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Thread: Image of difference of sets

  1. #1
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    Image of difference of sets

    Hey,

    I want to show that $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$

    I have done the following:

    Assume that $\displaystyle y \in f(A\backslash B)$.
    This means that $\displaystyle \exists x\in A\backslash B: f(x)=y$ (by definition).
    This means that $\displaystyle x\in A \land x\notin B$.
    This means that $\displaystyle f(x) \in f(A) \land f(x) \notin B$.
    This means that $\displaystyle f(x) \in f(A) \backslash f(B)$

    Therefore $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$.

    Is the above correct?
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  2. #2
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    Re: Image of difference of sets

    Quote Originally Posted by detalosi View Post
    Hey,

    I want to show that $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$

    I have done the following:

    Assume that $\displaystyle y \in f(A\backslash B)$.
    This means that $\displaystyle \exists x\in A\backslash B: f(x)=y$ (by definition).
    This means that $\displaystyle x\in A \land x\notin B$.
    This means that $\displaystyle f(x) \in f(A) \land f(x) \notin B$.
    This means that $\displaystyle f(x) \in f(A) \backslash f(B)$

    Therefore $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$.

    Is the above correct?
    Yes it is completely correct. Good work.

    One comment: I would have preferred using \subseteq $\subseteq$
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  3. #3
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    Re: Image of difference of sets

    The statement is not true for arbitrary functions. For example, let A be the set of positive reals, B the set of negative reals and f the squaring function. Your conclusion says $A\subseteq\emptyset$ ??

    One hypothesis to make the statement true is that f is one to one. in the 4th line of your proof, you need some reason why $f(x)\not\in f(B)$ is true; I hope you understand why f one to one guarantees this to be true.
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