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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 $\implies$ this one

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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, $(X,\mathcal{J})$ is compact if^[1]^[2]:

Every open covering of $X$ , $\{U_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}$ contains a finite sub-cover

Note that in this definition we'll actually have (if $\{U_\alpha\}_{\alpha\in I}$ is actually a covering) $X=\bigcup_{\alpha\in I}U_\alpha$ (notice equality rather than $\subseteq$ , this is because the union on the right cannot contain more than $X$ itself, The elements of $\mathcal{J}$ are subsets of $X$ and the $U_\alpha$ are elements of $\mathcal{J}$ after all.

Compactness of a subset

A subset, $S\subseteq X$ of a topological space $(X,\mathcal{J})$ is compact if^[1]^[2]:

The topology $(S,\mathcal{J}_\text{subspace})$ 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, $S\subseteq X$ of a topological space $(X,\mathcal{J})$ is compact if:

Every covering by sets open in $X$ of $S$ contains a finite sub-cover

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 $Y\subseteq X$ is a compact (def 1) in $(X,\mathcal{J})$ $\iff$ 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

[Expand]
Claim 1: A subspace $Y\subseteq X$ is a compact (def 1) in $(X,\mathcal{J})$ $\iff$ every covering of $Y$ by sets open in $X$ contains a finite subcovering (def 2).

Suppose that $(Y,\mathcal{J}_\text{subspace})$ is compact $\implies$ every covering consisting of open sets of $(X,\mathcal{J})$ contains a finite subcover.

Let

$\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}$ be a family of open sets in

$X$ with

$Y\subseteq\cup_{\alpha\in I}A_\alpha$

Take

$B_\alpha=A_\alpha\cap Y$ , then

$\{B_\alpha\}_{\alpha\in I}$ is an open (in

$Y$ ) covering of

$Y$ , that is

$Y\subseteq\cup_{\alpha\in I}B_\alpha$ (infact we have

$Y=\cup_{\alpha\in I}B_\alpha$ )

[Expand]

Proof of $Y\subseteq\cup_{\alpha\in I}B_\alpha$ (we actually have

$Y=\cup_{\alpha\in I}B_\alpha$ )

We wish to show that

$Y\subseteq\cup_{\alpha\in I}A_\alpha\implies Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)$ , using the Implies-subset relation we actually just want to show that:

Given $Y\subseteq\cup_{\alpha\in I}A_\alpha$ that $y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ - which is what we'll do.
Note additionally that $y\in\cup_{\alpha\in I}(A_i\cap Y)\iff \exists\beta\in I[y\in A_\beta\wedge y\in Y]$

Let

$y\in Y$ , then by hypothesis

$y\in\cup_{\alpha\in I}A_\alpha\iff\exists \beta\in I[y\in A_\beta]$

It is easily seen that

$y\in Y\wedge\exists\beta\in I[y\in A_\beta]\implies\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]$ simply by choosing

$\gamma:=\beta$ .

Lastly, note that

$\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]\iff y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$

We have shown that $y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ and by the Implies-subset relation we see
$Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)$ - as required.

[Expand]

I earlier claimed that actually

$Y=\cup_{\alpha\in I}(A_\alpha\cap Y)$ - this isn't important to the proof but it shows something else.

This shows that considering an open covering as a union of sets open in

$Y$ whose union is exactly

$Y$ is the same as a covering by open sets in

$X$ whose union contains (but need not be exactly equal to)

$Y$ . So we have shown so far that:

Compact in the subspace with equality for an open covering $\implies$ compact with the open cover of sets in $X$ whose union contains $Y$

Claim:

$\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y$

Let

$y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ then:

$\exists\beta\in I[y\in A_\beta\wedge y\in Y]\iff \exists\beta\in I[y\in(A_\beta\cap Y)]$

As the intersection of sets is a subset of each set we see that (it's trivial to show without this result too, but this uses a general result)

$(A_\beta\cap Y)\subseteq Y)$ by the implies-subset relation we see immediately that:

$y\in(A_\beta\cap Y)\implies y\in Y$

Thus we have shown that

$y\in\cup_{\alpha\in I}(A_\alpha\cap Y)\implies y\in Y$ and finally this means:

$\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y$

Combining this with

$Y\subseteq \cup_{\alpha\in I}(A_\alpha\cap Y)$ above we see that:

$Y=\cup_{\alpha\in I}(A_\alpha\cap Y)$

This completes the proof

By hypothesis,

$Y$ is compact, this means that

$\{B_\alpha\}_{\alpha\in I}$ contains a finite subcover

call this subcover $\{B'_i\}_{i=1}^n$ where each $B'_i\in\{B_\alpha\}_{\alpha\in I}$ , now we have $Y\subseteq\cup_{i=1}^n B'_i$ (we actually have equality, see the blue box in the yellow note box above)

As each

$B'_i=A'_i\cap Y$ (where

$A'_i$ is the corresponding

$A_\alpha$ for the

$B_\alpha$ that

$B'_i$ represents) we see that

$\{A_i\}_{i=1}^n$ is a finite subcover by sets open in

$X$

[Expand]

Proof of:

$Y\subseteq\cup_{i=1}^nB'_i\implies$

$Y\subseteq\cup_{i=1}^nA'_i$ (proving that

$\{A'_i\}_{i=1}^n$ is an open cover)

For each

$i$ we have

$B'_i:=A'_i\cap Y$ , by invoking the intersection of sets is a subset of each set we note that:

$B'_i\subseteq A'_i$

We now invoke Union of subsets is a subset of the union

This theorem states that given two families of sets, $\{A_\alpha\}_{\alpha\in I}$ and $\{B_\alpha\}_{\alpha\in I}$ with $\forall\alpha\in I[B_\alpha\subseteq A_\alpha]$ we have $\cup_{\alpha\in I}B_\alpha\subseteq\cup_{\alpha\in I}A_\alpha$

It follows that

$Y\subseteq\cup_{i=1}^nB'_i\subseteq\cup_{i=1}^nA'_i$ , in particular:

$Y\subseteq\cup_{i=1}^nA'_i$

This confirms that

$\{A'_i\}_{i=1}$ is an open cover by sets in

$X$

This completes this half of the proof.

$(Y,\mathcal{J}_\text{subspace})$ is compact every covering of $Y$ by sets open in $X$ contains a finite subcovering

Suppose that every covering of

$Y$ by sets open in

$X$ contains a finite subcollection covering

$Y$ . We need to show

$Y$ is compact.

Suppose we have a covering,

$\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}$ of

$Y$ by sets open in

$Y$

For each

$\alpha$ choose an open set

$A_\alpha$ open in

$X$ such that:

$A'_\alpha=A_\alpha\cap Y$

Then the collection

$\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$ covers

$Y$

By hypothesis we have a finite sub-collection from

$\mathcal{A}$ of things open in

$X$ that cover

$Y$

Thus the corresponding finite subcollection of

$\mathcal{A}'$ covers

$Y$

References

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 $[0,5]\subset\mathbb{R}$ ) 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 $X$ contains a finite sub-covering that also covers $X$ .

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

$\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$ such that each $A_\alpha$ is open in $X$ and
$X=\bigcup_{\alpha\in I}A_\alpha$ ^{[Note 2]}

The following is true:

$\exists \{i_1,\cdots,i_n\}\subset I$ such that $X=\bigcup_{\alpha\in\{i_1,\cdots,i_n\} }A_\alpha$

Then $X$ is compact^[1]

Lemma for a set being compact

Take a set

$Y\subset X$ in a topological space

$(X,\mathcal{J})$ . Then to say:

$Y$ is compact

Means

$Y$ satisfies the definition of compactness when considered as a subspace of

$(X,\mathcal{J})$

[Expand]
Theorem: A set $Y\subseteq X$ is a compact in $(X,\mathcal{J})$ if and only if every covering of $Y$ by sets open in $X$ contains a finite subcovering.

Suppose that $(Y,\mathcal{J}_\text{subspace})$ is compact $\implies$ every covering consisting of open sets of $(X,\mathcal{J})$ contains a finite subcover.

Let

$\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}$ be a family of open sets in

$X$ with

$Y\subseteq\cup_{\alpha\in I}A_\alpha$

Take

$B_\alpha=A_\alpha\cap Y$ , then

$\{B_\alpha\}_{\alpha\in I}$ is an open (in

$Y$ ) covering of

$Y$ , that is

$Y\subseteq\cup_{\alpha\in I}B_\alpha$ (infact we have

$Y=\cup_{\alpha\in I}B_\alpha$ )

[Expand]

Proof of $Y\subseteq\cup_{\alpha\in I}B_\alpha$ (we actually have

$Y=\cup_{\alpha\in I}B_\alpha$ )

We wish to show that

$Y\subseteq\cup_{\alpha\in I}A_\alpha\implies Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)$ , using the Implies-subset relation we actually just want to show that:

Given $Y\subseteq\cup_{\alpha\in I}A_\alpha$ that $y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ - which is what we'll do.
Note additionally that $y\in\cup_{\alpha\in I}(A_i\cap Y)\iff \exists\beta\in I[y\in A_\beta\wedge y\in Y]$

Let

$y\in Y$ , then by hypothesis

$y\in\cup_{\alpha\in I}A_\alpha\iff\exists \beta\in I[y\in A_\beta]$

It is easily seen that

$y\in Y\wedge\exists\beta\in I[y\in A_\beta]\implies\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]$ simply by choosing

$\gamma:=\beta$ .

Lastly, note that

$\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]\iff y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$

We have shown that $y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ and by the Implies-subset relation we see
$Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)$ - as required.

[Expand]

I earlier claimed that actually

$Y=\cup_{\alpha\in I}(A_\alpha\cap Y)$ - this isn't important to the proof but it shows something else.

This shows that considering an open covering as a union of sets open in

$Y$ whose union is exactly

$Y$ is the same as a covering by open sets in

$X$ whose union contains (but need not be exactly equal to)

$Y$ . So we have shown so far that:

Compact in the subspace with equality for an open covering $\implies$ compact with the open cover of sets in $X$ whose union contains $Y$

Claim:

$\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y$

Let

$y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ then:

$\exists\beta\in I[y\in A_\beta\wedge y\in Y]\iff \exists\beta\in I[y\in(A_\beta\cap Y)]$

As the intersection of sets is a subset of each set we see that (it's trivial to show without this result too, but this uses a general result)

$(A_\beta\cap Y)\subseteq Y)$ by the implies-subset relation we see immediately that:

$y\in(A_\beta\cap Y)\implies y\in Y$

Thus we have shown that

$y\in\cup_{\alpha\in I}(A_\alpha\cap Y)\implies y\in Y$ and finally this means:

$\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y$

Combining this with

$Y\subseteq \cup_{\alpha\in I}(A_\alpha\cap Y)$ above we see that:

$Y=\cup_{\alpha\in I}(A_\alpha\cap Y)$

This completes the proof

By hypothesis,

$Y$ is compact, this means that

$\{B_\alpha\}_{\alpha\in I}$ contains a finite subcover

call this subcover $\{B'_i\}_{i=1}^n$ where each $B'_i\in\{B_\alpha\}_{\alpha\in I}$ , now we have $Y\subseteq\cup_{i=1}^n B'_i$ (we actually have equality, see the blue box in the yellow note box above)

As each

$B'_i=A'_i\cap Y$ (where

$A'_i$ is the corresponding

$A_\alpha$ for the

$B_\alpha$ that

$B'_i$ represents) we see that

$\{A_i\}_{i=1}^n$ is a finite subcover by sets open in

$X$

[Expand]

Proof of:

$Y\subseteq\cup_{i=1}^nB'_i\implies$

$Y\subseteq\cup_{i=1}^nA'_i$ (proving that

$\{A'_i\}_{i=1}^n$ is an open cover)

For each

$i$ we have

$B'_i:=A'_i\cap Y$ , by invoking the intersection of sets is a subset of each set we note that:

$B'_i\subseteq A'_i$

We now invoke Union of subsets is a subset of the union

This theorem states that given two families of sets, $\{A_\alpha\}_{\alpha\in I}$ and $\{B_\alpha\}_{\alpha\in I}$ with $\forall\alpha\in I[B_\alpha\subseteq A_\alpha]$ we have $\cup_{\alpha\in I}B_\alpha\subseteq\cup_{\alpha\in I}A_\alpha$

It follows that

$Y\subseteq\cup_{i=1}^nB'_i\subseteq\cup_{i=1}^nA'_i$ , in particular:

$Y\subseteq\cup_{i=1}^nA'_i$

This confirms that

$\{A'_i\}_{i=1}$ is an open cover by sets in

$X$

This completes this half of the proof.

$(Y,\mathcal{J}_\text{subspace})$ is compact every covering of $Y$ by sets open in $X$ contains a finite subcovering

Suppose that every covering of

$Y$ by sets open in

$X$ contains a finite subcollection covering

$Y$ . We need to show

$Y$ is compact.

Suppose we have a covering,

$\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}$ of

$Y$ by sets open in

$Y$

For each

$\alpha$ choose an open set

$A_\alpha$ open in

$X$ such that:

$A'_\alpha=A_\alpha\cap Y$

Then the collection

$\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$ covers

$Y$

By hypothesis we have a finite sub-collection from

$\mathcal{A}$ of things open in

$X$ that cover

$Y$

Thus the corresponding finite subcollection of

$\mathcal{A}'$ covers

$Y$

Notes

Jump up ↑ To say statements are equivalent means we have one $\iff$ one of the other(s)
Jump up ↑ Note that we actually have $X\subseteq\bigcup_{\alpha\in I}A_\alpha$ 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 $X=\bigcup_{\alpha\in I}A_\alpha$

References

↑ ^{Jump up to: 1.0} ^1.1 Topology - James R. Munkres - Second Edition

[ITTGG-1] {Jump up to: 1.0} ^1.1 ^1.2 Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene

[ITTBM-2] {Jump up to: 2.0} ^2.1 Introduction to Topology - Bert Mendelson

[3] Jump up ↑ To say statements are equivalent means we have one $\iff$ one of the other(s)

[5] Jump up ↑ Note that we actually have $X\subseteq\bigcup_{\alpha\in I}A_\alpha$ 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 $X=\bigcup_{\alpha\in I}A_\alpha$

[Topology-4] {Jump up to: 1.0} ^1.1 Topology - James R. Munkres - Second Edition

[1]

[2]

[Note 1]

[1]

[Note 2]

@@ Line 1: / Line 1: @@
+'''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 {{M|\implies}} this one
+{{Refactor notice}}
+==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 [[subspace topology|topological subspaces]]
+===Definition 1===
+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]]
+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====
+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.
+This maintains compactness as a strictly ''topological'' property.
+===Definition 2===
+A subset, {{M|S\subseteq X}} of a topological space {{M|(X,\mathcal{J})}} is compact if:
+* Every [[covering]] by sets [[open set|open]] in {{M|X}} of {{M|S}} contains a ''finite'' [[sub-cover]]
+{{Todo|Find reference}}
+'''Note that: ''' when {{M|1=S=X}} we get definition 1.
+===Claim 1: The definitions are equivalent===
+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}}
+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'').
+{{Begin Proof}}
+{{:Compactness/Uniting covers proof}}
+{{End Proof}}
+{{End Theorem}}
+==References==
+<references/>
+{{Definition|Metric Space|Topology}}
+=OLD PAGE=
 Not to be confused with [[Sequential compactness]]
@@ Line 7: / Line 50: @@
 The difference comes into play when we cover a set (take the interval {{m|[0,5]\subset\mathbb{R} }}) with open sets. Suppose we have the covering {{M|\{(-1,3),(2,6)\} }} this is already finite and covers the interval. The corresponding sets in the subspace topology are {{M|\{[0,3),(2,5]\} }} which are both open in the subspace topology.
 ==Definition==
-A [[Topological space|topological space]] is compact if every [[Covering|open cover]] (often denoted <math>\mathcal{A}</math>) of <math>X</math> contains a finite sub-collection that also covers <math>X</math>
+*A [[Topological space|topological space]] is compact<ref name="Topology">Topology - James R. Munkres - Second Edition</ref> if every [[Covering|open cover]] of <math>X</math> contains a finite sub-covering that also covers <math>X</math>.
+That is to say that given an arbitrary collection of sets:
+* {{M|1=\mathcal{A}=\{A_\alpha\}_{\alpha\in I} }} such that each {{M|A_\alpha}} is [[Open set|open]] in {{M|X}} and
+* {{MM|1=X=\bigcup_{\alpha\in I}A_\alpha}}<ref group="Note">Note that we actually have {{M|X\subseteq\bigcup_{\alpha\in I}A_\alpha}} but as topologies are closed under arbitrary union and contain the set the open sets are subsets of we cannot "exceed {{M|X}}", so we must have {{M|1=X=\bigcup_{\alpha\in I}A_\alpha}}</ref>
+The following is true:
+* {{M|1=\exists \{i_1,\cdots,i_n\}\subset I}} such that {{MM|1=X=\bigcup_{\alpha\in\{i_1,\cdots,i_n\} }A_\alpha}}
+Then {{M|X}} is ''compact''<ref name="Topology"/>
 ==Lemma for a set being compact==
-Take a set <math>Y\subset X</math> in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math>.
+Take a set <math>Y\subset X</math> in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math>. Then to say:
+* <math>Y</math> is compact
-To say <math>Y</math> is compact is for <math>Y</math> to be compact when considered as a [[Subspace topology|subspace]] of <math>(X,\mathcal{J})</math>
+Means <math>Y</math> satisfies the definition of compactness when considered as a [[Subspace topology|subspace]] of <math>(X,\mathcal{J})</math>
-That is to say that <math>Y</math> is compact if and only if every covering of <math>Y</math> by sets open in <math>X</math> contains a finite subcovering covering <math>Y</math>
 {{Begin Theorem}}
-Theorem: A set {{M|Y\subseteq X}} is a compact space (considered with the subspace topology) of {{M|(X,\mathcal{J})}} ''if and only if'' every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering.
+Theorem: A set {{M|Y\subseteq X}} is a compact in {{M|(X,\mathcal{J})}} ''if and only if'' every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering.
 {{Begin Proof}}
-'''{{M|(Y,\mathcal{J}_\text{subspace})}} is compact {{M|\implies}} every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering'''
+{{:Compactness/Uniting covers proof}}
-:Suppose that the space <math>(Y,\mathcal{J}_\text{subspace})</math> is compact and that <math>\mathcal{A}=\{A_\alpha\}_{\alpha\in I}</math> (where each <math>A_\alpha\in\mathcal{J}</math> - that is each set is open in <math>X</math>) is an open covering (which is to say {{M|Y\subseteq\cup_{\alpha\in I}A_\alpha}})
+{{End Proof}}
+{{End Theorem}}
-:Then the collection <math>\{A_\alpha\cap Y|\alpha\in I\}</math> is a covering of <math>Y</math> by sets open in <math>Y</math> (by definition of [[Subspace topology|being a subspace]])
+==See also==
+* [[Subspace topology]]
-:By hypothesis <math>Y</math> is compact, hence a finite sub-collection <math>\{A_{\alpha_i}\cap Y\}^n_{i=1}</math> covers <math>Y</math> '''(as to be compact ''every'' open cover must have a finite subcover)'''
+==Notes==
+<references group="Note"/>
-:Then <math>\{A_{\alpha_i}\}^n_{i=1}</math> is a sub-collection of <math>\mathcal{A}</math> that covers <math>Y</math>.
+==References==
+<references/>
-'''Proof of details'''
-:As [[The intersection of sets is a subset of each set]] and <math>\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y</math> we see <br />
-:<math>x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y</math> <math>\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}</math><br />
-:The important part being <math>x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}</math><br />
-:then by the [[Implies and subset relation|implies and subset relation]] we have <math>Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}</math> and conclude <math>Y\subset\cup^n_{i=1}A_{\alpha_i}</math>
-:'''Warning:''' this next bit looks funny - do not count on!
-::Lastly, as <math>\mathcal{A}</math> was a covering <math>\cup_{\alpha\in I}A_\alpha=Y</math>.
-::It is clear that <math>x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha</math> so again  [[Implies and subset relation|implies and subset relation]] we have:<br />
-::<math>\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y</math> thus concluding <math>\cup^n_{i=1}A_{\alpha_i}\subset Y</math>
-::Combining <math>Y\subset\cup^n_{i=1}A_{\alpha_i}</math> and <math>\cup^n_{i=1}A_{\alpha_i}\subset Y</math> we see <math>\cup^n_{i=1}A_{\alpha_i}=Y</math>
-::Thus <math>\{A_{\alpha_i}\}^n_{i=1}</math> is a finite covering of <math>Y</math> consisting of '''open''' sets from <math>X</math>
-'''{{M|(Y,\mathcal{J}_\text{subspace})}} is compact <math>\impliedby</math> every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering'''
-:Suppose that every covering of <math>Y</math> by sets open in <math>X</math> contains a finite subcollection covering <math>Y</math>. We need to show <math>Y</math> is compact.
-:Suppose we have a covering, <math>\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}</math> of <math>Y</math> by sets open in <math>Y</math>
-:For each <math>\alpha</math> choose an open set <math>A_\alpha</math> open in <math>X</math> such that: <math>A'_\alpha=A_\alpha\cap Y</math>
-:Then the collection <math>\mathcal{A}=\{A_\alpha\}_{\alpha\in I}</math> covers <math>Y</math>
-:By hypothesis we have a finite sub-collection from {{M|\mathcal{A} }} of things open in <math>X</math> that cover <math>Y</math>
-:Thus the corresponding finite subcollection of <math>\mathcal{A}'</math> covers <math>Y</math>
-{{End Proof}}
-{{End Theorem}}
 {{Definition|Topology}}

Difference between revisions of "Compactness"

Latest revision as of 15:59, 1 December 2015

Contents

Definition

Definition 1

Compactness of a subset

Definition 2

Claim 1: The definitions are equivalent

Compactness of a metric space

Proof of claims

References

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Definition

Lemma for a set being compact

See also

Notes

References

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