Editing Subbase

{{short description|Collection of subsets that generate a topology}}
{{about|an object in mathematical topology|the term in highway engineering|Subbase (pavement)|the frequency range|Sub-bass}}

In [[topology]], a '''subbase''' (or '''subbasis''', '''prebase''', '''prebasis''') for the [[Topology (structure)|topology]] {{math|τ}} of a [[topological space]] {{math|(''X'', τ)}} is a subcollection <math>B</math> of <math>\tau</math> that generates <math>\tau,</math> in the sense that <math>\tau</math> is the smallest topology containing <math>B</math> as open sets. A slightly different definition is used by some authors, and there are other useful equivalent formulations of the definition; these are discussed below.

Subbase is a weaker notion than that of a [[Base (topology)|base]] for a topology.

==Definition==

Let <math>X</math> be a topological space with topology <math>\tau.</math> A '''subbase''' of <math>\tau</math> is usually defined as a subcollection <math>B</math> of <math>\tau</math> satisfying one of the three following equivalent conditions:

#<math>\tau</math> is the smallest topology containing <math>B</math>: any topology <math>\tau^\prime</math> on <math>X</math> containing <math>B</math> must also contain <math>\tau.</math>
#<math>\tau</math> is the [[Intersection (set theory)|intersection]] of all topologies on X containing <math>B.</math>
#The collection of open sets consisting of <math>X</math> and all finite [[Intersection (set theory)|intersections]] of elements of <math>B</math> forms a [[Basis (topology)|basis]] for <math>\tau.</math>{{sfn|Rudin|1991|p=392 Appendix A2}}<ref group=note>Rudin's definition is less general than ours, because it effectively requires that <math>B</math> covers <math>X</math> (see "Alternative definition" subsection below). We drop this requirement here, and assume that <math>B</math> is any subset of <math>\mathcal{P}(X)</math></ref> This means that every proper [[open set]] in <math>\tau</math> can be written as a [[Union (set theory)|union]] of finite intersections of elements of <math>B.</math> Explicitly, given a point <math>x</math> in an open set <math>U \subsetneq X,</math> there are finitely many sets <math>S_1, \ldots, S_n</math> of <math>B,</math> such that the intersection of these sets contains <math>x</math> and is contained in <math>U.</math>

If we additionally assume that <math>B</math> [[Cover (topology)|covers]] <math>X</math>, or if we use the [[nullary intersection]] convention, then there is no need to include <math>X</math> in the third definition.

If <math>B</math> is a subbase of <math>\tau</math>, we say that <math>B</math> '''generates''' the topology <math>\tau.</math> This terminology originates from the explicit construction of <math>\tau</math> from <math>B</math> using the second or third definition above.

Elements of subbase are called {{em|subbasic (open) sets}}. A [[Cover (topology)|cover]] composed of subbasic sets is called a {{em|subbasic (open) cover}}.

For {{em|any}} subcollection <math>S</math> of the [[power set]] <math>\wp(X),</math> there is a unique topology having <math>S</math> as a subbase; it is the intersection of all topologies on <math>X</math> containing <math>S</math>. In general, however, the converse is not true, i.e. there is no unique subbasis for a given topology.

Thus, we can start with a fixed topology and find subbases for that topology, and we can also start with an arbitrary subcollection of the power set <math>\wp(X)</math> and form the topology generated by that subcollection. We can freely use either equivalent definition above; indeed, in many cases, one of the three conditions is more useful than the others.

===Alternative definition===

Less commonly, a slightly different definition of subbase is given which requires that the subbase <math>\mathcal{B}</math> cover <math>X.</math>{{sfn | Munkres | 2000 | pp=82}}  In this case, <math>X</math> is the union of all sets contained in <math>\mathcal{B}.</math> This means that there can be no confusion regarding the use of nullary intersections in the definition.

However, this definition is not always equivalent to the three definitions above. There exist topological spaces <math>(X, \tau)</math> with subcollections <math>\mathcal{B} \subseteq \tau</math> of the topology such that <math>\tau</math> is the smallest topology containing <math>\mathcal{B}</math>, yet <math>\mathcal{B}</math> does not cover <math>X</math>. For example, consider a topological space <math>(X,\tau)</math> with <math>\tau=\{\varnothing, \{p\}, X\}</math> and <math>\mathcal{B}=\{\{p\}\}</math> for some <math>p\in X.</math> Clearly, <math>\mathcal{B}</math> is a subbase of <math>\tau</math>, yet <math>\mathcal{B}</math> doesn't cover <math>X</math> as long as <math>X</math> has at least <math>2</math> elements. In practice, this is a rare occurrence. E.g. a subbase of a space that has at least two points and satisfies the [[T1 separation axiom|T<sub>1</sub> separation axiom]] must be a cover of that space.

==Examples==

The topology generated by any subset <math>\mathcal{S} \subseteq \{\varnothing, X\}</math> (including by the empty set <math>\mathcal{S} := \varnothing</math>) is equal to the trivial topology <math>\{\varnothing, X\}.</math>

If <math>\tau</math> is a topology on <math>X</math> and <math>\mathcal{B}</math> is a basis for <math>\tau</math> then the topology generated by <math>\mathcal{B}</math> is <math>\tau.</math> Thus any basis <math>\mathcal{B}</math> for a topology <math>\tau</math> is also a subbasis for <math>\tau.</math> 
If <math>\mathcal{S}</math> is any subset of <math>\tau</math> then the topology generated by <math>\mathcal{S}</math> will be a subset of <math>\tau.</math>

The usual topology on the [[real number]]s <math>\R</math> has a subbase consisting of all [[semi-infinite]] open intervals either of the form <math>(-\infty, a)</math> or <math>(b, \infty),</math> where <math>a</math> and <math>b</math> are real numbers. Together, these generate the usual topology, since the intersections <math>(a,b) = (-\infty, b) \cap (a, \infty)</math> for <math>a \leq b</math> generate the usual topology. A second subbase is formed by taking the subfamily where <math>a</math> and <math>b</math> are [[Rational number|rational]]. The second subbase generates the usual topology as well, since the open intervals <math>(a, b)</math> with <math>a,</math> <math>b</math> rational, are a basis for the usual Euclidean topology.

The subbase consisting of all semi-infinite open intervals of the form <math>(-\infty, a)</math> alone, where <math>a</math> is a real number, does not generate the usual topology. The resulting topology does not satisfy the [[T1 space|T<sub>1</sub> separation axiom]], since if <math>a < b</math> every [[open set]] containing <math>b</math> also contains <math>a.</math>

The [[initial topology]] on <math>X</math> defined by a family of functions <math>f_i : X \to Y_i,</math> where each <math>Y_i</math> has a topology, is the coarsest topology on <math>X</math> such that each <math>f_i</math> is [[Continuous function (topology)|continuous]]. Because continuity can be defined in terms of the [[Inverse image|inverse images]] of open sets, this means that the initial topology on <math>X</math> is given by taking all <math>f_i^{-1}(U),</math>
where <math>U</math> ranges over all open subsets of <math>Y_i,</math> as a subbasis.

Two important special cases of the initial topology are the [[product topology]], where the family of functions is the set of projections from the product to each factor, and the [[subspace topology]], where the family consists of just one function, the [[inclusion map]].

The [[compact-open topology]] on the space of continuous functions from <math>X</math> to <math>Y</math> has for a subbase the set of functions
<math display=block>V(K,U) = \{f : X \to Y \mid f(K) \subseteq U\}</math>
where <math>K \subseteq X</math> is [[Compact space|compact]] and <math>U</math> is an open subset of <math>Y.</math>

Suppose that <math>(X, \tau)</math> is a [[Hausdorff space|Hausdorff]] topological space with <math>X</math> containing two or more elements (for example, <math>X = \R</math> with the [[Euclidean topology]]). Let <math>Y \in \tau</math> be any non-empty {{em|open}} subset of <math>(X, \tau)</math> (for example, <math>Y</math> could be a non-empty bounded open interval in <math>\R</math>) and let <math>\nu</math> denote the [[subspace topology]] on <math>Y</math> that <math>Y</math> inherits from <math>(X, \tau)</math> (so <math>\nu \subseteq \tau</math>). Then the topology generated by <math>\nu</math> '''on <math>X</math>''' is equal to the union <math>\{X\} \cup \nu</math> (see the footnote for an explanation),
<ref group=note>Since <math>\nu</math> is a topology on <math>Y</math> and <math>Y</math> is an open subset of <math>(X, \tau),</math>, it is easy to verify that <math>\{X\} \cup \nu</math> is a topology on <math>X</math>. In particular, <math>\nu</math> is closed under unions and finite intersections because <math>\tau</math> is. But since <math> X \not\in \nu </math>, <math>\nu</math> is not a topology on <math>X</math> an <math>\{X\} \cup \nu</math> is clearly the smallest topology on <math>X</math> containing <math>\nu</math>).</ref>
where <math>\{X\} \cup \nu \subseteq \tau</math> (since <math>(X, \tau)</math> is Hausdorff, equality will hold if and only if <math>Y = X</math>). Note that if <math>Y</math> is a [[proper subset]] of <math>X,</math> then <math>\{X\} \cup \nu</math> is the smallest topology ''on <math>X</math>'' containing <math>\nu</math> yet <math>\nu</math> does not cover <math>X</math> (that is, the union <math>\bigcup_{V \in \nu} V = Y</math> is a proper subset of <math>X</math>).

==Results using subbases==

One nice fact about subbases is that [[continuity (topology)|continuity]] of a function need only be checked on a subbase of the range. That is, if <math>f : X \to Y</math> is a map between topological spaces and if <math>\mathcal{B}</math> is a subbase for <math>Y,</math> then <math>f : X \to Y</math> is continuous [[if and only if]] <math>f^{-1}(B)</math> is open in <math>X</math> for every <math>B \in \mathcal{B}.</math> 
A [[Net (mathematics)|net]] (or [[sequence]]) <math>x_{\bull} = \left(x_i\right)_{i \in I}</math> converges to a point <math>x</math> if and only if every {{em|sub}}basic neighborhood of <math>x</math> contains all <math>x_i</math> for sufficiently large <math>i \in I.</math>

===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>
{{collapse bottom}}

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>
{{collapse bottom}}

==See also==

* {{annotated link|Base (topology)}}

==Notes==
{{reflist|group=note}}

==Citations==
{{reflist}}

==References==

* {{Bourbaki General Topology Part I Chapters 1-4}} <!--{{sfn|Bourbaki|1998|p=}}-->
* {{Dugundji Topology}} <!--{{sfn|Dugundji|1966|p=}}-->
* {{Munkres Topology|edition=2}}
* {{Rudin Walter Functional Analysis|edition=2}} <!-- {{sfn|Rudin|1991|p=}} -->
* {{Willard General Topology}} <!--{{sfn|Willard|2004|p=}}-->

[[Category:Articles containing proofs]]
[[Category:General topology]]