Is there a continuous function $f:X\rightarrow Y$ such that $f\big(\cap_{i\in I} A_i) \neq \cap_{i\in I}f\big(A_i)$ Is there a continuous function $f:X\rightarrow Y$ such that $$f\big(\cap_{i\in I} A_i) \neq \cap_{i\in I}f\big(A_i)$$
Where $A_i\subseteq X$ and $I$ is an arbitrary index set.
I can easily find a discontinuous function, for example $f(x) = 1$ when $x = 0$ and $f(x) =x$ otherwise. Then taking subsets $[-\frac{1}{n}, \frac{1}{n}]$ for $n\in\mathbb{N}$ works. 
However I am struggling to find a continous $f$. Does one exist?

Take $f$ constant, let $A_i\neq\varnothing$ for each $i\in I$ and let $\bigcap_{i\in I}A_i=\varnothing$.
