How to prove this? "For all sets $A,B\subseteq D$ and functions $f:D\mapsto R$, we have $f(A\cap B)\subseteq(f(A)\cap f(B))$." Here's my attempt:

*

*$f(A∩B) = f(\{x|x∈A∧x∈B\}) = \{f(x)|x∈\{x|x∈A∧x∈B\}\}$

*$f(A)∩f(B) = f(\{x|x∈A\}) ∩ f(\{x|x∈B\}) = \{f(x)|x∈\{x|x∈A\}\} ∩ \{f(x)|x∈\{x|x∈B\}\} = \{x|x∈\{f(x)|x∈\{x|x∈A\}\}∧x∈\{f(x)|x∈\{x|x∈B\}\}\}$
And now I'm stuck. Please help.

You have
$$
A\cap B\subset A\implies f(A\cap B)\subset f(A),\\
A\cap B\subset B\implies f(A\cap B)\subset f(B)
$$
so it follows that $f(A\cap B)\subset f(A)\cap f(B)$.
