In topology, a subbase (or subbasis, prebase, prebasis) for a topological space X with topology \tau is a subcollection B of \tau that generates \tau, in the sense that \tau is the smallest topology containing B 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.

Definition

Let X be a topological space with topology \tau. A subbase of \tau is usually defined as a subcollection B of \tau satisfying one of the two following equivalent conditions: (If we use the nullary intersection convention, then there is no need to include X in the second definition.) For subcollection S of the power set \wp(X), there is a unique topology having S 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 \wp(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

Less commonly, a slightly different definition of subbase is given which requires that the subbase \mathcal{B} cover X. In this case, X is the union of all sets contained in 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 two definitions above. There exist topological spaces (X, \tau) with subcollections of the topology such that \tau is the smallest topology containing \mathcal{B}, yet \mathcal{B} does not cover X. (An example is given at the end of the next section.) 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. But as seen below, to prove the Alexander subbase theorem, one must assume that \mathcal{B} covers X.

Examples

The topology generated by any subset (including by the empty set ) is equal to the trivial topology If \tau is a topology on X and \mathcal{B} is a basis for \tau then the topology generated by \mathcal{B} is \tau. Thus any basis \mathcal{B} for a topology \tau is also a subbasis for \tau. If \mathcal{S} is any subset of \tau then the topology generated by \mathcal{S} will be a subset of \tau. The usual topology on the real numbers \R has a subbase consisting of all semi-infinite open intervals either of the form or where a and b are real numbers. Together, these generate the usual topology, since the intersections for a \leq 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 alone, where a is a real number, does not generate the usual topology. The resulting topology does not satisfy the T1 separation axiom, since if a < b every open set containing b also contains a. The initial topology on X defined by a family of functions 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 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 where is compact and U is an open subset of Y. Suppose that (X, \tau) is a Hausdorff topological space with X containing two or more elements (for example, X = \R with the Euclidean topology). Let Y \in \tau be any non-empty subset of (X, \tau) (for example, Y could be a non-empty bounded open interval in \R) and let \nu denote the subspace topology on Y that Y inherits from (X, \tau) (so ). Then the topology generated by \nu on X is equal to the union (see the footnote for an explanation), where (since (X, \tau) is Hausdorff, equality will hold if and only if Y = X). Note that if Y is a proper subset of X, then is the smallest topology on X containing \nu yet \nu does not cover X (that is, the union is a proper subset of X).

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 f : X \to Y is a map between topological spaces and if \mathcal{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 A net (or sequence) converges to a point x if and only if every basic neighborhood of x contains all x_i for sufficiently large i \in I.

Alexander subbase theorem

The Alexander Subbase Theorem is a significant result concerning subbases that is due to James Waddell Alexander II. The corresponding result for basic (rather than subbasic) open covers is much easier to prove. The converse to this theorem also holds and it is proven by using (since every topology is a subbasis for itself). Suppose for the sake of contradiction that the space X is not compact (so X is an infinite set), yet every subbasic cover from \mathcal{S} has a finite subcover. Let \mathbb{S} denote the set of all open covers of X that do not have any finite subcover of X. Partially order \mathbb{S} by subset inclusion and use Zorn's Lemma to find an element that is a maximal element of \mathbb{S}. Observe that: We will begin by showing that is a cover of X. Suppose that was a cover of X, which in particular implies that is a cover of X by elements of The theorem's hypothesis on \mathcal{S} implies that there exists a finite subset of that covers X, which would simultaneously also be a finite subcover of X by elements of \mathcal{C} (since ). But this contradicts which proves that does not cover X. Since does not cover X, there exists some x \in X that is not covered by (that is, x is not contained in any element of ). But since \mathcal{C} does cover X, there also exists some such that x \in U. Since \mathcal{S} is a subbasis generating X's topology, from the definition of the topology generated by there must exist a finite collection of subbasic open sets such that We will now show by contradiction that for every If i was such that then also so the fact that x \in S_i would then imply that x is covered by which contradicts how x was chosen (recall that x was chosen specifically so that it was not covered by ). As mentioned earlier, the maximality of \mathcal{C} in \mathbb{S} implies that for every there exists a finite subset of \mathcal{C} such that forms a finite cover of X. Define which is a finite subset of Observe that for every is a finite cover of X so let us replace every with Let denote the union of all sets in (which is an open subset of X) and let Z denote the complement of in X. Observe that for any subset covers X if and only if In particular, for every the fact that covers X implies that Since i was arbitrary, we have Recalling that we thus have which is equivalent to being a cover of X. Moreover, is a finite cover of X with Thus \mathcal{C} has a finite subcover of X, which contradicts the fact that Therefore, the original assumption that X is not compact must be wrong, which proves that X is compact. 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. 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. The product topology on has, by definition, a subbase consisting of cylinder sets that are the inverse projections of an open set in one factor. Given a family C of the product that does not have a finite subcover, we can partition into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space. By assumption, if then C_i does have 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 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

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