Compactness is weakly hereditary

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
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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

Given: $X$ a compact space, $A$ a closed subset (given the subspace topology)

To prove: $A$ is compact

Proof: 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 contains $A$.

Since $A$ is closed, we can throw in the open set $X\setminus A$, 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 $X\setminus A$, covers the whole of $X$. By throwing out $X\setminus A$, 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 $A$.

Proof in terms of finite intersection property

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References

Textbook references

• Topology (2nd edition) by James R. Munkres, ^{More info}, Page 165, Theorem 26.2, Chapter 3, Section 26
• Lecture Notes on Elementary Topology and Geometry (Undergraduate Texts in Mathematics) by I. M. Singer and J. A. Thorpe, ^{More info}, Page 12 (Theorem 4)