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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?

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

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$f(x)=x^2$; $A_1=\{-1\}$; $A_2=\{1\}$.

$g(x)=0$; $A_i=\{i\}$, $i\in\mathbb R$.

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