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topology In mathematics, topology (from the Greek words , and ) is concerned with the properties of a geometric object that are preserved under continuous deformations, such as stretching, twisting, crumpling, and bending; that is, without closing ho ...
, a subbase (or subbasis, prebase, prebasis) for a topological space X with
topology In mathematics, topology (from the Greek words , and ) is concerned with the properties of a geometric object that are preserved under continuous deformations, such as stretching, twisting, crumpling, and bending; that is, without closing ho ...
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

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: #The subcollection B ''generates'' the topology T. This means that T is the smallest topology containing B: any topology T^\prime on X containing B must also contain T. #The collection of open sets consisting of all finite
intersections In mathematics, the intersection of two or more objects is another object consisting of everything that is contained in all of the objects simultaneously. For example, in Euclidean geometry, when two lines in a plane are not parallel, their ...
of elements of B, together with the set X, forms a basis for T. This means that every 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 open set U \subseteq X, there are finitely many sets S_1, \ldots, S_n of B, such that the intersection of these sets contains x and is contained in U. (If we use the
nullary intersection In set theory, the intersection of two sets A and B, denoted by A \cap B, is the set containing all elements of A that also belong to B or equivalently, all elements of B that also belong to A. Notation and terminology Intersection is writt ...
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 cover X. In this case, X is the union of all sets contained in \mathcal. 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. In other words, there exist topological spaces (X, \tau) with a subset \mathcal \subseteq \tau, such that \tau is the smallest topology containing \mathcal, yet \mathcal does not cover X (such an example is given below). 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 topology generated by any subset \mathcal \subseteq \ (including by the empty set \mathcal := \varnothing) is equal to the trivial topology \. If \tau is a topology on X and \mathcal is a basis for \tau then the topology generated by \mathcal is \tau. Thus any basis \mathcal for a topology \tau is also a subbasis for \tau. If \mathcal is any subset of \tau then the topology generated by \mathcal will be a subset of \tau. The usual topology on the
real number In mathematics, a real number is a number that can be used to measurement, measure a ''continuous'' one-dimensional quantity such as a distance, time, duration or temperature. Here, ''continuous'' means that values can have arbitrarily small var ...
s \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 \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 (-\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 if a < b every open set containing b also contains a. 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^(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 In topology and related areas of mathematics, a subspace of a topological space ''X'' is a subset ''S'' of ''X'' which is equipped with a topology induced from that of ''X'' called the subspace topology (or the relative topology, or the induced t ...
, 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) = \ where K \subseteq X 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 In topology and related areas of mathematics, a subspace of a topological space ''X'' is a subset ''S'' of ''X'' which is equipped with a topology induced from that of ''X'' called the subspace topology (or the relative topology, or the induced t ...
on Y that Y inherits from (X, \tau) (so \nu \subseteq \tau). Then the topology generated by \nu on X is equal to the union \ \cup \nu (see this footnote for an explanation),Since \nu is a topology on Y and Y is an open subset of (X, \tau), it is easy to verify that \ \cup \nu is a topology on X. Since \nu is not a topology on X, \ \cup \nu is clearly the smallest topology on X containing \nu). where \ \cup \nu \subseteq \tau (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 \ \cup \nu is the smallest topology ''on X'' containing \nu yet \nu does not cover X (that is, the union \bigcup_ V = Y 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 is a subbase for Y, then f : X \to Y is continuous if and only if f^(B) is open in X for every B \in \mathcal. A net (or sequence) x_ = \left(x_i\right)_ 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. :Alexander Subbase Theorem: Let (X, \tau) be a topological space. If X has a subbasis \mathcal such that every cover of X by elements from \mathcal has a finite subcover, then X is compact. The converse to this theorem also holds and it is proven by using \mathcal = \tau (since every topology is a subbasis for itself). :If X is compact and \mathcal is a subbasis for X, every cover of X by elements from \mathcal has a finite subcover. 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 has a finite subcover. Let \mathbb denote the set of all open covers of X that do not have any finite subcover of X. Partially order \mathbb by subset inclusion and use Zorn's Lemma to find an element \mathcal \in \mathbb that is a maximal element of \mathbb. Observe that: # Since \mathcal \in \mathbb, by definition of \mathbb, \mathcal is an open cover of X and there does not exist any finite subset of \mathcal that covers X (so in particular, \mathcal is infinite). # The maximality of \mathcal in \mathbb implies that if V is an open set of X such that V \not\in \mathcal then \mathcal \cup \ has a finite subcover, which must necessarily be of the form \ \cup \mathcal_V for some finite subset \mathcal_V of \mathcal (this finite subset depends on the choice of V). We will begin by showing that \mathcal \cap \mathcal is a cover of X. Suppose that \mathcal \cap \mathcal was a cover of X, which in particular implies that \mathcal \cap \mathcal is a cover of X by elements of \mathcal. The theorem's hypothesis on \mathcal implies that there exists a finite subset of \mathcal \cap \mathcal that covers X, which would simultaneously also be a finite subcover of X by elements of \mathcal (since \mathcal \cap \mathcal \subseteq \mathcal). But this contradicts \mathcal \in \mathbb, which proves that \mathcal \cap \mathcal does not cover X. Since \mathcal \cap \mathcal does not cover X, there exists some x \in X that is not covered by \mathcal \cap \mathcal (that is, x is not contained in any element of \mathcal \cap \mathcal). But since \mathcal does cover X, there also exists some U \in \mathcal such that x \in U. Since \mathcal is a subbasis generating X's topology, from the definition of the topology generated by \mathcal, there must exist a finite collection of subbasic open sets S_1, \ldots, S_n \in \mathcal such that x \in S_1 \cap \cdots \cap S_n \subseteq U. We will now show by contradiction that S_i \not\in \mathcal for every i = 1, \ldots, n. If i was such that S_i \in \mathcal, then also S_i \in \mathcal \cap \mathcal so the fact that x \in S_i would then imply that x is covered by \mathcal \cap \mathcal, which contradicts how x was chosen (recall that x was chosen specifically so that it was not covered by \mathcal \cap \mathcal). As mentioned earlier, the maximality of \mathcal in \mathbb implies that for every i = 1, \ldots, n, there exists a finite subset \mathcal_ of \mathcal such that\left\ \cup \mathcal_ forms a finite cover of X. Define \mathcal_F := \mathcal_ \cup \cdots \cup \mathcal_, which is a finite subset of \mathcal. Observe that for every i = 1, \ldots, n, \left\ \cup \mathcal_F is a finite cover of X so let us replace every \mathcal_ with \mathcal_F. Let \cup \mathcal_F denote the union of all sets in \mathcal_F (which is an open subset of X) and let Z denote the complement of \cup \mathcal_F in X. Observe that for any subset A \subseteq X, \ \cup \mathcal_F covers X if and only if Z \subseteq A. In particular, for every i = 1, \ldots, n, the fact that \left\ \cup \mathcal_F covers X implies that Z \subseteq S_i. Since i was arbitrary, we have Z \subseteq S_1 \cap \cdots \cap S_n. Recalling that S_1 \cap \cdots \cap S_n \subseteq U, we thus have Z \subseteq U, which is equivalent to \ \cup \mathcal_F being a cover of X. Moreover, \ \cup \mathcal_F is a finite cover of X with \ \cup \mathcal_F \subseteq \mathcal. Thus \mathcal has a finite subcover of X, which contradicts the fact that \mathcal \in \mathbb. Therefore, the original assumption that X is not compact must be wrong, which proves that X is compact. \blacksquare 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 \prod_ 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 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, if C_i \neq \varnothing 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 \left(x_i\right)_i \in \prod_ X_i is not covered by C. \blacksquare Note, that in the last step we implicitly used the
axiom of choice In mathematics, the axiom of choice, or AC, is an axiom of set theory equivalent to the statement that ''a Cartesian product of a collection of non-empty sets is non-empty''. Informally put, the axiom of choice says that given any collection ...
(which is actually equivalent to Zorn's lemma) to ensure the existence of \left(x_i\right)_i.


See also

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Notes


References

* * * * Articles containing proofs General topology