TITLE: $ f(A) \cap f(B) = \emptyset \, \forall \, A,B \subseteq S, $ and $ A \cap B \, = \emptyset \implies f(A-B)=f(A)-f(B) $ for $B \subseteq A$
QUESTION [1 upvotes]: This is the full question:

$f: S \to T$ is a function, then the following statements are equivalent:
a) $f$ is one-one on $S$
b) $f(A \cap B) = f(A)\cap f(B) \, \forall \, A,B \subseteq S $
c) $f^{-1}(f(A))=A  \hspace{4mm} \forall \, A\subseteq S $
d) $f(A) \cap f(B) = \emptyset \, \forall \, A,B \subseteq S, $ and $ A \cap B \, = \emptyset$
e) $f(A-B)=f(A)-f(B) \,\, \forall \, A,B \subseteq S$ and $B \subseteq A$

I am having trouble is understanding a step in the proof of $\, d \implies e$ , Here is the proof

If $B \subseteq A \text{ then } (A-B)\cap B = \emptyset \\ \implies f(A-B) \cap f(B) = \emptyset \\ \implies f(A-B) \subseteq f(A)-f(B) \hspace{5mm} (*) \\ \text{if } \, y \in f(A-B) \text{ then } y=f(x) \text{ for some } x \in A \text{ and } x \notin B \\ x\in A-B \implies f(x)=y \in f(A-B) \\ \implies f(A)-f(B) \subseteq f(A-B)  $

How did we get statement (*)? I didn't understand that step.

REPLY [1 votes]: We know that $f(A - B) \subset f(A)$. Then, it must also be that, as $f(B) \subset f(A)$, because $f(A) = f(A - B) \cup f(B)$. Thus, we know that the set of $f(A-B)$ must be contained in $f(A)$.
However, the statement preceding (*) tells us that $f(A - B) \cup f(B) = \emptyset$. Or, in other words, it tells us that $f(A - B)$, which is contained in $f(A)$, does not intersect $f(B)$ at all. Thus, by definition, $f(A - B)$ is contained within (and is thus a subset of) that subset of $f(A)$ which contains no points from $f(B)$. This subset of $f(A)$ is, by definition, $f(A) - f(B)$.
{"set_name": "stack_exchange", "score": 1, "question_id": 3331719}