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Proving equivalence for a collection of subsets?

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f:X→Y where X and Y are sets. Prove that if {S[itex]\alpha[/itex]}[itex]\alpha\in[/itex]I is a collection of subsets of Y, then f-1([itex]\cup[/itex][itex]\alpha\in[/itex]IS[itex]\alpha[/itex])=[itex]\cup[/itex][itex]\alpha\in[/itex]If-1(S[itex]\alpha[/itex])


    Would I prove this by showing set inclusion both ways? And any hints on how to begin?

    Thanks.
     
  2. jcsd
  3. Dec 8, 2012 #2

    SammyS

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    OK. So if [itex]\displaystyle x\in f^{-1}\left(\cup_{\alpha\in I}\,S_\alpha \right)[/itex], what does that say?
     
  4. Dec 8, 2012 #3
    That f(x) is in the union of the subsets?
     
  5. Dec 8, 2012 #4

    SammyS

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    and that means ...
     
  6. Dec 8, 2012 #5
    This is what i'm thinking:
    If you take [itex]\cup[/itex]S[itex]\alpha[/itex] in Y, then that's the collection of subsets of Y, and "map" it back to X, then it would include every x in X such that f(x) is in one of the subsets, correct?
    I'm not sure if i'm interpreting the other side of the equivalence correctly, but here's my attempt: If I take a particular S[itex]\alpha[/itex] and I map it back to X, then that will encompass every x value such that f(x) is in that particular subset. Then, I do this for each S[itex]\alpha[/itex] and take the union of all the x values in the preimage...is this a correct line of thinking?
     
  7. Dec 8, 2012 #6
    That f(x) must be in at least one particular subset...?
     
  8. Dec 8, 2012 #7
    How about giving that subset a name, say, [itex]S_a[/itex]? Now, if [itex]f(x)\in S_a[/itex], where does [itex]x[/itex] lie?
     
  9. Dec 8, 2012 #8
    Then x would be in the preimage of S[itex]\alpha[/itex]?
     
  10. Dec 8, 2012 #9
    Okay, I see how this shows the left-hand side of the equivalence is a subset of the right-hand side. But when writing my proof, will I have to generalize it so that f(x) can be in more than one S[itex]\alpha[/itex]?
     
  11. Dec 8, 2012 #10
    No, one is enough. If some element ##e\in S_k##, then ##e\in\bigcup_{k\in I}S_k##.
     
  12. Dec 9, 2012 #11

    SammyS

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    I would be inclined to say, [itex]\displaystyle \dots\text{ "then }f(x)\in S_{\alpha_0}\,, \text{ for some }\alpha_0\in I/ .\text{"}[/itex]
     
  13. Dec 9, 2012 #12
    Got it, thank you both for your help!
     
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