Difference between revisions of "Compactness"

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For 1) we may talk about the compactness of subsets if we consider them as [[subspace topology|topological subspaces]]
 
For 1) we may talk about the compactness of subsets if we consider them as [[subspace topology|topological subspaces]]
 
===Definition 1===
 
===Definition 1===
A [[topological space]], {{M|(X,\mathcal{J})}} is ''compact'' if{{rITTGG}}
+
A [[topological space]], {{M|(X,\mathcal{J})}} is ''compact'' if{{rITTGG}}{{rITTBM}}:
 
* Every [[open covering]] of {{M|X}}, {{M|\{U_\alpha\}_{\alpha\in I}\subseteq\mathcal{J} }} contains a ''finite'' [[sub-cover]]
 
* Every [[open covering]] of {{M|X}}, {{M|\{U_\alpha\}_{\alpha\in I}\subseteq\mathcal{J} }} contains a ''finite'' [[sub-cover]]
 
Note that in this definition we'll actually have (if {{M|\{U_\alpha\}_{\alpha\in I} }} is actually a covering) {{M|1=X=\bigcup_{\alpha\in I}U_\alpha}} (notice equality rather than {{M|\subseteq}}, this is because the union on the right cannot contain more than {{M|X}} itself, The elements of {{M|\mathcal{J} }} are subsets of {{M|X}} and the {{M|U_\alpha}} are elements of {{M|\mathcal{J} }} after all.  
 
Note that in this definition we'll actually have (if {{M|\{U_\alpha\}_{\alpha\in I} }} is actually a covering) {{M|1=X=\bigcup_{\alpha\in I}U_\alpha}} (notice equality rather than {{M|\subseteq}}, this is because the union on the right cannot contain more than {{M|X}} itself, The elements of {{M|\mathcal{J} }} are subsets of {{M|X}} and the {{M|U_\alpha}} are elements of {{M|\mathcal{J} }} after all.  
 
====Compactness of a subset====
 
====Compactness of a subset====
A subset, {{M|S\subseteq X}} of a topological space {{M|(X,\mathcal{J})}} is compact if<ref name="ITTGG"/>:
+
A subset, {{M|S\subseteq X}} of a topological space {{M|(X,\mathcal{J})}} is compact if<ref name="ITTGG"/><ref name="ITTBM"/>:
 
* The topology {{M|(S,\mathcal{J}_\text{subspace})}} is compact (the [[subspace topology]] on {{M|S}} inherited from {{M|X}}) as per the definition above.
 
* The topology {{M|(S,\mathcal{J}_\text{subspace})}} is compact (the [[subspace topology]] on {{M|S}} inherited from {{M|X}}) as per the definition above.
 
This maintains compactness as a strictly ''topological'' property.
 
This maintains compactness as a strictly ''topological'' property.
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{{Todo|Find reference}}
 
{{Todo|Find reference}}
 
'''Note that: ''' when {{M|1=S=X}} we get definition 1.
 
'''Note that: ''' when {{M|1=S=X}} we get definition 1.
==Claim 1: The definitions are equivalent==
+
===Claim 1: The definitions are equivalent===
 
These 2 definitions are the same, that is:
 
These 2 definitions are the same, that is:
 +
* Claim 1: A subspace {{M|Y\subseteq X}} is a compact (''def 1'') in {{M|(X,\mathcal{J})}} {{M|\iff}} every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering (''def 2'').
 +
==[[Equivalent statements to compactness of a metric space|Compactness of a metric space]]==
 +
{{:Equivalent statements to compactness of a metric space/Statement}}
 +
(see [[Equivalent statements to compactness of a metric space]] for proof)
 +
==Proof of claims==
 
{{Begin Theorem}}
 
{{Begin Theorem}}
 
Claim 1: A subspace {{M|Y\subseteq X}} is a compact (''def 1'') in {{M|(X,\mathcal{J})}} {{M|\iff}} every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering (''def 2'').
 
Claim 1: A subspace {{M|Y\subseteq X}} is a compact (''def 1'') in {{M|(X,\mathcal{J})}} {{M|\iff}} every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering (''def 2'').

Latest revision as of 15:59, 1 December 2015

See Notes:Compactness and sequences - I think there's a different definition for metric spaces, I have not seen a proof that the metric one this one

This page is currently being refactored (along with many others)
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Definition

There are 2 distinct definitions of compactness, however they are equivalent:

  1. We may only say a topological space is compact, we may not speak of the compactness of subsets. Compactness is strictly a property of topological spaces.
  2. Sure talk about the compactness of subsets of a space.

For 1) we may talk about the compactness of subsets if we consider them as topological subspaces

Definition 1

A topological space, (X,J) is compact if[1][2]:

  • Every open covering of X, {Uα}αIJ contains a finite sub-cover

Note that in this definition we'll actually have (if {Uα}αI is actually a covering) X=αIUα (notice equality rather than , this is because the union on the right cannot contain more than X itself, The elements of J are subsets of X and the Uα are elements of J after all.

Compactness of a subset

A subset, SX of a topological space (X,J) is compact if[1][2]:

  • The topology (S,Jsubspace) is compact (the subspace topology on S inherited from X) as per the definition above.

This maintains compactness as a strictly topological property.

Definition 2

A subset, SX of a topological space (X,J) is compact if:


TODO: Find reference


Note that: when S=X we get definition 1.

Claim 1: The definitions are equivalent

These 2 definitions are the same, that is:

  • Claim 1: A subspace YX is a compact (def 1) in (X,J) every covering of Y by sets open in X contains a finite subcovering (def 2).

Compactness of a metric space

Given a metric space (X,d), the following are equivalent[1][Note 1]:

  1. X is compact
  2. Every sequence in X has a subsequence that converges (AKA: having a convergent subsequence)
  3. X is totally bounded and complete

(see Equivalent statements to compactness of a metric space for proof)

Proof of claims

[Expand]

Claim 1: A subspace YX is a compact (def 1) in (X,J) every covering of Y by sets open in X contains a finite subcovering (def 2).


References

  1. Jump up to: 1.0 1.1 1.2 Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
  2. Jump up to: 2.0 2.1 Introduction to Topology - Bert Mendelson

OLD PAGE

Not to be confused with Sequential compactness


There are two views here.

  1. Compactness is a topological property and we cannot say a set is compact, we say it is compact and implicitly consider it with the subspace topology
  2. We can say "sure that set is compact".

The difference comes into play when we cover a set (take the interval [0,5]R) with open sets. Suppose we have the covering {(1,3),(2,6)} this is already finite and covers the interval. The corresponding sets in the subspace topology are {[0,3),(2,5]} which are both open in the subspace topology.

Definition

That is to say that given an arbitrary collection of sets:

  • A={Aα}αI such that each Aα is open in X and
  • X=αIAα
    [Note 2]

The following is true:

  • {i1,,in}I such that X=α{i1,,in}Aα

Then X is compact[1]

Lemma for a set being compact

Take a set YX

in a topological space (X,J)
. Then to say:

  • Y
    is compact

Means Y

satisfies the definition of compactness when considered as a subspace of (X,J)

[Expand]

Theorem: A set YX is a compact in (X,J) if and only if every covering of Y by sets open in X contains a finite subcovering.


See also

Notes

  1. Jump up To say statements are equivalent means we have one one of the other(s)
  2. Jump up Note that we actually have XαIAα but as topologies are closed under arbitrary union and contain the set the open sets are subsets of we cannot "exceed X", so we must have X=αIAα

References

  1. Jump up to: 1.0 1.1 Topology - James R. Munkres - Second Edition