# $f(\bigcap K_n)=\bigcap f(K_n)$? Where $K_n$ are compact

$f:X\rightarrow Y$ is continous map between metric spaces, $K_n$ are non empty nested sequence of compact subsets of $X$, then we need to show the title above.

Please tell me which result I should apply here? regarding cont map and compact set I know that image of compact set is compact, attains bounds, uniformly continous etc. please help. well, we can start by taking $y\in \bigcap f(K_n)$ and then show that it is also in the left side?

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Let $y \in \bigcap f(K_n)$. This means there is $x_n$ so that $f(x_n) = y$ and $x_n \in K_n$ for all $n$.
If $\{x_n\}$ is finite, we're done. If it's infinite, it has a limit point $x$ (why?). Use the continuity of $f$ to find $f(x)$ and conclude the proof.
What means nested ? Is it $K_i \subseteq K_{i+1}$ and / or $K_{i+1} \subseteq K_i$ ? –  André Jan 14 '13 at 8:07
You can take any $x_n\in K_n$ such that $f(x_n)=y$ and that take a converging subsequance.