Compactness is weakly hereditary: Difference between revisions
No edit summary |
(→Proof) |
||
| Line 34: | Line 34: | ||
'''Proof''': | '''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== | ||
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
- 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
Fill this in later