For $f: A \to B$ with $S, T \subset A $, show that $f(S \cap T) \subset f(S) \cap f(T) $.

Question

Let $f:A\mapsto B$ be given and let $S\subseteq A$ and $T\subseteq A$.

Show that,

$$f(S\cap T)\subseteq f(S)\cap f(T)$$

Answer 1

For $s \in S \cap T$ we have $f(s)\in f(S)$ and $f(s) \in f(T)$ impy $f(s) \in f(S) \cap f(T)$. Hence $f(S \cap T) \subseteq f(S) \cap f(T)$

Answer 2

Let $y \in f(S \cap T )$, this means by definition that there is some $x \in S \cap T $ so that $y = f(x) $. Hence, there is some $x \in S$ so that $y = f(x) \in f(S) $ and there is some $x \in T$ so that $y = f(x) \in S$. In particular, $y \in f(S) \cap f(T) $