Prove $f(S \cap T) \subseteq f(S) \cap f(T)$ $f(S \cap T) \subseteq f(S) \cap f(T)$
$x$ lies in ($S \cap T$), which means the domain has fewer elements than the domain of $S$ and $T$, since $x$ must be in $S$ and $T$. All $f(x)$ values of $x$, which resides in ($S \cap T$) is also a member of $f(S) \cap f(T)$, because $f(S)$ encompasses all $x$ in $S$ even those in ($S \cap T$) and the same can be said about $f(T)$.
Can you give me the solution?
 A: For this one you take an element $x\in f(S\cap T)$. You want to prove that this element is also in $f(S)\cap f(T)$. That is, you want to prove that $x\in f(S)$ and also $x\in f(T)$. Now since $x\in f(S\cap T)$ you know that there is some $y\in S\cap T$ such that $x = f(y)$. Now $y\in S\cap T$ so in particular $y\in S$, so that means $x = f(y) \in f(S)$. The same argument works to show that $x\in f(T)$. 
Hence in all $x$ is both an element of $f(S)$ and $f(T)$, so $x\in f(S)\cap f(T)$.
A: $y \in f(S\cap T)$ means $y=f(x)$ for some $x\in S\cap T$.
That means $y=f(x)$ for some $x$ for which $x\in S$ and $x\in T$.
That implies $y=f(x)$ for some $x\in S$, so $y\in f(S)$, and for the same reason, $y\in f(T)$.
Since $y \in f(S)$ and $y\in f(T)$, we have $y\in f(S)\cap f(T)$.
So we have proved that if $y\in f(S\cap T)$ then $y\in f(S)\cap f(T)$.
That is true of every value of $y$.
Thus every member of $f(S\cap T)$ is a member of $f(S)\cap f(T)$.
That's what it means to say $f(S\cap T)\subseteq f(S)\cap f(T)$.
A: Let $f(x) \in f(S \cap T)$. This means that $x \in S \cap T$, so $x \in S$ and $x \in T$.  You fill in this rest.
