Prove that $f(A\cap {B})\subseteq {f(A)\cap {f(B)}}$

Question

Let $f:S\to{T}$ be a function. If $A$ and $B$ are two arbitary subsets of $S$ prove that $f(A\cap {B})\subseteq {f(A)\cap{f(B)}}$

Can you prove that $f(A\cap B)\subseteq f(A)$? Then you are done. — Did, Aug 25 '14 at 10:41

score 5 · Answer 1 · answered Aug 25 '14 at 10:43

5

Let $y \in f(A\cap B)$. Then there is an $x \in A\cap B$ so that $f(x) = y$

But $x\in A$ so $f(x)\in f(A)$ and $x\in B$ so $f(x)\in f(B)$ therefore $f(x)\in f(A)\cap f(B)$ i.e. $y\in f(A)\cap f(B)$

answered Aug 25 '14 at 10:43

Darth Geek

But converse is also true so equality should hold!? – Satvik Mashkaria Aug 25 '14 at 11:11
Think $f(2)=f(1)=2$ and $A={1}$, $B={2}$. – Maddy Aug 25 '14 at 11:22
The if $A\cap B\neq \emptyset$ then equality only holds for $A\subseteq B$ or $B\subseteq A$. If $A\cap B = \emptyset$ then it will only hold for $f_{A\cup B}$ injective. – Darth Geek Aug 25 '14 at 11:41

1 Answers1