Editing Subbase (section)

===Alexander subbase theorem===

The Alexander Subbase Theorem is a significant result concerning subbases that is due to [[James Waddell Alexander II]].<ref name="Muger2020" /> The corresponding result for basic (rather than subbasic) open covers is much easier to prove.

:'''Alexander subbase theorem''':<ref name="Muger2020">{{cite book|last=Muger|first= Michael|title=Topology for the Working Mathematician|year=2020}}</ref>{{sfn|Rudin|1991|p=392 Appendix A2}} Let <math>(X, \tau)</math> be a topological space, and <math>\mathcal{S}</math> be a subbase of <math>\tau.</math> If every cover of <math>X</math> by elements from <math>\mathcal{S}</math> has a finite subcover, then <math>X</math> is [[compact space|compact]].

The converse to this theorem also holds (because every cover of <math>X</math> by elements of <math>\mathcal{S}</math> is an open cover of <math>X</math>) 
:Let <math>(X, \tau)</math> be a topological space, and <math>\mathcal{S}</math> be a subbase of <math>\tau.</math> If <math>X</math> is compact, then every cover of <math>X</math> by elements from <math>\mathcal{S}</math> has a finite subcover.

{{collapse top|title=Proof|left=true}}
Suppose for the sake of contradiction that the space <math>X</math> is not compact (so <math>X</math> is an infinite set), yet every subbasic cover from <math>\mathcal{S}</math> has a finite subcover. 
Let <math>\mathbb{S}</math> denote the set of all open covers of <math>X</math> that do not have any finite subcover of <math>X.</math> 
Partially order <math>\mathbb{S}</math> by subset inclusion and use [[Zorn's Lemma]] to find an element <math>\mathcal{C} \in \mathbb{S}</math> that is a maximal element of <math>\mathbb{S}.</math> 
Observe that:

# Since <math>\mathcal{C} \in \mathbb{S},</math> by definition of <math>\mathbb{S},</math> <math>\mathcal{C}</math> is an open cover of <math>X</math> and there does not exist any finite subset of <math>\mathcal{C}</math> that covers <math>X</math> (so in particular, <math>\mathcal{C}</math> is infinite).
# The maximality of <math>\mathcal{C}</math> in <math>\mathbb{S}</math> implies that if <math>V</math> is an open set of <math>X</math> such that <math>V \not\in \mathcal{C}</math> then  <math>\mathcal{C} \cup \{V\}</math> has a finite subcover, which must necessarily be of the form <math>\{V\} \cup \mathcal{C}_V</math> for some finite subset <math>\mathcal{C}_V</math> of <math>\mathcal{C}</math> (this finite subset depends on the choice of <math>V</math>).

We will begin by showing that <math>\mathcal{C} \cap \mathcal{S}</math> is {{em|not}} a cover of <math>X.</math> 
Suppose that <math>\mathcal{C} \cap \mathcal{S}</math> was a cover of <math>X,</math> which in particular implies that <math>\mathcal{C} \cap \mathcal{S}</math> is a cover of <math>X</math> by elements of <math>\mathcal{S}.</math> 
The theorem's hypothesis on <math>\mathcal{S}</math> implies that there exists a finite subset of <math>\mathcal{C} \cap \mathcal{S}</math> that covers <math>X,</math> which would simultaneously also be a finite subcover of <math>X</math> by elements of <math>\mathcal{C}</math> (since <math>\mathcal{C} \cap \mathcal{S} \subseteq \mathcal{C}</math>). 
But this contradicts <math>\mathcal{C} \in \mathbb{S},</math> which proves that <math>\mathcal{C} \cap \mathcal{S}</math> does not cover <math>X.</math>

Since <math>\mathcal{C} \cap \mathcal{S}</math> does not cover <math>X,</math> there exists some <math>x \in X</math> that is not covered by <math>\mathcal{C} \cap \mathcal{S}</math> (that is, <math>x</math> is not contained in any element of <math>\mathcal{C} \cap \mathcal{S}</math>). 
But since <math>\mathcal{C}</math> does cover <math>X,</math> there also exists some <math>U \in \mathcal{C}</math> such that <math>x \in U.</math> 
It follows that <math>U\neq X</math>, because otherwise it would imply <math>\mathcal{C}</math> has a finite subcover of <math>X</math>, namely the subcover <math>\{U\}=\{X\},</math> contradicting <math>\mathcal{C}\in \mathbb{S}.</math> Since <math>U\neq X,</math> and <math>\mathcal{S}</math> is a subbasis generating <math>X</math>'s topology (together with <math>X</math>), from the definition of the topology generated by <math>\mathcal{S},</math> there must exist a finite collection of subbasic open sets <math>S_1, \ldots, S_n \in \mathcal{S}</math> with <math>n\ge 1</math> such that
<math display=block>x \in S_1 \cap \cdots \cap S_n \subseteq U.</math>

We will now show by contradiction that <math>S_i \not\in \mathcal{C}</math> for every <math>i = 1, \ldots, n.</math> 
If <math>i</math> was such that <math>S_i \in \mathcal{C},</math> then also <math>S_i \in \mathcal{C} \cap \mathcal{S}</math> so the fact that <math>x \in S_i</math> would then imply that <math>x</math> is covered by <math>\mathcal{C} \cap \mathcal{S},</math> which contradicts how <math>x</math> was chosen (recall that <math>x</math> was chosen specifically so that it was not covered by <math>\mathcal{C} \cap \mathcal{S}</math>).

As mentioned earlier, the maximality of <math>\mathcal{C}</math> in <math>\mathbb{S}</math> implies that for every <math>i = 1, \ldots, n,</math> there exists a finite subset <math>\mathcal{C}_{S_i}</math> of <math>\mathcal{C}</math> such that<math>\left\{S_i\right\} \cup \mathcal{C}_{S_i}</math> forms a finite cover of <math>X.</math> 
Define
<math display=block>\mathcal{C}_F := \mathcal{C}_{S_1} \cup \cdots \cup \mathcal{C}_{S_n},</math>
which is a finite subset of <math>\mathcal{C}.</math> 
Observe that for every <math>i = 1, \ldots, n,</math> <math>\left\{S_i\right\} \cup \mathcal{C}_F</math> is a finite cover of <math>X</math> so let us replace every <math>\mathcal{C}_{S_i}</math> with <math>\mathcal{C}_F.</math>

Let <math>\cup \mathcal{C}_F</math> denote the union of all sets in <math>\mathcal{C}_F</math> (which is an open subset of <math>X</math>) and let <math>Z</math> denote the complement of <math>\cup \mathcal{C}_F</math> in <math>X.</math> 
Observe that for any subset <math>A \subseteq X,</math> <math>\{A\} \cup \mathcal{C}_F</math> covers <math>X</math> if and only if <math>Z \subseteq A.</math> 
In particular, for every <math>i = 1, \ldots, n,</math> the fact that <math>\left\{S_i\right\} \cup \mathcal{C}_F</math> covers <math>X</math> implies that <math>Z \subseteq S_i.</math> 
Since <math>i</math> was arbitrary, we have <math>Z \subseteq S_1 \cap \cdots \cap S_n.</math> 
Recalling that <math>S_1 \cap \cdots \cap S_n \subseteq U,</math> we thus have <math>Z \subseteq U,</math> which is equivalent to <math>\{U\} \cup \mathcal{C}_F</math> being a cover of <math>X.</math> 
Moreover, <math>\{U\} \cup \mathcal{C}_F</math> is a finite cover of <math>X</math> with <math>\{U\} \cup \mathcal{C}_F \subseteq \mathcal{C}.</math> 
Thus <math>\mathcal{C}</math> has a finite subcover of <math>X,</math> which contradicts the fact that <math>\mathcal{C} \in \mathbb{S}.</math> 
Therefore, the original assumption that <math>X</math> is not compact must be wrong, which proves that <math>X</math> is compact. <math>\blacksquare</math>
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Although this proof makes use of [[Zorn's Lemma]], the proof does not need the full strength of choice. 
Instead, it relies on the intermediate [[Ultrafilter principle]].<ref name="Muger2020" />

Using this theorem with the subbase for <math>\R</math> above, one can give a very easy proof that bounded closed intervals in <math>\R</math> are compact.  
More generally, [[Tychonoff's theorem]], which states that the product of non-empty compact spaces is compact, has a short proof if the Alexander Subbase Theorem is used.

{{collapse top|title=Proof|left=true}}
The product topology on <math>\prod_{i} X_i</math> has, by definition, a subbase consisting of ''cylinder'' sets that are the inverse projections of an open set in one factor.  
Given a {{em|subbasic}} family <math>C</math> of the product that does not have a finite subcover, we can partition <math>C = \cup_i C_i</math> into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space.  
By assumption, if <math>C_i \neq \varnothing</math> then <math>C_i</math> does {{em|not}} have a finite subcover.  
Being cylinder sets, this means their projections onto <math>X_i</math> have no finite subcover, and since each <math>X_i</math> is compact, we can find a point <math>x_i \in X_i</math> that is not covered by the projections of <math>C_i</math> onto <math>X_i.</math> But then <math>\left(x_i\right)_i \in \prod_{i} X_i</math> is not covered by <math>C.</math> <math>\blacksquare</math>

Note, that in the last step we implicitly used the [[axiom of choice]] (which is actually equivalent to [[Zorn's lemma]]) to ensure the existence of <math>\left(x_i\right)_i.</math>
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