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Given a locally compact hausdorff space $X$, and some compact subset $K$. Does there exist a countable collection $\{V_n\}$ of open subsets of $X$, such that $$K=\bigcap_{n=1}^{\infty}V_n$$

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    If you don't know the terminology, a set which can be written as the countable intersection of open sets is called a $G_{\delta}$ set. (So you could rephrase your question to ask whether compact sets are $G_\delta$ in a locally compact Hausdorff space.) – 2017-01-23
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    in metric spaces they are because compact spaces are separable. – 2017-01-23

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No. For example, in $2^{\bf R}$, every point is compact, but every intersection of a countable family of open sets depends only on countably many coordinates, so it has more than one point (in fact, it must have many more points).

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    Can I ask for a place to find more information on $2^{\bf R}$? I have not personally encounter that topology, and would like to know more about it. – 2017-01-24
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    This is just the space of all $\{0,1\}$-valued real functions with pointwise convergence (product) topology. You can put instead of ${\bf R}$ any uncountable set at all, really, I just thought that ${\bf R}$ might be more familiar. – 2017-01-24
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    I assume that $\{0,1\}$ is imbued with the topology $\{\emptyset, \{1\},\{0\},\{0,1\}\}$ – 2017-01-24
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    @user160110: Naturally, yes. Otherwise, the topology wouldn't be Hausdorff. – 2017-01-24