# How to prove this? “For all sets A,B⊆D and functions f:D→R, we have f(A∩B)⊆(f(A)∩f(B)).” [duplicate]

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}}}

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 $f(A\cap B)\subset f(A)\cap f(B)$.
• This is great. But can you explain or prove why if $X \subset Y$, then $f(X) \subset f(Y)$? – FutureTrillionaire Sep 21 '14 at 3:41
• If $y\in f(X)$ then by definition $y=f(s)$ for $s\in X$, but $X\subset Y$, so $s\in Y$ as well, which, together with $y=f(s)$, implies that $y\in f(Y)$. – Kim Jong Un Sep 21 '14 at 3:43