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Injective function proof

  1. Sep 23, 2007 #1
    This is something I understood before, but for some reason I forgot it. How do you prove this inequality holds, if f is injective?

    [tex] A_0 \subset f^{-1}(f(A_0)) [/tex]
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2
    If x is not in A, then f(x) is not in f(A) (injectivity) so x is not in f^-1(f(A))
  4. Sep 23, 2007 #3


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    Proofs of set algebra identities tend to be rather formulaic. If you look at the definition of "subset", then two proofs should immediately suggest themselves:
    Let x be an element of A_0 ... Therefore x is in f^{-1}(f(A_0))​
    Suppose x is not an element of f^{-1}(f(A_0)) ... Therefore x is not in A_0​

    And from there, you simply have to fill in the missing steps. And again, the missing steps are usually obvious from unwinding the definitions.
    Last edited: Sep 23, 2007
  5. Sep 23, 2007 #4


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    Yes, but that doesn't say anything about what happens if x IS A, which is the whole point.

    waht, the standard way of proving "[itex]A\subset B[/itex] is to start "If x is in A" and then conclude "then x is in B".
    If x is in A_0, you know that f(x) is in f(A_0). Now, what does the fact that f is injective say about x and f-1(f(A_0)).
  6. Sep 23, 2007 #5

    I'm not sure what you mean. I showed [tex] f^{-1}(f(A_0)) \subset A_0 [/tex] which is the other half of the equality waht was asking for.
  7. Sep 23, 2007 #6


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    Yes, I understood that. That was why I said, "If x is in A_0"- because that's the direction you want to prove.
  8. Sep 23, 2007 #7
    Am I reading this incorrectly?

  9. Sep 24, 2007 #8
    I'll assume you're starting from: f:A -> B and A_0 is a subset of A.

    The inclusion relation you've written holds regardless of whether f is injective or not.
    However, if f is injective, then the relation can be written as an equality.
    Proof is nothing more than working the definitions, as has already been suggested.
    Last edited: Sep 24, 2007
  10. Sep 24, 2007 #9


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    Finally, it dawns on me. I was reading the whole thing backwards. I thought the question was to prove that if f is injective, then.... Sorry, everyone.
  11. Sep 24, 2007 #10
    Thanks I get it now,

    I should have been more clearer.
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