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Function and subsets - inverses

  1. Aug 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f:X \rightarrow Y[/itex] and [itex]B_1, B_2 \in P(Y)[/itex] where P(Y) is the power set.

    Prove that [itex]f^{-1}(B_1\cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)[/itex]

    2. Relevant equations
    The book gives this definition:
    Suppose [itex]f:X \rightarrow Y[/itex] is a function.

    The function [itex]f^{-1}:P(Y) \rightarrow P(X)[/itex] is defined by [tex]f^{-1}(B)=\begin{cases} x\in X | f(x)\in B \end{cases}\}[/tex] for [itex]B\in P(Y)[/itex]


    3. The attempt at a solution
    All I can do is just basically rewrite the definition:

    Say [itex]y_0 \in B_1\cap B_2[/itex]. Then [tex]f^{-1}(\{y_0\})=\{x \in X ~|~ y_0=f(x)\}[/tex]

    Then I make magic leap concluding that since [itex]y_0 \in B_1\cap B_2[/itex], [itex]y\in B_1[/itex] and [itex]y\in B_2[/itex]. Hence, these sets will be equal and so [itex]f^{-1}(B_1\cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)[/itex] . But I realize that a lot of grammar is missing. Can somebody help me out?
     
  2. jcsd
  3. Aug 3, 2012 #2

    HallsofIvy

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    The "magic leap" can be given in more detail. You are basically trying to prove "[itex]A= B[/itex]". To prove that for sets, first prove "[itex]A\subset B[/itex] and the prove [itex]B\subset A[/itex]. And to prove [itex]A\subset B[/itex], start with "if [itex]x\in A[/itex]" and use the properties of A and B to conclude "then [itex]x\in B[/itex]".

    Here, you want to prove that [itex]f^{-1}(B_1\cap B_2)= f^{-1}(B_1)\cap f^{-1}(B_2)[/itex]. So start "if [itex]x\in f^{-1}(B_1\cap B_2)[/itex]". Then, by definition of [itex]f^{-1}[/itex], there exist [itex]y\in B_1\cap B_2[/itex] such that f(x)= y. Since y is in [itex]B_1\cap B_2[/itex], y is in [itex]B_1[/itex] and y is in [itex]B_2[/itex].
    1) If y is in [itex]B_1[/itex] and f(x)= y, then [itex]x\in f^{-1}(B_1)[/itex].
    2) If y is in [itex]B_2[/itex] and f(x)= y, then [itex]x\in f^{-1}(B_2)[/itex].

    Since both are true, [itex]x\in f^{-1}(B_1)\cap f^{-1}(B_2)[/itex].
    That is, [itex]f^{-1}(B_1\cap B_2)\subset f^{-1}(B_1)\cap f^{-1}(B_2)[/itex].

    Now, to prove the other way, that [itex]f^{-1}(B_1)\cap f^{-1}(B_2)\subset f^{-1}(B_1\cap B_2)/itex], start:
    If [itex]x\in f^{-1}(B_1)\cap f^{-1}(B_2)[/itex] and I will let you try the rest.
     
  4. Aug 4, 2012 #3
    Lets try it:

    Say [itex]x\in f^{-1}(B_1)\cap f^{-1}(B_2)[/itex]. Then there exists [itex]y\in B_1[/itex] and it also [itex]y\in B_2[/itex] such that [itex]y=f(x)[/itex]. Because of [itex]x[/itex] belonging to both sets, and such [itex]y[/itex] existence, it also belongs to [itex]y\in B_1\cap B_2[/itex] such that [itex]y=f(x)[/itex]. Hence [itex]f^{-1}(B_1)\cap f^{-1}(B_2)[/itex] is a subset of [itex]f^{-1}(B_1\cap B_2)[/itex]

    right?
     
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