Compactness is weakly hereditary: Difference between revisions

Revision as of 19:51, 20 July 2008

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
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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

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 $X ∖ A$ , 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 $A$ .

Proof in terms of finite intersection property

Fill this in later

References

Textbook references

Topology (2nd edition) by James R. Munkres^{More info}, Page 165 (Theorem 26.2)
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)

@@ Line 15: / Line 15: @@
 ===Proof in terms of open covers===
-Let <math>X</math> be a [[compact space]] and <math>A</math> be a [[closed subset]] (given the subspace topology). We need to prove that <math>A</math> is compact.
+'''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology)
-Let's start with an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. Our goal is to exhibit a finite subcover.
+'''To prove''': <math>A</math> is compact
-By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contaiins <math>A</math>.
+'''Proof''': Let's start with an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. Our goal is to exhibit a finite subcover.
+By the definition of subspace topology, we can find open sets <math>V_i</math> of <math>X</math> such that <math>V_i \cap A = U_i</math>, thus the union of the <math>V_i</math>s contains <math>A</math>.
 Since <math>A</math> is closed, we can ''throw in'' the open set <math>X \setminus A</math>, 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 <math>V_i</math>s, that, possibly along with <math>X \setminus A</math>, covers the whole of <math>X</math>. By ''throwing out'' <math>X \setminus A</math>, we get a finite collection of <math>V_i</math>s whose union contains <math>A</math>. The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>.
+===Proof in terms of finite intersection property===
+{{fillin}}
 ==References==