From Academic Kids

In topology, a subbase (or subbasis) for a topological space X with topology T is a subcollection B of T which 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.



Let X be a topological space with topology T. A subbase of T is usually defined as a subcollection B of T satisfying one of the two following equivalent conditions:

  1. The subcollection B generates the topology T. This means that T is the smallest topology containing B: any topology U on X containing B must also contain T.
  2. The collection of open sets containing the empty set, X, and all finite intersections of elements of B forms a basis for T. This means that every non-empty proper open set in T can be written as a union of finite intersections of elements of B. Explicitly, given a point x in an proper open set U, there are finitely many sets S1, …, Sn of B, such that the intersection of these sets contains x and is contained in U.

(Note that if we use the nullary intersection convention, then there is no need to include X in the second definition.)

For any subcollection S of the power set P(X), there is a unique topology having B as a subbase. In particular, the intersection of all topologies on X containing S satisfies this condition. 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 P(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.

Alternate definition

Sometimes, a slightly different definition of subbase is given which requires that the subbase B cover X. In this case, X is an open set in the topology generated, because it is the union of all the {Bi} as Bi ranges over 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 spaces X with topology T, such that there exists a subcollection B of T such that T is the smallest topology containing B, yet B does not cover X. In practice, this a rare occurrence; e.g. a subbase of a space satisfying the T1 axiom must be a cover of that space.


The usual topology on the real numbers R has a subbase consisting of all semi-infinite open intervals either of the form (−∞,a) or (a,∞), where a is a real number. A second subbase is formed by taking the subfamily where a is rational.

The initial topology defined by a family of functions fi : XYi, where each Yi has a topology, is the coarsest topology on X such that each fi is continuous. Because continuity can be defined by the inverse images of open sets, this means that the weak topology on X is given by taking all fi−1(Ui), where Ui ranges over all open subsets of Yi, 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

<math>V(K,U) = \{f\colon X\to Y \mid f(K)\sub U\}<math>

where K is compact and U is open 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. That is, if B is a subbase for Y, a function f : XY is continuous iff f−1(U) is open in X for each U in B.

Alexander subbase theorem

There is one significant result concerning subbases, due to J. W. Alexander.

Theorem: If every subbasic cover has a finite subcover, then the space is compact.

(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 a cover C without finite subcover that is maximal amongst such covers. That means that if V is not in C, then C∪{V} has a finite subcover, necessarily of the form C0∪{V}.

Consider CB, that is, the subbasic subfamily of C. If it covered X, then by hypothesis, it would have a finite subcover. But C does not have such, so CB does not cover X. Let xX that is not covered. C covers X, so for UC, xU. B is a subbasis, so for some S1, … ,SnB, xS1∩…∩SnU.

Since x is uncovered, SiC. As noted above, this means that for each i, Si along with a finite subfamily Ci of C, covers X. But then U and all the Ci’s cover X, so C has a finite subcover after all. QED

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

Tychonoff's theorem, that the product of compact spaces is compact, also has a short proof. The product topology on ∏iXi 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=∪iCi into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space. By assumption, no Ci has a finite subcover. Being cylinder sets, this means their projections onto Xi have no finite subcover, and since each Xi is compact, we can find a point xiXi that is not covered by the projections of Ci onto Xi. But then ‹xi› is not covered by C.

See also


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