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

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

  1. The subcollection   generates the topology   This means that   is the smallest topology containing  : any topology   on   containing   must also contain  
  2. The collection of open sets consisting of all finite intersections of elements of   together with the set   forms a basis for  [1] This means that every proper open set in   can be written as a union of finite intersections of elements of   Explicitly, given a point   in an open set   there are finitely many sets   of   such that the intersection of these sets contains   and is contained in  

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

For any subcollection   of the power set   there is a unique topology having   as a subbase. In particular, the intersection of all topologies on   containing   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   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 edit

Less commonly, a slightly different definition of subbase is given which requires that the subbase   cover  [2] In this case,   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   with subcollections   of the topology such that   is the smallest topology containing  , yet   does not cover  . (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,[3] one must assume that   covers  

Examples edit

The topology generated by any subset   (including by the empty set  ) is equal to the trivial topology  

If   is a topology on   and   is a basis for   then the topology generated by   is   Thus any basis   for a topology   is also a subbasis for   If   is any subset of   then the topology generated by   will be a subset of  

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

The subbase consisting of all semi-infinite open intervals of the form   alone, where   is a real number, does not generate the usual topology. The resulting topology does not satisfy the T1 separation axiom, since if   every open set containing   also contains  

The initial topology on   defined by a family of functions   where each   has a topology, is the coarsest topology on   such that each   is continuous. Because continuity can be defined in terms of the inverse images of open sets, this means that the initial topology on   is given by taking all   where   ranges over all open subsets of   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   to   has for a subbase the set of functions

where   is compact and   is an open subset of  

Suppose that   is a Hausdorff topological space with   containing two or more elements (for example,   with the Euclidean topology). Let   be any non-empty open subset of   (for example,   could be a non-empty bounded open interval in  ) and let   denote the subspace topology on   that   inherits from   (so  ). Then the topology generated by   on   is equal to the union   (see the footnote for an explanation), [note 1] where   (since   is Hausdorff, equality will hold if and only if  ). Note that if   is a proper subset of   then   is the smallest topology on   containing   yet   does not cover   (that is, the union   is a proper subset of  ).

Results using subbases edit

One nice fact about subbases is that continuity of a function need only be checked on a subbase of the range. That is, if   is a map between topological spaces and if   is a subbase for   then   is continuous if and only if   is open in   for every   A net (or sequence)   converges to a point   if and only if every subbasic neighborhood of   contains all   for sufficiently large  

Alexander subbase theorem edit

The Alexander Subbase Theorem is a significant result concerning subbases that is due to James Waddell Alexander II.[3] The corresponding result for basic (rather than subbasic) open covers is much easier to prove.

Alexander Subbase Theorem:[3][1] Let   be a topological space. If   has a subbasis   such that every cover of   by elements from   has a finite subcover, then   is compact.

The converse to this theorem also holds and it is proven by using   (since every topology is a subbasis for itself).

If   is compact and   is a subbasis for   every cover of   by elements from   has a finite subcover.

Suppose for the sake of contradiction that the space   is not compact (so   is an infinite set), yet every subbasic cover from   has a finite subcover. Let   denote the set of all open covers of   that do not have any finite subcover of   Partially order   by subset inclusion and use Zorn's Lemma to find an element   that is a maximal element of   Observe that:

  1. Since   by definition of     is an open cover of   and there does not exist any finite subset of   that covers   (so in particular,   is infinite).
  2. The maximality of   in   implies that if   is an open set of   such that   then   has a finite subcover, which must necessarily be of the form   for some finite subset   of   (this finite subset depends on the choice of  ).

We will begin by showing that   is not a cover of   Suppose that   was a cover of   which in particular implies that   is a cover of   by elements of   The theorem's hypothesis on   implies that there exists a finite subset of   that covers   which would simultaneously also be a finite subcover of   by elements of   (since  ). But this contradicts   which proves that   does not cover  

Since   does not cover   there exists some   that is not covered by   (that is,   is not contained in any element of  ). But since   does cover   there also exists some   such that   Since   is a subbasis generating  '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   was such that   then also   so the fact that   would then imply that   is covered by   which contradicts how   was chosen (recall that   was chosen specifically so that it was not covered by  ).

As mentioned earlier, the maximality of   in   implies that for every   there exists a finite subset   of   such that  forms a finite cover of   Define

which is a finite subset of   Observe that for every     is a finite cover of   so let us replace every   with  

Let   denote the union of all sets in   (which is an open subset of  ) and let   denote the complement of   in   Observe that for any subset     covers   if and only if   In particular, for every   the fact that   covers   implies that   Since   was arbitrary, we have   Recalling that   we thus have   which is equivalent to   being a cover of   Moreover,   is a finite cover of   with   Thus   has a finite subcover of   which contradicts the fact that   Therefore, the original assumption that   is not compact must be wrong, which proves that   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.[3]

Using this theorem with the subbase for   above, one can give a very easy proof that bounded closed intervals in   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 subbasic family   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   does not have a finite subcover. Being cylinder sets, this means their projections onto   have no finite subcover, and since each   is compact, we can find a point   that is not covered by the projections of   onto   But then   is not covered by    

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  

See also edit

Notes edit

  1. ^ Since   is a topology on   and   is an open subset of  , it is easy to verify that   is a topology on  . In particular,   is closed under unions and finite intersections because   is. But since  ,   is not a topology on   an   is clearly the smallest topology on   containing  ).

Citations edit

  1. ^ a b Rudin 1991, p. 392 Appendix A2.
  2. ^ Munkres 2000, pp. 82.
  3. ^ a b c d Muger, Michael (2020). Topology for the Working Mathematician.

References edit