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Question: So, f: X → Y is the same as f(x) = y We know that A,B are both contained in X. So f(A) and f(B) are both equal to X? I know f(A) not f(B) then means that there's no such element found in f(B). Then how do I actually connect the dots to show that f(A\B) is true.

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    "$f:X\to Y$ is the same as $f(x)=y$"... *arguably*. For sure, it is **not** the same as "$f(X)=Y$". – 2017-02-12

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If $y \in f(A) \setminus f(B)$, there exists $x \in A$ such that $y = f(x)$. Moreover, for all $z \in B$, $f(z) \neq y$, because $y \not\in f(B)$.

This means that $x \not\in B$; since it is in $A$, we conclude that $x \in A \setminus B$, and since $y = f(x)$, it follows that $y \in f(A \setminus B)$.

Taking stock, we have shown that a generic $y$ in $f(A) \setminus f(B)$ is also contained in $f(A \setminus B)$. Hence $f(A) \setminus f(B) \subseteq f(A \setminus B)$.

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    why do you introduce z? Is it because of X --> Y --> Z map? – 2017-02-13
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    Because I want to give a straightforward translation of the formula $y \not\in f(B)$. We say that $y$ is in $f(B)$ if there exists some $z \in B$ such that $y = f(z)$. If $y$ is not in $f(B)$, then for all possible choices of $z \in B$ we find that $y \neq f(z)$. – 2017-02-13
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    @Paul24 No, it's not because of an $X \to Y \to Z$ map. You can call that $z$ what you want. It's just a variable to designate a generic member of $B$. – 2017-02-13
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No, $f\colon X\to Y$ means $f$ is a function with domain $X$ and codomain $Y$. If $A$ is a subset of $X$, then $$ f(A)=\{f(x):x\in A\} $$ or, if you want to be fussy, $$ f(A)=\{y\in Y:\text{there exists $x\in A$ with $f(x)=y$}\} $$

Thus you have to start with an element $y\in f(A)\setminus f(B)$ and prove it is an element of $f(A\setminus B)$.

Saying $y\in f(A)\setminus f(B)$ means $y=f(a)$, for some $a\in A$, but $y\ne f(b)$, for all $b\in B$.

If you now prove $a\in A\setminus B$, you are done. Can $a\in B$?