View source for Compactness/Uniting covers proof

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==Theorem statement==
{{Yellow Note|
*A subset {{M|Y\subseteq X}} of a [[Topological space|topological space]] {{M|(X,\mathcal{J})}} is [[Compactness|compact]] (when {{M|Y}} is imbued with the [[Subspace topology|subspace topology]])
''if and only if''
* Every [[Covering|cover]] by sets open in {{M|X}} has a finite subcover. }}
This is very important in uniting the two definitions of "open cover" used with compactness. These are:
# A covering is a collection of sets whose union ''contains'' {{M|Y}}
#* That is a collection {{M|1=\{A_\alpha\}_{\alpha\in I} }} where {{M|Y\subseteq\cup_{\alpha\in I}A_\alpha}}
# A covering is a collection of sets whose union ''is exactly'' {{M|Y}}, by sets open in {{M|Y}} when {{M|Y}} has the subspace topology.
#* That is a collection {{M|1=\{A_\alpha\}_{\alpha\in I} }} where {{M|1=Y=\cup_{\alpha\in I}A_\alpha}}
Lastly recall:
*A subcovering is a collection of elements of a cover whose union also cover {{M|Y}}

I use the definition of "contains" because to cover something is to overlap it, like cover a dog with a blanket, or cover a face with a pillow. 
==Proof==
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'''Suppose that {{M|(Y,\mathcal{J}_\text{subspace})}} is compact {{M|\implies}} every covering consisting of open sets of {{M|(X,\mathcal{J})}} contains a finite subcover.
: Let {{M|1=\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{J} }} be a family of open sets in {{M|X}} with {{M|Y\subseteq\cup_{\alpha\in I}A_\alpha}}
: Take {{M|1=B_\alpha=A_\alpha\cap Y}}, then {{M|\{B_\alpha\}_{\alpha\in I} }} is an open (in {{M|Y}}) covering of {{M|Y}}, that is {{M|Y\subseteq\cup_{\alpha\in I}B_\alpha}} (infact we have {{M|1=Y=\cup_{\alpha\in I}B_\alpha}})
{{Begin Notebox}}
:: '''Proof of {{M|Y\subseteq\cup_{\alpha\in I}B_\alpha}}''' (we actually have {{M|1=Y=\cup_{\alpha\in I}B_\alpha}})
{{Begin Notebox Content}}
::: We wish to show that {{M|1=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 {{M|Y\subseteq\cup_{\alpha\in I}A_\alpha}} that {{M|1=y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)}} - which is what we'll do.
:::* Note additionally that {{M|1=y\in\cup_{\alpha\in I}(A_i\cap Y)\iff \exists\beta\in I[y\in A_\beta\wedge y\in Y]}}
::: Let {{M|1=y\in Y}}, then by hypothesis {{M|y\in\cup_{\alpha\in I}A_\alpha\iff\exists \beta\in I[y\in A_\beta]}}
::: It is easily seen that {{M|1=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 {{M|1=\gamma:=\beta}}.
::: Lastly, note that {{M|1=\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 {{M|y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)}} and by the [[Implies-subset relation]] we see
:::* {{M|1=Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)}} - as required.
{{Blue Notebox|
:: I earlier claimed that actually {{M|1=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 {{M|Y}} whose union is exactly {{M|Y}} is the same as a covering by open sets in {{M|X}} whose union contains (but need not be exactly equal to) {{M|Y}}. So we have shown so far that:
::* Compact in the subspace with equality for an open covering {{M|\implies}} compact with the open cover of sets in {{M|X}} whose union contains {{M|Y}}
:: '''Claim:''' {{M|1=\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y}}
::: Let {{M|y\in\cup_{\alpha\in I}(A_\alpha\cap Y)}} then:
:::* {{M|\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)
:::* {{M|(A_\beta\cap Y)\subseteq Y)}} by the [[implies-subset relation]] we see immediately that:
::: {{M|y\in(A_\beta\cap Y)\implies y\in Y}}
::: Thus we have shown that {{M|1=y\in\cup_{\alpha\in I}(A_\alpha\cap Y)\implies y\in Y}} and finally this means:
:::* {{M|\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y}}
::: Combining this with {{M|Y\subseteq \cup_{\alpha\in I}(A_\alpha\cap Y)}} above we see that:
:::* {{M|1=Y=\cup_{\alpha\in I}(A_\alpha\cap Y)}}
::: This completes the proof
}}
{{End Notebox Content}}{{End Notebox}}
: By hypothesis, {{M|Y}} is compact, this means that {{M|\{B_\alpha\}_{\alpha\in I} }} contains a finite subcover
:* call this subcover {{M|1=\{B'_i\}_{i=1}^n}} where each {{M|B'_i\in\{B_\alpha\}_{\alpha\in I} }}, now we have {{M|1=Y\subseteq\cup_{i=1}^n B'_i}} (we actually have equality, see the blue box in the yellow note box above)
: As each {{M|1=B'_i=A'_i\cap Y}} (where {{M|A'_i}} is the corresponding {{M|A_\alpha}} for the {{M|B_\alpha}} that {{M|B'_i}} represents) we see that {{M|1=\{A_i\}_{i=1}^n}} is a finite subcover by sets open in {{M|X}}
{{Notebox|
:: '''Proof of:''' {{M|1=Y\subseteq\cup_{i=1}^nB'_i\implies}} {{M|1=Y\subseteq\cup_{i=1}^nA'_i}} (proving that {{M|1=\{A'_i\}_{i=1}^n}} is an open cover)
|
: For each {{M|i}} we have {{M|1=B'_i:=A'_i\cap Y}}, by invoking [[the intersection of sets is a subset of each set]] we note that:
:* {{M|B'_i\subseteq A'_i}}
: We now invoke [[Union of subsets is a subset of the union]]
{{Note|This theorem states that given two families of sets, {{M|\{A_\alpha\}_{\alpha\in I} }} and {{M|\{B_\alpha\}_{\alpha\in I} }} with {{M|\forall\alpha\in I[B_\alpha\subseteq A_\alpha]}} we have {{M|\cup_{\alpha\in I}B_\alpha\subseteq\cup_{\alpha\in I}A_\alpha}} }}
: It follows that {{M|1=Y\subseteq\cup_{i=1}^nB'_i\subseteq\cup_{i=1}^nA'_i}}, in particular:
:* {{M|1=Y\subseteq\cup_{i=1}^nA'_i}}
: This confirms that {{M|1=\{A'_i\}_{i=1} }} is an open cover by sets in {{M|X}}
}}
: This completes this half of the proof.



'''{{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>
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==Notes==
This theorem shows that the two are the same, that is:
* A set {{M|Y\subseteq X}} is compact if every open cover of {{M|Y}} by sets open in {{M|X}} has a finite subcover (that is {{M|Y\subset\text{the covering} }})
and
* A set {{M|Y\subseteq X}} is compact if, when given the [[Subspace topology]], every covering of {{M|Y}} by sets open in {{M|Y}} has a finite subcover.
Are the same, (an ''if and only if'' relationship)

{{Todo|Make the second part of the proof a bit more hand holdy, it's valid but some of the set theoretic stuff could use proofs}}
{{Todo|References for the definitions and stuff - they are CERTAINLY valid}}
{{Theorem Of|Topology}}
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