Compactness is weakly hereditary: Difference between revisions

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'''Proof''':  
'''Proof''':  


# 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>.
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# 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>.  
! Step no. !! Assertion/construction !! Given data used !! Facts used !! Previous steps used !! Explanation
# 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>.  
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# 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>.
| 1 ||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>. || <math>A</math> is a subspace of <math>X</math> || -- || --
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| 2 || The <math>V_i</math>s along with <math>X \setminus A</math> form an open cover of <math>X</math> || <math>A</math> is closed in <math>X</math> || -- || Step (1) || <toggledisplay><math>X \setminus A</math> is open because <math>A</math> is closed. Further, since the union of all the <math>V_i</math>s contains <math>A</math>, that along with <math>X \setminus A</math> covers all of <math>X</math>.</toggledisplay>
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| 3 || The open cover from step (2) has a finite subcover. In other words, there is a finite subcollection of the <math>V_i</math>s, that, along with <math>X \setminus A</math>, covers <math>X</math>. || <math>X</math> is compact || -- || Step (2) ||
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| 4 || 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> || -- || -- || Step (3) ||
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| 5 || The corresponding <math>U_i</math> now form a finite subcover of the original cover of <math>A</math>. || -- || -- || Steps (1), (4) ||
|}


===Proof in terms of finite intersection property===
===Proof in terms of finite intersection property===
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==References==
==References==



Revision as of 14:55, 15 September 2010

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)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about compact space |Get facts that use property satisfaction of compact space | Get facts that use property satisfaction of compact space|Get more facts about weakly hereditary property of topological spaces

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

Weakly hereditary for properties related to compactness


Proof

Proof in terms of open covers

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

To prove: Consider an open cover of by open sets with , an indexing set. The 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 of such that , thus the union of the s contains . is a subspace of -- --
2 The s along with form an open cover of is closed in -- 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 s, that, along with , covers . is compact -- Step (2)
4 By throwing out , we get a finite collection of s whose union contains -- -- Step (3)
5 The corresponding now form a finite subcover of the original cover of . -- -- 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)