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So right now I'm working through Topology by Munkres and reviewing some basic set theory. I came across the following problem:

Show $f(A-B) \supset f(A)-f(B)$ with equality iff $f$ is injective.

The problem I'm having here is that I feel I can show equality without $f$ being injective. My proof to show the reverse inclusion goes as follows: take $y \in f(A-B)$. This means that $y = f(a)$ for some $a \in A-B$. So, as $a \in A$ and $a \notin B$, $f(a) \in f(A)$ and $f(a) \notin f(B)$. Since $y=f(a)$ then $y \in f(A)-f(B)$. I feel like this shows equality without injectivity but I know from looking up other answers I am wrong.

Andrés E. Caicedo
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Mike P.
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    $a \not\in B$ does not imply $f(a) \not\in f(B)$ in general. – user332239 Nov 03 '16 at 20:40
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    Let $f(x) = x^2$, $A = \mathbf{R}$, and $B = (0, \infty)$. Then $f(-1) = 1$ and $f(1) = 1$, so in fact $f(-1) \in f(B)$. Your statement "So, as $a \in A$, and $a \notin B$, $f(a) \in f(A)$ and $f(a) \notin f(B)$" is not correct, unless $f$ is injective. – dannum Nov 03 '16 at 20:41
  • Thanks for the example! That clears a lot up for me. – Mike P. Nov 03 '16 at 20:55
  • if f is not injective f(a) could very easily equal f(b) if f(B)$. You need injectivity to claim f(a) \in A-B means $f(a) \ne f(b) \in f(B)$. – fleablood Nov 04 '16 at 04:18

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Since $y=f(a)$ and $a\in A$, $y\in f(A)$. But you haven't proven that $y\not\in f(B)$. By definition, $f(B)$ is the set of $x$ such that $x=f(b)$ for some $b\in B$. So to prove $y\not\in f(B)$, you have to prove $y\neq f(b)$ for all $b\in B$. But all you know about $y$ is that $y=f(a)$, where $a$ happens to not be an element of $B$. For all you know, there might be some different element $b\in B$ such that $y$ is also equal to $f(b)$. Then $y=f(b)$ would be an element of $f(B)$.

Eric Wofsey
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An example let $ A=\{1,2,3\}, B=\{-1\} $ Let $ f (x)=x^2$ you can see that $ f (A)=\{1,4,9\}, f (B)=\{1\} $ hence $ f (A)-f (B)=\{4,9\} $ but $ A-B= A $ no intersection hence $ f (A-B)= \{1,4,9\} $ . So you will not have equality if we have an element in $ B $ which is not belong to $ A $ but has a same image to an element in $ A $

IrbidMath
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