Compactness is weakly hereditary: Difference between revisions

Revision as of 02:01, 17 July 2009

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).

Related facts

Hausdorff implies KC: In other words, every compact subset of a Hausdorff space is a closed subset.

Weakly hereditary for properties related to compactness

Paracompactness is weakly hereditary: Every closed subset of a paracompact space is paracompact.
Orthocompactness is weakly hereditary
Metacompactness is weakly hereditary

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:

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 $X$ .
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

Fill this in later

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)

@@ Line 12: / Line 12: @@
 Any [[closed subset]] of a [[compact space]] is compact (when given the [[subspace topology]]).
+==Related facts==
+* [[Hausdorff implies KC]]: In other words, every compact subset of a Hausdorff space is a [[closed subset]].
+===Weakly hereditary for properties related to compactness===
+* [[Paracompactness is weakly hereditary]]: Every [[closed subset]] of a paracompact space is paracompact.
+* [[Orthocompactness is weakly hereditary]]
+* [[Metacompactness is weakly hereditary]]
 ==Proof==
@@ Line 19: / Line 30: @@
 '''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology)
-'''To prove''': <math>A</math> is compact
+'''To prove''': Consider an open cover of <math>A</math> by open sets <math>U_i</math> with <math>i \in I</math>, an indexing set. The <math>U_i</math> have a finite subcover.
-'''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>.
+'''Proof''':
-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>.
+# 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'' <math>X</math>.
+# 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===