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Sets and equivalence between images of sets

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Let f be a function from E to F . Prove that f is an injective function if and only if for all A and B subsets of P(E)^2.

    [tex]f(A\cap B)=f(A)\cap f(B)[/tex]

    3. The attempt at a solution

    Well since we have "if and only if" that means we have an equivalences so for.

    [tex]\Rightarrow[/tex]

    If f is an injective function so it's trivial to say that

    [tex]f(A\cap B)=f(A)\cap f(B)[/tex]

    For: [tex]\Leftarrow[/tex]

    We have to show a double inclusion so since:

    [tex]A\cap B\subset A[/tex] and [tex]A\cap B\subset B[/tex] then:

    [tex]f(A\cap B)\subset f(A)[/tex] and [tex]f(A\cap B)\subset f(B)[/tex]

    so therefore: [tex]f(A\cap B)\subset f(A)\cap f(B)[/tex]

    And the other way around:

    let [tex]y\in f(A)\cap f(B)[/tex] so there exists [tex]x\in A\cap B[/tex] such that f(x)=y then by the definition of an image we get that [tex]f(x)=y\in f(A\cap B)[/tex] so therefore:

    [tex]f(A)\cap f(B)\subset f(A\cap B)[/tex]

    So finally : [tex]f(A\cap B)=f(A)\cap f(B)[/tex]

    Hence f has to be an injective function. Any help or any remarks would be very well appreciated.
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2

    LCKurtz

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    No, it may be easy but it isn't trivial. You have to prove it. Your assumption is ##f## is injective and you have to show for any ##A,B## that ##f(A\cap B)=f(A)\cap f(B)##. In particular you argument must use the fact that ##f## is 1-1 somewhere.

    Here you you should be assuming that for any ##A,B## that ##f(A\cap B)=f(A)\cap f(B)## and you must show that ##f## is 1-1. So what's the argument below about? Are you trying to show what you are assuming?

    But that is given.

    You can't just claim it. You have to prove it.
     
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