Subbase

In topology, a subbase (or subbasis) for a topological space $X$ with topology $T$ is a subcollection $B$ of $T$ that generates $T$ , in the sense that $T$ is the smallest topology containing $B$ . A slightly different definition is used by some authors, and there are other useful equivalent formulations of the definition; these are discussed below.

Definition

Throughout, $X$ will be be a set.

Notation: Let

℘(X)

denote the power set of a set

X

.

Definition: If

𝒞, ℱ \subseteq ℘(X)

then we say that

ℱ

is finer than

𝒞

and that

𝒞

is coarser than

ℱ

if

𝒞 \subseteq ℱ

.

Generating a topology $τ$

The intersection of any collection of topologies on $X$ is a topology on $X$ , where the finest topology is the discrete topology $℘(X)$ and the coarsest topology is the trivial (or indiscrete) topology ${ \emptyset, X$ }.

Definition: If

𝒮

is any collection of subsets of

X

then the topology generated by $𝒮$ on $X$ is the topology on

X

, denoted by

τ 𝒮

, defined in any of the following equivalent ways:

$τ 𝒮$ is equal to the intersection of all topologies on $X$ that contain $𝒮$ .
$τ 𝒮$ is the (necessarily unique) coarsest topology on $X$ containing $𝒮$ (as open subsets)
$τ 𝒮$ is the (necessarily unique) minimal topology containing $𝒮$ , which means that $𝒮 \subseteq τ 𝒮$ and if $ν$ is a topology on $X$ satisfying $𝒮 \subseteq ν$ , then $ν$ must also contain $τ 𝒮$ (i.e. $τ 𝒮 \subseteq ν$ ).
τ_𝒮 is the (necessarily unique) topology on X generated by the basis ℬ ∪ { X }, where ℬ denotes the set of all possible finite intersections of elements of 𝒮 (where elements of 𝒮 are subsets of X).
- The set $ℬ \cup {X$ } is always a basis for some topology on $X$ .
- Recall that the topology on $X$ generated by a basis $𝒜$ on $X$ consists of all possible unions of elements in $𝒜$ .
- If we use the nullary intersection convention then there is no need to include the " ∪ { X }" in this definition.
  - The nullary intersection convention is that the empty intersection (the intersection of no subsets of $X$ ) satisfies $X = \cap B \in \emptyset B$ , where note that the empty intersection is the intersection of 0 sets and is thus a finite intersection of sets. So under this convention, the collection $ℬ$ of all finite intersections of sets in $𝒮$ would always contain the set $X = \cap B \in \emptyset B$ and thus $ℬ = ℬ \cup {X$ } would be a basis on $X$ .

in this case we say that

𝒮

is a subbase or subbasis for $τ 𝒮$ .

Note that it is possible for two different collections of subsets of $X$ to generate the same topology.

Basic examples

The topology generated by any subset $𝒮 \subseteq { \emptyset, X$ } (including by the empty set $𝒮 := \emptyset$ ) is equal to the trivial topology ${ \emptyset, X$ }.
Let (X, τ) be any Hausdorff topological space with X containing two or more elements (e.g. X = ℝ with the Euclidean topology). Let Y ∈ τ be any non-empty open subset of (X, τ) (e.g. Y could be a non-empty bounded open interval in ℝ) and let ν denote the subspace topology on Y that Y inherits from (X, τ) (so ν ⊆ τ). Then the topology generated by ν on $X$ is equal to the union { X } ∪ ν (see this footnote for an explanation),^[1] where { X } ∪ ν ⊆ τ (since (X, τ) is Hausdorff, equality will hold if and only if Y = X).
- Note that if $Y$ is a proper subset of $X$ , then ${X} \cup ν$ is the smallest topology on $X$ containing $ν$ yet $ν$ does not cover $X$ (i.e. the union $\cup V \in ν V = Y$ is a proper subset of $X$ ).
If $τ$ is a topology on $X$ and $ℬ$ is a basis for $τ$ then the topology generated by $ℬ$ is $τ$ . Thus any basis $ℬ$ for a topology $τ$ is also a subbasis for $τ$ . If $𝒮$ is any subsets of $τ$ then the topology generated by $𝒮$ will be a subset of $τ$ .

Generating the topology $τ$

Let $τ$ be a topology on $X$ and let $𝒮$ be a collection of subsets of $X$ . Usually, $𝒮$ is a subbase for $τ$ is defined to mean that any one of the following equivalent conditions is true:

Definition: We say that $𝒮$ is a subbase (or subbasis) for $τ$ if any of the following equivalent conditions hold:

The topology on $X$ generated by $𝒮$ is equal to $τ$ .
The basis ℬ ∪ { X } generates the topology τ, where ℬ denotes the set of all possible finite intersectionss of elements of 𝒮. This means that every proper open set in τ can be written as a union of finite intersections of elements of 𝒮.
- If we use the nullary intersection convention, then there is no need to include $X$ in this definition.
𝒮 ⊆ τ and given any open subset U ∈ τ of (X, τ) such that U ≠ X, for every x ∈ U there must exist finitely many sets S₁, ..., S_n in 𝒮 such that the intersection of these sets contains x and is contained in U i.e. x ∈ S₁ ∩ ⋅⋅⋅ ∩ S_n ⊆ U.
- Note that the condition " $U \neq X$ " is necessarily since otherwise, for instance, $𝒮 := \emptyset$ and $𝒮 := { \emptyset$ } wouldn't be subbases for the trivial topology (which they are). Similar problems would also occur for other less trivial topologies.

In general, however, 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 $℘(X)$ and form the topology generated by that subcollection. We can freely use either equivalent definition above; indeed, in many cases, one of the two conditions is more useful than the other.

Alternative definition

Sometimes, a slightly different definition of subbase is given which requires that the subbase $ℬ$ cover $X$ .^[2] In this case, $X$ is the union of all sets contained in $ℬ$ . This means that there can be no confusion regarding the use of nullary intersections in the definition.

However, with this definition, the two definitions above are not always equivalent. In other words, there exist topological spaces $(X, τ)$ with a subset $ℬ \subseteq τ$ , such that $τ$ is the smallest topology containing $ℬ$ , yet $ℬ$ does not cover $X$ (such an example is given above). 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 must be a cover of that space.

Examples

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

The subbase consisting of all semi-infinite open intervals of the form $(-\infty, a)$ alone, where $a$ is a real number, does not generate the usual topology. The resulting topology does not satisfy the T₁ separation axiom, since all open sets have a non-empty intersection.

The initial topology on $X$ defined by a family of functions $f i : X \to Y i$ , where each $Y i$ has a topology, is the coarsest topology on $X$ such that each $f i$ is continuous. Because continuity can be defined in terms of the inverse images of open sets, this means that the initial topology on $X$ is given by taking all $f i -1 (U)$ , where $U$ ranges over all open subsets of $Y i$ , 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 $X$ to $Y$ has for a subbase the set of functions

V(K,U)=\{f\colon X\to Y\mid f(K)\subseteq U\}

where $K \subseteq X$ is compact and $U$ is an open subset of $Y$ .

Results using subbases

One nice fact about subbases is that continuity of a function need only be checked on a subbase of the range.

Proposition — If $f : X \to Y$ is a map between topological spaces and if $ℬ$ is a subbase for $Y$ , then $f : X \to Y$ is continuous if and only if $f -1 (B)$ is open in $X$ for every $B \in ℬ$ .

Alexander subbase theorem

There is one significant result concerning subbases, due to James Waddell Alexander II.

Alexander Subbase Theorem — Let $X$ be a topological space with a subbasis $B$ . If every cover by elements from $B$ has a finite subcover, then the space is compact.

Note that the corresponding result for basic covers is trivial.

Proof Outline

Assume by way of contradiction that the space $X$ is not compact, yet every subbasic cover from $B$ has a finite subcover. Use Zorn's Lemma to find an open cover $C$ without finite subcover that is maximal amongst such covers. That means that if $V$ is an open set of $X$ which is not in $C$ , then $C \cup {V}$ has a finite subcover, necessarily of the form ${V} \cup C V$ , where the choice of the finite subset $C V$ of the cover $C$ depends on the picked additional set $V$ .

Consider $C \cap B$ , that is, the subbasic subfamily of $C$ . We claim $C \cap B$ does not cover $X$ . If it covered $X$ , then it would be a cover from elements of $B$ and by hypothesis on $B$ , it would have a finite subcover from $C \cap B$ which is at the same time also a finite subcover from $C$ . But from definition of $C$ , $C$ does not have a finite subcover of $X$ , so $C \cap B$ does not cover $X$ . So there exists an element $x$ from $X$ but uncovered by $C \cap B$ . $C$ covers $X$ (with infinite number of open sets), so $x \in U$ for some $U \in C$ . $B$ is a subbasis, so for some $S 1, ..., S n \in B$ , we have: $x \in S 1 \cap \cdot\cdot\cdot \cap S n \subseteq U$ .

Since $x$ is uncovered by $C \cap B$ , $S i \notin C$ for each $i$ . (If $S i \in C$ for some i, then it would hold $S i \in C \cap B$ and since $x \in S i$ , $C \cap B$ would also cover point $x$ , contrary to its choice). As noted above from the maximality of the cover $C$ , for each $i$ there exists a finite subset $C S i$ of cover $C$ such that ${S i} \cup C S i$ forms a finite cover of $X$ . Let's denote $C F$ the finite union of the finite sets $C S i$ where $i$ iterates from 1 to n. Then for each $i$ the former finite cover of X can be replaced by a new bigger and still finite cover ${S i} \cup C F$ of X. The finite set ${S i} \cup C F$ covers $X$ for each $i$ , so also ${S 1 \cap \cdot\cdot\cdot \cap S n} \cup C F$ covers $X$ . The intersection in the cover can be replaced by the single bigger open set $U$ from cover $C$ . So ${U}\cup C F$ is also a finite cover of X and made of the open sets only from $C$ . Thus $C$ has a finite subcover of X, in contradiction to the choice of $C$ . Therefore the original assumption of X not being compact is wrong due to a contradiction we reached. Therefore X is compact. Q.E.D.

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.

Using this theorem with the subbase for $R$ above, one can give a very easy proof that bounded closed intervals in $R$ are compact.

Tychonoff's theorem, that the product of compact spaces is compact, also has a short proof. The product topology on $\prod i X i$ has, by definition, a subbase consisting of cylinder sets that are the inverse projections of an open set in one factor. Given a subbasic family $C$ of the product that does not have a finite subcover, we can partition $C = \cup i C i$ into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space. By assumption, no $C i$ has a finite subcover. Being cylinder sets, this means their projections onto $X i$ have no finite subcover, and since each $X i$ is compact, we can find a point $x i \in X i$ that is not covered by the projections of $C i$ onto $X i$ . But then $(x i) i \in \prod i X i$ is not covered by $C$ .

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 $(x i) i$ .

References

^ Since $ν$ is a topology on $Y$ and $Y$ is an open subset of $(X, τ)$ , it is easy to verify that ${X} \cup ν$ is a topology on $X$ . Since $ν$ isn't a topology on $X$ , ${X} \cup ν$ is clearly the smallest topology on $X$ containing $ν$ ).
^ Merrifield, Richard E.; Simmons, Howard E. (1989). Topological Methods in Chemistry. John Wiley & Sons. p. 17. ISBN 0-471-83817-9. Retrieved 13 June 2013. A collection $S$ of subsets that satisfies criterion (i) is called a subbasis for a topology on $X$ .

Willard, Stephen (2004), General topology, New York: Dover Publications, ISBN 978-0-486-43479-7, MR 2048350

[1] Since $ν$ is a topology on $Y$ and $Y$ is an open subset of $(X, τ)$ , it is easy to verify that ${X} \cup ν$ is a topology on $X$ . Since $ν$ isn't a topology on $X$ , ${X} \cup ν$ is clearly the smallest topology on $X$ containing $ν$ ).

[2] Merrifield, Richard E.; Simmons, Howard E. (1989). Topological Methods in Chemistry. John Wiley & Sons. p. 17. ISBN 0-471-83817-9. Retrieved 13 June 2013. A collection $S$ of subsets that satisfies criterion (i) is called a subbasis for a topology on $X$ .

[1]

[2]