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

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

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

    My attempt: Let $\displaystyle y \in f(A-B).$ Then $\displaystyle y \in f(A)-f(B)$, by assumption. Then, $\displaystyle \exists x \in A - B$ such that $\displaystyle y = f(x)$. Similarly, $\displaystyle \exists x_1 \in A$ such that $\displaystyle y = f(x_1)$.
    Since $\displaystyle y \in f(A)-f(B)$, $\displaystyle y=f(x_1) \in f(A)$, but $\displaystyle y=f(x_1) \notin f(B)$. Then $\displaystyle x_1 \notin B$. Since $\displaystyle x \in A-B$, $\displaystyle x \in A$, but $\displaystyle x \notin B$.
    So we have that $\displaystyle x,x_1 \in A$ and $\displaystyle x,x_1 \notin B$. Since $\displaystyle y = f(x)$ and $\displaystyle y = f(x_1)$, $\displaystyle f(x) = y = f(x_1)$.
    This is equivalent to saying that for $\displaystyle (x,y) \in f$ and $\displaystyle (x_1, y) \in f$, we have that $\displaystyle (x, y) = (x_1, y)$.
    By theorem 2.2, it must be that $\displaystyle x=x_1$. So $\displaystyle f(x) = f(x_1)$ implies that $\displaystyle x=x_1$. 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 $\displaystyle f: X \rightarrow Y$. Prove that if $\displaystyle f(A-B) = f(A) - f(B) \text {for all } A,B \subseteq X \text{, then } f \text{ is injective.}$
    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 $\Large {f(\{s\}\setminus \{t\})\text{ and }f(\{s\})\setminus f(\{t\})~?}$
    $\displaystyle f(\{s\}) \setminus f(\{t\}) = \emptyset$, since $\displaystyle f(s) = f(t)$. But $\displaystyle f(\{s\}\setminus \{t\})$ cannot be empty since $\displaystyle s \ne t$
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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
    $\displaystyle f(\{s\}) \setminus f(\{t\}) = \emptyset$, since $\displaystyle f(s) = f(t)$. But $\displaystyle f(\{s\}\setminus \{t\})$ cannot be empty since $\displaystyle s \ne t$
    Is that a proof by contradiction?
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