# Finite union of compact sets is compact using a different def of open cover

I am trying to prove from definition that a finite union of compact sets is compact given that the definition of an open cover I have from my lecture notes is:

An open cover $\cal U$ of a space $M$ is a collection of open subsets of $M$ s.t. their union is $M$.

P.S. I have proved the statement using sequential compactness but it's rather long. I've also seen many proofs of the fact but they all seem to use a different definition of an open cover.

• What different definition of open cover have you seen? – David C. Ullrich Jan 1 '16 at 16:25

Let $\{K_{i}\}$ be a finite collection of compact sets and $K=\cup_{i}K_{i}$. Let $\{B_{\alpha}\}$ cover $K$. For each $i$, it follows that $\{B_{\alpha}\}$ also covers $K_{i}$, and hence admits a finite subcover $\{B_{\alpha}^{i}\}$. Can you figure out the rest?

• That's exactly the proof I've already seen. But notice that in my definition, the cover for K can't be a cover for any of its subsets. – Stargazer Jan 1 '16 at 17:04

You note that an open cover for the union of compact sets $K_1\cup K_2\cup \ldots\cup K_n$ is union of open covers for the sets $K_1 , K_2, \ldots,K_n$.

• That;s the thing I've been struggling to prove. Can you elaborate? – Stargazer Jan 1 '16 at 17:07

Oh, I finally see what the problem is! The problem is talking about an openn cover of a space as opposed to an open cover of a subset of that space. If $M$ is a topological space (or metric space if that's what you're talking about) then an open cover of $M$ is what you say. An open cover of a subset $A\subset M$ is a collection of open sets whose union contains $A$.

• Oh, dear! You're right. Thanks for explaining, I mixed up the definitions – Stargazer Jan 1 '16 at 17:36
• This is cleared up via the subspace topology. When finding an open cover of a subset $A$ of a space $M$ it is enough to find a collection of open sets whose union contains $A$, since as a topological subspace, the open sets in $A$ are precisely open sets in $M$ intersected with $A$. – Morgan Rogers Aug 4 '16 at 10:57

Let $K_1,\cdots, K_t$ be compact sets, and let $\{U_i\}_{i\in I}$ be an open cover of $\bigcup_{j=1}^t K_j$. Then $\{U_i\}_{i\in I}$ is an open cover of every $K_j$, so we get $I_1,\ldots,I_t \subseteq I$ finite sets such that $K_j \subseteq \bigcup_{i \in I_j} U_i$. Now it follows that $$\bigcup_{j=1}^t K_j \subseteq \bigcup_{i \in \bigcup_{j=1}^t I_j} U_i,$$ but $\bigcup_{j=1}^t I_j$ is finite, hence $\{U_i\}_{i\in \bigcup_{j=1}^tI_j}$ is the finite subcover you want.

• Using my definition of an open cover, why do you think {Ui}i∈I is an open cover for ⋃Kj as well as for every Kj? – Stargazer Jan 1 '16 at 17:06
• I didn't understand your question. When one wants to prove that a set is compact, one takes an open cover of a set and finds a finite subcover – Ivo Terek Jan 1 '16 at 17:08
• I just took an open cover of the union and found a finite subcover – Ivo Terek Jan 1 '16 at 17:09
• Since {Ui}i∈I is an open cover, using my definition, K=⋃Ui. Then it's clear that {Ui}i∈I can't be an open cover for any Kj. – Stargazer Jan 1 '16 at 17:25
• Ah, ok. Then intersect each $U_i$ with each $K_j$ - now the sets $U_i \cap K_j$ are open in $K_j$ and we'll have $=$ instead of $\subseteq$ – Ivo Terek Jan 1 '16 at 17:42