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Math Help - Prove that f is injective

  1. #1
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    Prove that f is injective

    Hello,
    I need to prove that function defined like this  f:X->Y for any two given sets X,Y is one-to-one if and only if  f(A \cap B)= f(A) \cap f(B) , where  A,B \subseteq X.

    What do I need to do to prove this?

    EDIT:

    I've figured out how to prove that  f(A \cap B)= f(A) \cap f(B) when f is injective. I still do not know how to conclude that f is injective when  f(A \cap B)= f(A) \cap f(B) .
    Last edited by MachinePL1993; December 4th 2012 at 07:40 AM.
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  2. #2
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    Re: Prove that f is injective

    Quote Originally Posted by MachinePL1993 View Post
    Hello,
    I need to prove that function defined like this  f:X->Y for any two given sets X,Y is one-to-one if and only if  f(A \cap B)= f(A) \cap f(B) , where  A,B \subseteq X.
    What do I need to do to prove this?

    It is well known that for any function f(A\cap B)\subseteq f(A)\cap f(B).

    So you must show that equality holds if and only if the function is injective.

    So assume it is injective and show equality holds. Then the converse.
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  3. #3
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    Re: Prove that f is injective

    Now how to prove that f is injective if the above equality holds?
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  4. #4
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    Re: Prove that f is injective

    In general, we always have f(AnB) included in f(A)nf(B), proof by contradiction:

    let A,B subsets of X. Since f(AnB) =/= f(A)nf(B) there is an element (let's call it x) in AnB such that f(x) does not belong to f(A)nf(B). Because f(x) does not belong to f(A)nf(B), x is not in A or x is not in B. So we have an element x in AnB such that f(x) is not in A or not in B this is a contradiction.

    Now let f be injective. let c be an element of f(A)nf(B). There exists c_a in A such that f(c_a)=c. And there exists c_b in B such that f(c_b)=c. Because f is injective c_a=c_b. So c is the image of an element in both A and B, so he belongs to f(AnB).
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  5. #5
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    Re: Prove that f is injective

    Is this correct?
    Let's assume indirectly that the function is not injective

    Let's choose some y such that f(a)=y where a \in A and f(b)=y where b \in B, and A \cap B=\emptyset. Then f(A \cap B)={\emptyset}, because A \cap B=\emptyset, and f(A) \cap f(B) = {y}, which is a contradiction with the assumption because  f(A \cap B)\not= f(A) \cap f(B) , so the function must be injective.
    Last edited by MachinePL1993; December 4th 2012 at 08:01 AM.
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  6. #6
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    Re: Prove that f is injective

    it seems correct.
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    Re: Prove that f is injective

    Quote Originally Posted by MachinePL1993 View Post
    Is this correct?
    Let's assume indirectly that the function is not injective
    That proof is a bit rough. But it is the correct idea.
    If we assume that equality holds, but the function is not injective then \exists \{t,s\} such that  f(t)=f(s) but  t \not= s.

    Define A=X\setminus\{t\} ~\&~ B=\{t\}.

    Then f(A\cap B)=\emptyset but f(t)=f(s)\in f(A)\cap f(B).

    There is the contradiction.
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