Subbase#Alexander subbase theorem

{{short description|Collection of subsets that generate a topology}}

{{about|an object in mathematical topology|the term in highway engineering|Subbase (pavement)|the frequency range|Sub-bass}}

In topology, a subbase (or subbasis, prebase, prebasis) for the topology {{math|τ}} of a topological space {{math|(X, τ)}} 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.

Subbase is a weaker notion than that of a base for a topology.

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 three following equivalent conditions:

  1. \tau is the smallest topology containing B: any topology \tau^\prime on X containing B must also contain \tau.
  2. \tau is the intersection of all topologies on X containing B.
  3. The collection of open sets consisting of X and all finite intersections of elements of B forms a basis for \tau.{{sfn|Rudin|1991|p=392 Appendix A2}}Rudin's definition is less general than ours, because it effectively requires that B covers X (see "Alternative definition" subsection below). We drop this requirement here, and assume that B is any subset of \mathcal{P}(X) This means that every proper open set in \tau can be written as a union of finite intersections of elements of B. Explicitly, given a point x in an open set U \subsetneq 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 additionally assume that B covers X, or if we use the nullary intersection convention, then there is no need to include X in the third definition.

If B is a subbase of \tau, we say that B generates the topology \tau. This terminology originates from the explicit construction of \tau from B using the second or third definition above.

Elements of subbase are called {{em|subbasic (open) sets}}. A cover composed of subbasic sets is called a {{em|subbasic (open) cover}}.

For {{em|any}} subcollection S of the power set \wp(X), there is a unique topology having S as a subbase; it is the intersection of all topologies on X containing S. In general, however, the converse is not true, i.e. 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 three conditions is more useful than the others.

=Alternative definition=

Less commonly, a slightly different definition of subbase is given which requires that the subbase \mathcal{B} cover X.{{sfn | Munkres | 2000 | pp=82}} In this case, X is the union of all sets contained in \mathcal{B}. 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 three definitions above. There exist topological spaces (X, \tau) with subcollections \mathcal{B} \subseteq \tau of the topology such that \tau is the smallest topology containing \mathcal{B}, yet \mathcal{B} does not cover X. For example, consider a topological space (X,\tau) with \tau=\{\varnothing, \{p\}, X\} and \mathcal{B}=\{\{p\}\} for some p\in X. Clearly, \mathcal{B} is a subbase of \tau, yet \mathcal{B} doesn't cover X as long as X has at least 2 elements. 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{S} \subseteq \{\varnothing, X\} (including by the empty set \mathcal{S} := \varnothing) is equal to the trivial topology \{\varnothing, X\}.

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 (-\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^{-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 : X \to Y \mid f(K) \subseteq 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 {{em|open}} 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 \nu \subseteq \tau). Then the topology generated by \nu on X is equal to the union \{X\} \cup \nu (see the 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 \{X\} \cup \nu is a topology on X. In particular, \nu is closed under unions and finite intersections because \tau is. But since X \not\in \nu , \nu is not a topology on X an \{X\} \cup \nu is clearly the smallest topology on X containing \nu).

where \{X\} \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 \{X\} \cup \nu is the smallest topology on X containing \nu yet \nu does not cover X (that is, the union \bigcup_{V \in \nu} 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{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 B \in \mathcal{B}.

A net (or sequence) x_{\bull} = \left(x_i\right)_{i \in I} converges to a point x if and only if every {{em|sub}}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:{{cite book|last=Muger|first= Michael|title=Topology for the Working Mathematician|year=2020}}{{sfn|Rudin|1991|p=392 Appendix A2}} Let (X, \tau) be a topological space, and \mathcal{S} be a subbase of \tau. If every cover of X by elements from \mathcal{S} has a finite subcover, then X is compact.

The converse to this theorem also holds (because every cover of X by elements of \mathcal{S} is an open cover of X)

:Let (X, \tau) be a topological space, and \mathcal{S} be a subbase of \tau. If X is compact, then every cover of X by elements from \mathcal{S} has a finite subcover.

{{collapse top|title=Proof|left=true}}

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 \mathcal{C} \in \mathbb{S} that is a maximal element of \mathbb{S}.

Observe that:

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

We will begin by showing that \mathcal{C} \cap \mathcal{S} is {{em|not}} a cover of X.

Suppose that \mathcal{C} \cap \mathcal{S} was a cover of X, which in particular implies that \mathcal{C} \cap \mathcal{S} is a cover of X by elements of \mathcal{S}.

The theorem's hypothesis on \mathcal{S} implies that there exists a finite subset of \mathcal{C} \cap \mathcal{S} that covers X, which would simultaneously also be a finite subcover of X by elements of \mathcal{C} (since \mathcal{C} \cap \mathcal{S} \subseteq \mathcal{C}).

But this contradicts \mathcal{C} \in \mathbb{S}, which proves that \mathcal{C} \cap \mathcal{S} does not cover X.

Since \mathcal{C} \cap \mathcal{S} does not cover X, there exists some x \in X that is not covered by \mathcal{C} \cap \mathcal{S} (that is, x is not contained in any element of \mathcal{C} \cap \mathcal{S}).

But since \mathcal{C} does cover X, there also exists some U \in \mathcal{C} such that x \in U.

It follows that U\neq X, because otherwise it would imply \mathcal{C} has a finite subcover of X, namely the subcover \{U\}=\{X\}, contradicting \mathcal{C}\in \mathbb{S}. Since U\neq X, and \mathcal{S} is a subbasis generating X's topology (together with X), from the definition of the topology generated by \mathcal{S}, there must exist a finite collection of subbasic open sets S_1, \ldots, S_n \in \mathcal{S} with n\ge 1 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{C} for every i = 1, \ldots, n.

If i was such that S_i \in \mathcal{C}, then also S_i \in \mathcal{C} \cap \mathcal{S} so the fact that x \in S_i would then imply that x is covered by \mathcal{C} \cap \mathcal{S}, which contradicts how x was chosen (recall that x was chosen specifically so that it was not covered by \mathcal{C} \cap \mathcal{S}).

As mentioned earlier, the maximality of \mathcal{C} in \mathbb{S} implies that for every i = 1, \ldots, n, there exists a finite subset \mathcal{C}_{S_i} of \mathcal{C} such that\left\{S_i\right\} \cup \mathcal{C}_{S_i} forms a finite cover of X.

Define

\mathcal{C}_F := \mathcal{C}_{S_1} \cup \cdots \cup \mathcal{C}_{S_n},

which is a finite subset of \mathcal{C}.

Observe that for every i = 1, \ldots, n, \left\{S_i\right\} \cup \mathcal{C}_F is a finite cover of X so let us replace every \mathcal{C}_{S_i} with \mathcal{C}_F.

Let \cup \mathcal{C}_F denote the union of all sets in \mathcal{C}_F (which is an open subset of X) and let Z denote the complement of \cup \mathcal{C}_F in X.

Observe that for any subset A \subseteq X, \{A\} \cup \mathcal{C}_F covers X if and only if Z \subseteq A.

In particular, for every i = 1, \ldots, n, the fact that \left\{S_i\right\} \cup \mathcal{C}_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 \{U\} \cup \mathcal{C}_F being a cover of X.

Moreover, \{U\} \cup \mathcal{C}_F is a finite cover of X with \{U\} \cup \mathcal{C}_F \subseteq \mathcal{C}.

Thus \mathcal{C} has a finite subcover of X, which contradicts the fact that \mathcal{C} \in \mathbb{S}.

Therefore, the original assumption that X is not compact must be wrong, which proves that X is compact. \blacksquare

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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.

{{collapse top|title=Proof|left=true}}

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 {{em|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, if C_i \neq \varnothing then C_i does {{em|not}} 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_{i} X_i is not covered by C. \blacksquare

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 \left(x_i\right)_i.

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See also

  • {{annotated link|Base (topology)}}

Notes

{{reflist|group=note}}

Citations

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References

  • {{Bourbaki General Topology Part I Chapters 1-4}}
  • {{Dugundji Topology}}
  • {{Munkres Topology|edition=2}}
  • {{Rudin Walter Functional Analysis|edition=2}}
  • {{Willard General Topology}}

Category:Articles containing proofs

Category:General topology