Let $K_1$,$K_2$,...,$K_N$ be compact subsets of the metric space $(X,d)$. Show that

a) $K_1\cap K_2\,\cap\,...\cap\,K_N$ is compact.

b) $K_1\cup K_2\,\cup\,...\cup\,K_N$ is compact.

My own idea is to show that

a) $K_1\cap K_2\,\cap\,...\cap\,K_N$ is closed and $K_1\cap K_2\,\cap\,...\cap\,K_N$ is bounded, which then will give that the set is compact.

b) The same idea as a).

Is that the right way to do it?

  • 4
    $\begingroup$ It is certainly not true that closed and bounded sets are compact... $\endgroup$ Feb 14 '16 at 19:55
  • $\begingroup$ No, because it may depend on the metric $d$. You can have metrics on $X$ such that every set is bounded. Your argument only works in $\Bbb R$ endowed with the usual metric. $\endgroup$
    – Watson
    Feb 14 '16 at 19:55
  • $\begingroup$ You should use the definition of compactness for this. $\endgroup$
    – Thomas
    Feb 14 '16 at 19:56
  • $\begingroup$ Are yes you are right. It is not a general rule $\endgroup$
    – 309324
    Feb 14 '16 at 19:56

Let $(U_\alpha)_{\alpha\in I}$ be an open cover of $\bigcup_{i=1}^n K_i$.

Then clearly $(U_\alpha)_{\alpha\in I}$ is an open cover of $K_k$ for each index $k$. So for each $k$, a finite subcover $(U^k_\alpha)_{\alpha\in I_k}$ exist. The union of these covers with respect to $k$ is still finite and covers $\bigcup_{i=1}^n K_i$, so the latter set is compact.

The case of finite intersections is similar but easier.


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