Properties of the pre-image of a function
From Maths
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Tidy up and then demote
Contents
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:
- 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]
- 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]
- For [ilmath]A,B\in\mathcal{P}(Y)[f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)[/ilmath]
- 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
- [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)
- thus [ilmath]x\in f^{-1}(A)-f^{-1}(B)[/ilmath], by definition of relative complement.
- 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)
- 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])
- [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]
- So [ilmath]f(x)\in A-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)
- 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)
- 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]
