Compactness is weakly hereditary

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
|

Property "Page" (as page type) with input value "{{{property}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.Property "Page" (as page type) with input value "{{{metaproperty}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.

Statement

Property-theoretic statement

The property of topological spaces of being compact satisfies the metaproperty of being weakly hereditary: in other words, it is inherited by closed subsets.

Verbal statement

Any closed subset of a compact space is compact (when given the subspace topology).

Proof

Proof in terms of open covers

Let $X$ be a compact space and $A$ be a closed subset (given the subspace topology). We need to prove that $A$ is compact.

Let's start with an open cover of $A$ by open sets $U_{i}$ with $i \in I$ , an indexing set. Our goal is to exhibit a finite subcover.

By the definition of subspace topology, we can find open sets $V_{i}$ of $X$ such that $V_{i} \cap A = U_{i}$ , thus the union of the $V_{i}$ s contaiins $A$ .

Since $A$ is closed, we can throw in the open set $X ∖ A$ , and get an open cover of the whole space. But since the whole space is compact, this open cover has a finite subcover. In other words, there is a finite subcollection of the $V_{i}$ s, that, possibly along with $X ∖ A$ , covers the whole of $X$ . By throwing out $X ∖ A$ , we get a finite collection of $V_{i}$ s whose union contains $A$ . The corresponding $U_{i}$ now form a finite subcover of the original cover of $A$ .