, then $f(x)\in B$, which we've established is not the case)
       • thus $x\in f^{-1}(A)-f^{-1}(B)$, by definition of relative complement.
2. $f^{-1}(A)-f^{-1}(B)\subseteq f^{-1}(A-B)$ (we use the implies-subset relation to see this is equivalent to $\forall x\in f^{-1}(A)-f^{-1}(B)[x\in f^{-1}(A-B)]$
   • Let $x\in f^{-1}(A)-f^{-1}(B)$ be given. Then $x\in f^{-1}(A)$ and $x\notin f^{-1}(B)$ (by definition of relative complement)
     • Then $f(x)\in A$ and $f(x)\notin B$ (as if $f(x)\in B$ then $x\in f^{-1}(B)$ which we've established is not the case)
       • So $f(x)\in A-B$ (by definition of relative complement)
         • thus $x\in f^{-1}(A-B)$
3. We combine $f^{-1}(A)-f^{-1}(B)\subseteq f^{-1}(A-B)$ and $f^{-1}(A-B)\subseteq f^{-1}(A)-f^{-1}(B)$ to see:
   • $f^{-1}(A)-f^{-1}(B)=f^{-1}(A-B)$