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

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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 $K_1 , K_2, \ldots,K_n$.

Answer 4

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 $A$.