# Properties of the pre-image of a function

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Tidy up and then demote

## Statement

Let [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be sets and let [ilmath]f:X\rightarrow Y[/ilmath] be a function between them. Then:

1. For [ilmath]\{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)][/ilmath]
2. For [ilmath]\{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)][/ilmath]
3. For [ilmath]A,B\in\mathcal{P}(Y)[f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)[/ilmath]
4. For [ilmath]A\in\mathcal{P}(Y)[f^{-1}(Y-A)=X-f^{-1}(A)][/ilmath] - corollary to 3

## Proof

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These are easy and routine

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### 3

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