# Couter example for set inequality

I read the result that $f:X\rightarrow Y$ and {$A_i$} collection of subsets of $X$. Then $f(\bigcup A_i)=\bigcup f(A_i)$ and $f(\bigcap A_i)\subset \bigcap f(A_i)$.

I can prove these results. But the latter one is only a subset. Why not equality?

Just as an easy example, consider a constant function (i.e. $f(x) = y_0$ for some particular $y_0$ and all $x$) and a collection of disjoint $A_i$'s.