elementary set theory - How to prove this? "For all sets A,B⊆D and functions f:D→R, we have f(A∩B)⊆(f(A)∩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.

Answer

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 $f(A\cap B)\subset f(A)\cap f(B)$.