# Compactness is weakly hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., compact space) satisfying a topological space metaproperty (i.e., weakly hereditary property of topological spaces)
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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: Consider an open cover of $A$ by open sets $U_i$ with $i \in I$, an indexing set. The $U_i$ have a finite subcover.

Proof:

Step no. Assertion/construction Given data used Facts used Previous steps used Explanation
1 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 $A$. $A$ is a subspace of $X$ -- --
2 The $V_i$s along with $X \setminus A$ form an open cover of $X$ $A$ is closed in $X$ -- Step (1) [SHOW MORE]
3 The open cover from step (2) has a finite subcover. In other words, there is a finite subcollection of the $V_i$s, that, along with $X \setminus A$, covers $X$. $X$ is compact -- Step (2)
4 By throwing out $X \setminus A$, we get a finite collection of $V_i$s whose union contains $A$ -- -- Step (3)
5 The corresponding $U_i$ now form a finite subcover of the original cover of $A$. -- -- Steps (1), (4)

### 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)