Does an arbitrary function preserve arbitrary intersections? Let $S,T$ be sets and let $\Lambda$ be an index-set. Let $f:S\to T$ and $\{A_\lambda\}_{\lambda\in\Lambda}$ be a collection of subsets in $S$ (i.e. $A\lambda\subset S$ for all $\lambda\in\Lambda$). Does $f$ satisfy $$f\left[\bigcap_{\lambda\in\Lambda} A_\lambda\right]\subset\bigcap_{\lambda\in\Lambda} f[A_\lambda]\text{ and }f\left[\bigcup_{\lambda\in\Lambda} A_\lambda\right]=\bigcup_{\lambda\in\Lambda} f[A_\lambda]?$$
 A: Sure; the arguments are exactly the same as for the finite case.
Suppose that $x\in\bigcap_{\lambda\in\Lambda}A_\lambda$; then $x\in A_\lambda$ for each $\lambda\in\Lambda$, so $f(x)\in f[A_\lambda]$ for each $\lambda\in\Lambda$, and therefore $f(x)\in\bigcap_{\lambda\in\Lambda}f[A_\lambda]$. It follows immediately that
$$f\left[\bigcap_{\lambda\in\Lambda}A_\lambda\right]\subseteq\bigcap_{\lambda\in\Lambda}f[A_\lambda]\;.$$
Now suppose that $x\in\bigcup_{\lambda\in\Lambda}A_\lambda$; then there is some $\lambda_0\in\Lambda$ such that $x\in A_{\lambda_0}$. Clearly $f(x)\in f[A_{\lambda_0}]\subseteq\bigcup_{\lambda\in\Lambda}A_\lambda$, so 
$$f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]\subseteq\bigcup_{\lambda\in\Lambda}f[A_\lambda]\;.$$
Conversely, if $y\in\bigcup_{\lambda\in\Lambda}f[A_\lambda]$, then $y\in f[A_{\lambda_0}]$ for some $\lambda_0\in\Lambda$, and therefore there is an $x\in A_{\lambda_0}$ such that $f(x)=y$. Clearly $x\in\bigcup_{\lambda\in\Lambda}A_\lambda$, so $f(x)\in f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]$, and it follows that
$$f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]\supseteq\bigcup_{\lambda\in\Lambda}f[A_\lambda]$$ and hence that 
$$f\left[\bigcup_{\lambda\in\Lambda}A_\lambda\right]=\bigcup_{\lambda\in\Lambda}f[A_\lambda]\;.$$
A: We can do some element-chasing and show this is indeed true:
Suppose $y\in f\left[\bigcap A_\lambda\right]$, then there is some $x\in\bigcap A_\lambda$ such that $f(x)=y$. This means that for all $\lambda$, $x\in A_\lambda$ so $y\in f\left[A_\lambda\right]$ for all $\lambda$, and thus the inclusion holds.
For the union, let $y\in f\left[\bigcup A_\lambda\right]$, then there is some $\lambda$ and $x\in A_\lambda$ such that $y=f(x)$, therefore $y\in f\left[A_\lambda\right]$ and therefore $y\in\bigcup f\left[A_\lambda\right]$. The other direction is similar.
A: The first holds since $f\left(\bigcap_\lambda A_\lambda\right) \subseteq f(A_\lambda)$
for all $\lambda\in\Lambda$.  This inequality can be strict.
The second is easy to prove. It is true.  
