Compactness is weakly hereditary: Difference between revisions
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property = compact space| | property = compact space| | ||
metaproperty = weakly hereditary property of topological spaces}} | metaproperty = weakly hereditary property of topological spaces}} | ||
[[Difficulty level::1| ]] | |||
==Statement== | ==Statement== | ||
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Any [[closed subset]] of a [[compact space]] is compact (when given the [[subspace topology]]). | 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== | ||
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'''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology) | '''Given''': <math>X</math> a [[compact space]], <math>A</math> a [[closed subset]] (given the subspace topology) | ||
'''To prove''': | '''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''': | |||
{| class="sortable" border="1" | |||
! 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 <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> || -- || -- || | |||
|- | |||
| 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> | |||
|- | |||
| 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) || | |||
|- | |||
| 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) || | |||
|- | |||
| 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=== | ||
{{fillin}} | {{fillin}} | ||
==References== | ==References== | ||
Latest revision as of 04:26, 30 January 2014
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
- Hausdorff implies KC: In other words, every compact subset of a Hausdorff space is a closed subset.
- 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: 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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