Subbase

For the term in highway engineering, see Subbase (pavement).

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

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 $T'$ on $X$ containing $B$ must also contain $T$ .
The collection of open sets consisting of all finite intersections 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 $S 1, \dots, S n$ 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 $S$ 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 $P(X)$ 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.

Alternative definition

Sometimes, a slightly different definition of subbase is given which requires that the subbase $B$ cover $X$ .^[1] 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 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 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 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. That is, if $B$ is a subbase for $Y$ , a function $f : X \to Y$ 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 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 not in

C

, then

C \cup {V}

has a finite subcover, necessarily of the form

C 0 \cup {V}.

Consider

C \cap B

, 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

C \cap B

does not cover

X

. Let

x

X

be uncovered.

C

covers

X

, 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 ... \cap S n \subseteq U

Since

x

is uncovered,

S i \notin C

. As noted above, this means that for each

i

S i

along with a finite subfamily

C i

C

, covers

X

. But then

U

and all the

C i

cover

X

, so

C

has a finite subcover after all. 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

↑ 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

This article is issued from Wikipedia - version of the Saturday, April 16, 2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.