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Thread: Prove: If f(A-B) = f(A)-f(B), then f is injective.

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    Prove: If f(A-B) = f(A)-f(B), then f is injective.

    Book: Mathematical Analysis by Apostol

    Theorem 2.2: (a,b) = (c,d) \iff a = c \text{ and } b = d.

    A - B = \{ x: x\in A, \text{ but } x \notin B\}

    Let f: X \rightarrow Y. Prove that if f(A-B) = f(A) - f(B) \text {for all } A,B \subseteq X \text{, then } f \text{ is injective.}

    My attempt: Let y \in f(A-B). Then y \in f(A)-f(B), by assumption. Then, \exists x \in A - B such that y = f(x). Similarly, \exists x_1 \in A such that y = f(x_1).
    Since y \in f(A)-f(B), y=f(x_1) \in f(A), but y=f(x_1) \notin f(B). Then x_1 \notin B. Since x \in A-B, x \in A, but x \notin B.
    So we have that x,x_1 \in A and x,x_1 \notin B. Since y = f(x) and y = f(x_1), f(x) = y = f(x_1).
    This is equivalent to saying that for (x,y) \in f and (x_1, y) \in f, we have that (x, y) = (x_1, y).
    By theorem 2.2, it must be that x=x_1. So f(x) = f(x_1) implies that x=x_1. Therefore, f is injective.

    Have I made any errors in my argument?
    Last edited by amthomasjr; Sep 18th 2016 at 12:33 PM.
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    Re: Prove: If f(A-B) = f(A)-f(B), then f is injective.

    Quote Originally Posted by amthomasjr View Post
    Book: Mathematical Analysis by Apostol
    Let f: X \rightarrow Y. Prove that if f(A-B) = f(A) - f(B) \text {for all } A,B \subseteq X \text{, then } f \text{ is injective.}
    I simply do not follow any of that. You are told to $f: X\to Y$ is injective if $f(A\setminus B)=f(A)\setminus f(B)$ for all $A~\&~B$ subsets of $X$.
    Suppose $f$ is not injective. Then $\exists \{s,t\}\subset X,~s\ne t\text{ BUT }f(s)=f(t)$.

    What can you say about $\Large {f(\{s\}\setminus \{t\})\text{ and }f(\{s\})\setminus f(\{t\})~?}$
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    Re: Prove: If f(A-B) = f(A)-f(B), then f is injective.

    Quote Originally Posted by Plato View Post
    What can you say about $\Large {f(\{s\}\setminus \{t\})\text{ and }f(\{s\})\setminus f(\{t\})~?}$
    f(\{s\}) \setminus f(\{t\}) = \emptyset, since f(s) = f(t). But f(\{s\}\setminus \{t\}) cannot be empty since s \ne t
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    Re: Prove: If f(A-B) = f(A)-f(B), then f is injective.

    Quote Originally Posted by amthomasjr View Post
    f(\{s\}) \setminus f(\{t\}) = \emptyset, since f(s) = f(t). But f(\{s\}\setminus \{t\}) cannot be empty since s \ne t
    Is that a proof by contradiction?
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