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

  • 4
    $\begingroup$ This is tagged with general topology. So, I assume you are not working in a metrizable context. It is NOT true that the intersection of compact subsets is compact, even for finite intersections. Take a look at non-Hausdorff spaces. $\endgroup$ – user370967 Jun 1 '18 at 18:21
  • $\begingroup$ It is, however, true whenever compact sub-spaces are also closed subspaces. $\endgroup$ – Thomas Andrews Jun 1 '18 at 18:22
  • $\begingroup$ @Math_QED i edited my question I'm on a Hausdorff space $\endgroup$ – Vrouvrou Jun 1 '18 at 18:39

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.

  • 2
    $\begingroup$ because a compac subset is not always closed. $\endgroup$ – Tsemo Aristide Jun 1 '18 at 18:19
  • $\begingroup$ How to find that $K_i $ is covered ? $\endgroup$ – Vrouvrou Jun 1 '18 at 18:44
  • $\begingroup$ if $x$ in $K_1$ is not in $\cap K_i$, it is in $X-\cap K_i$. $\endgroup$ – Tsemo Aristide Jun 1 '18 at 18:45
  • $\begingroup$ $K_i= K\cup (K_{i}\setminus K)$ you use this ? then $K_i \subset \bigcup\Omega_i\cup (K_i\setminus K)$ $\endgroup$ – Vrouvrou Jun 1 '18 at 18:49
  • $\begingroup$ yes, it is the idea. $\endgroup$ – Tsemo Aristide Jun 1 '18 at 18:51

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.


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