Suppose for a compact topological space(if you want, we can assume Hausdorff) $X$, $X$ is a disjoint union of compact subsets,that is $X=\bigcup_{i\in I} X_i$ such that $X_i\cap X_j=\emptyset$ for $i\neq j$ and $X_i$ is compact for any $i\in I$. For convenience, call it a compact disjoint cover. Is it true that we can find a finite subcover of $X$ for any compact disjoint cover?

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    $\begingroup$ So $[0,1]=\bigcup_{x\in [0,1]} \{x\}$ would be a compact disjoint cover, and it obviously doesn't have a finite subcover. Is this really what you mean? $\endgroup$ – Florian Sep 18 '11 at 19:35

As mentioned by Florian this is clearly not true. However we can say a bit more.

Suppose $X=\bigcup_{i\in I}X_i$ is a disjoint union of compact spaces, then $X$ is compact if and only if $I$ is finite.

Suppose $I$ is finite, take a $\mathcal U$ to be a cover of $X$, then you can consider $\mathcal U_i=\{U\cap X_i\mid U\in\mathcal U\}$ as a cover for $X_i$. Each $\mathcal U_i$ admits a finite subcover for $X_i$ and the union of the finite subcovers is finite, since $I$ is finite, and covers $X$.

Now suppose that $X$ is not compact, take $\mathcal U$ which is a covering without any finite subcover, and consider as above $\mathcal U_i$ as a covering of $X_i$ induced by $\mathcal U$. Since we have a finite subcover $\mathcal V_i\subseteq\mathcal U_i$, and $\mathcal V=\bigcup_{i\in I}\mathcal V_i$ is a cover of $X$ which is a subcover of $\mathcal U$ we have that $\mathcal V$ is infinite, therefore $I$ is infinite.

Now, if you require that the elements of the cover will be compact subsets then you have that $X$ has a finite subcover for any compact cover if and only if $X$ is finite.

The proof is quite simple, if $X$ is finite then it is compact since there are only finitely sets to begin with and every cover is finite.

If $X$ is such that every disjoint compact cover has a finite subcover then take the cover by singletons, it is a disjoint compact cover and it cannot be refined at all. By the assumption on $X$ this is a finite cover, therefore $X$ is a finite union of singletons, thus finite itself.

  • $\begingroup$ Thanks to both of you. Can we define a continuous (complex valued) function $f$ on $X$ such that $f$ on $X_i$ is constant? $\endgroup$ – user16283 Sep 18 '11 at 20:36
  • $\begingroup$ @user16283: (1) This is a whole different question. (2) If you fix the partition $X_i$ then it is possible, of course you can take $x\mapsto 0$. For a less degenerated form, I would have to check the details. However there is a concept of perfect map, which is a map $f\colon X\to Y$ such that for every $y$, $f^{-1}[\{y\}]$ is a compact subset of $X$. So I see no reason why this should not be possible. $\endgroup$ – Asaf Karagila Sep 18 '11 at 20:40

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