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
Contents
[hide]Definition
There are 2 distinct definitions of compactness, however they are equivalent:
- 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.
- 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}α∈I⊆J contains a finite sub-cover
Note that in this definition we'll actually have (if {Uα}α∈I is actually a covering) X=⋃α∈IUα (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, S⊆X 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, S⊆X 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 Y⊆X 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]:
- X is compact
- Every sequence in X has a subsequence that converges (AKA: having a convergent subsequence)
- X is totally bounded and complete
(see Equivalent statements to compactness of a metric space for proof)
Proof of claims
Claim 1: A subspace Y⊆X is a compact (def 1) in (X,J) ⟺ every covering of Y by sets open in X contains a finite subcovering (def 2).
References
- ↑ Jump up to: 1.0 1.1 1.2 Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene
- ↑ 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.
- 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
- 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
- A topological space is compact[1] if every open cover of Xcontains a finite sub-covering that also covers X.
That is to say that given an arbitrary collection of sets:
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 Y⊂X
- Yis compact
Means Y
Theorem: A set Y⊆X 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
- Jump up ↑ To say statements are equivalent means we have one ⟺ one of the other(s)
- Jump up ↑ Note that we actually have X⊆⋃α∈IAα 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=⋃α∈IAα
References
- ↑ Jump up to: 1.0 1.1 Topology - James R. Munkres - Second Edition