Compactness/Uniting covers proof
Theorem statement
- A subset [ilmath]Y\subseteq X[/ilmath] of a topological space [ilmath](X,\mathcal{J})[/ilmath] is compact (when [ilmath]Y[/ilmath] is imbued with the subspace topology)
if and only if
- Every cover by sets open in [ilmath]X[/ilmath] 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 [ilmath]Y[/ilmath]
- That is a collection [ilmath]\{A_\alpha\}_{\alpha\in I}[/ilmath] where [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath]
- A covering is a collection of sets whose union is exactly [ilmath]Y[/ilmath], by sets open in [ilmath]Y[/ilmath] when [ilmath]Y[/ilmath] has the subspace topology.
- That is a collection [ilmath]\{A_\alpha\}_{\alpha\in I}[/ilmath] where [ilmath]Y=\cup_{\alpha\in I}A_\alpha[/ilmath]
Lastly recall:
- A subcovering is a collection of elements of a cover whose union also cover [ilmath]Y[/ilmath]
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
Suppose that [ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [ilmath]\implies[/ilmath] every covering consisting of open sets of [ilmath](X,\mathcal{J})[/ilmath] contains a finite subcover.
- Let [ilmath]\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}[/ilmath] be a family of open sets in [ilmath]X[/ilmath] with [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath]
- Take [ilmath]B_\alpha=A_\alpha\cap Y[/ilmath], then [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] is an open (in [ilmath]Y[/ilmath]) covering of [ilmath]Y[/ilmath], that is [ilmath]Y\subseteq\cup_{\alpha\in I}B_\alpha[/ilmath] (infact we have [ilmath]Y=\cup_{\alpha\in I}B_\alpha[/ilmath])
- Proof of [ilmath]Y\subseteq\cup_{\alpha\in I}B_\alpha[/ilmath] (we actually have [ilmath]Y=\cup_{\alpha\in I}B_\alpha[/ilmath])
- We wish to show that [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha\implies Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath], using the Implies-subset relation we actually just want to show that:
- Given [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath] that [ilmath]y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] - which is what we'll do.
- Note additionally that [ilmath]y\in\cup_{\alpha\in I}(A_i\cap Y)\iff \exists\beta\in I[y\in A_\beta\wedge y\in Y][/ilmath]
- Let [ilmath]y\in Y[/ilmath], then by hypothesis [ilmath]y\in\cup_{\alpha\in I}A_\alpha\iff\exists \beta\in I[y\in A_\beta][/ilmath]
- It is easily seen that [ilmath]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][/ilmath] simply by choosing [ilmath]\gamma:=\beta[/ilmath].
- Lastly, note that [ilmath]\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]\iff y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath]
- We have shown that [ilmath]y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] and by the Implies-subset relation we see
- [ilmath]Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] - as required.
- We wish to show that [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha\implies Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath], using the Implies-subset relation we actually just want to show that:
- I earlier claimed that actually [ilmath]Y=\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] - 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 [ilmath]Y[/ilmath] whose union is exactly [ilmath]Y[/ilmath] is the same as a covering by open sets in [ilmath]X[/ilmath] whose union contains (but need not be exactly equal to) [ilmath]Y[/ilmath]. So we have shown so far that:
- Compact in the subspace with equality for an open covering [ilmath]\implies[/ilmath] compact with the open cover of sets in [ilmath]X[/ilmath] whose union contains [ilmath]Y[/ilmath]
- Claim: [ilmath]\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y[/ilmath]
- Let [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] then:
- [ilmath]\exists\beta\in I[y\in A_\beta\wedge y\in Y]\iff \exists\beta\in I[y\in(A_\beta\cap Y)][/ilmath]
- 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)
- [ilmath](A_\beta\cap Y)\subseteq Y)[/ilmath] by the implies-subset relation we see immediately that:
- [ilmath]y\in(A_\beta\cap Y)\implies y\in Y[/ilmath]
- Thus we have shown that [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)\implies y\in Y[/ilmath] and finally this means:
- [ilmath]\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y[/ilmath]
- Combining this with [ilmath]Y\subseteq \cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] above we see that:
- [ilmath]Y=\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath]
- This completes the proof
- Let [ilmath]y\in\cup_{\alpha\in I}(A_\alpha\cap Y)[/ilmath] then:
- This shows that considering an open covering as a union of sets open in [ilmath]Y[/ilmath] whose union is exactly [ilmath]Y[/ilmath] is the same as a covering by open sets in [ilmath]X[/ilmath] whose union contains (but need not be exactly equal to) [ilmath]Y[/ilmath]. So we have shown so far that:
- By hypothesis, [ilmath]Y[/ilmath] is compact, this means that [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] contains a finite subcover
- call this subcover [ilmath]\{B'_i\}_{i=1}^n[/ilmath] where each [ilmath]B'_i\in\{B_\alpha\}_{\alpha\in I} [/ilmath], now we have [ilmath]Y\subseteq\cup_{i=1}^n B'_i[/ilmath] (we actually have equality, see the blue box in the yellow note box above)
- As each [ilmath]B'_i=A'_i\cap Y[/ilmath] (where [ilmath]A'_i[/ilmath] is the corresponding [ilmath]A_\alpha[/ilmath] for the [ilmath]B_\alpha[/ilmath] that [ilmath]B'_i[/ilmath] represents) we see that [ilmath]\{A_i\}_{i=1}^n[/ilmath] is a finite subcover by sets open in [ilmath]X[/ilmath]
- Proof of: [ilmath]Y\subseteq\cup_{i=1}^nB'_i\implies[/ilmath] [ilmath]Y\subseteq\cup_{i=1}^nA'_i[/ilmath] (proving that [ilmath]\{A'_i\}_{i=1}^n[/ilmath] is an open cover)
- For each [ilmath]i[/ilmath] we have [ilmath]B'_i:=A'_i\cap Y[/ilmath], by invoking the intersection of sets is a subset of each set we note that:
- [ilmath]B'_i\subseteq A'_i[/ilmath]
- We now invoke Union of subsets is a subset of the union
This theorem states that given two families of sets, [ilmath]\{A_\alpha\}_{\alpha\in I} [/ilmath] and [ilmath]\{B_\alpha\}_{\alpha\in I} [/ilmath] with [ilmath]\forall\alpha\in I[B_\alpha\subseteq A_\alpha][/ilmath] we have [ilmath]\cup_{\alpha\in I}B_\alpha\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath]
- It follows that [ilmath]Y\subseteq\cup_{i=1}^nB'_i\subseteq\cup_{i=1}^nA'_i[/ilmath], in particular:
- [ilmath]Y\subseteq\cup_{i=1}^nA'_i[/ilmath]
- This confirms that [ilmath]\{A'_i\}_{i=1}[/ilmath] is an open cover by sets in [ilmath]X[/ilmath]
- This completes this half of the proof.
[ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [math]\impliedby[/math] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] 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 [ilmath]\mathcal{A} [/ilmath] 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]
Notes
This theorem shows that the two are the same, that is:
- A set [ilmath]Y\subseteq X[/ilmath] is compact if every open cover of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] has a finite subcover (that is [ilmath]Y\subset\text{the covering} [/ilmath])
and
- A set [ilmath]Y\subseteq X[/ilmath] is compact if, when given the Subspace topology, every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]Y[/ilmath] 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