Let $K_i, i\in I$ compact sets in a Hausdorff space $X$ and $K=\bigcap_{i\in I} K_i$ and $\Omega_i,i\in I$ a family of open sets such that $$K\subset \bigcup_{i\in I}\Omega_i$$
How to find a finite sub cover ?
Thank you
Let $K_i, i\in I$ compact sets in a Hausdorff space $X$ and $K=\bigcap_{i\in I} K_i$ and $\Omega_i,i\in I$ a family of open sets such that $$K\subset \bigcup_{i\in I}\Omega_i$$
How to find a finite sub cover ?
Thank you
I'm not sure about whether there is a direct way of finding it. But you can (assuming that your space is Hausdorff) say that $\bigcap_{i\in I}K$ is a closed subset of $K_i$ (for some $i\in I$ chosen by you) and that closed subsets of compact spaces are compact.
If the space is separated, then $K_1$ is covered by $U_i$ and $X-\cap K_i$, take the finite cover induced.
Given that $\{\Omega_i\}_{i\in I}$ covers $K$, one has \begin{align*} X= (X\setminus K)\cup\bigcup_{i\in I}\Omega_i.\tag{$\star$} \end{align*} Take your favorite index $i^*\in I$. It follows from ($\star$) that $$K_{i^*}\subseteq (X\setminus K)\cup\bigcup_{i\in I}\Omega_i,$$ so that $\{\Omega_i\}_{i\in I}$ and $X\setminus K$ together form an open cover of $K_{i^*}$ (the extra assumption that $X$ is Hausdorff guarantees that each of $\{K_i\}_{i\in I}$ is closed, so that $X\setminus K$ is open, right?).
Since $K_{i^*}$ is compact, and $\{\Omega_i\}_{i\in I}$ and $X\setminus K$ together form an open cover of it, there must exist a finite subcover, say $\Omega_{i_1},\ldots,\Omega_{i_n}$ for some positive integer $n$ and $i_1,\ldots,i_n\in I$, and $X\setminus K$ may or may not be included in this subcover. One may assume that $X\setminus K$ is in this subcover (if not, adding $X\setminus K$ to the finite subcover $\{\Omega_{i_1},\ldots,\Omega_{i_n}\}$ would still result in a finite subcover), so that $$K_{i^*}\subseteq(X\setminus K)\cup\Omega_{i_1}\cup\cdots\cup\Omega_{i_n}.$$
Now, $K$ is a subset of $K_{i^*}$, so one has also that $$K\subseteq(X\setminus K)\cup\Omega_{i_1}\cup\cdots\cup\Omega_{i_n}.$$ Since $X\setminus K$ has no common element with $K$, one may remove it from this cover to end up with $$K\subseteq\Omega_{i_1}\cup\cdots\cup\Omega_{i_n},$$ which is a finite subcover of the original cover, as sought.
Note that the exact composition of the subcover $\{\Omega_{i_m}\}_{m=1}^n$ may depend on the favorite index $i^*\in I$ you have chosen (different choices may result in different subcovers), but the conclusion that one always ends up with a finite subcover still holds in all cases.