Subbase

This article is about an object in mathematical topology. For the term in highway engineering, see Subbase (pavement). For the frequency range, see Sub-bass.

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:

1. The subcollection $B$ generates the topology $\tau .$ This means that $\tau$ is the smallest topology containing $B$: any topology $\tau ^{\prime }$ on $X$ containing $B$ must also contain $\tau .$
2. The collection of open sets consisting of all finite intersections of elements of $B$ forms a basis for $\tau .$ 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 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 $\wp (X),$ 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 ${\mathcal {B}}.$ 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 two 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 $X$. (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 ${\mathcal {S}}\subseteq \{\varnothing ,X\}$ (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 $\mathbb {R}$ 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, $X=\mathbb {R}$ with the Euclidean topology). Let $Y\in \tau$ be any non-empty open subset of $(X,\tau )$ (for example, $Y$ could be a non-empty bounded open interval in $\mathbb {R}$) 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), 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle f^{-1}(B)} is open in $X$ for every $B\in {\mathcal {B}}.$ A net (or sequence) $x_{\bullet }=\left(x_{i}\right)_{i\in I}$ converges to a point $x$ if and only if every subbasic 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 ${\mathcal {S}}$ such that every cover of $X$ by elements from ${\mathcal {S}}$ has a finite subcover, then $X$ is compact.

The converse to this theorem also holds and it is proven by using ${\mathcal {S}}=\tau$ (since every topology is a subbasis for itself).

If $X$ is compact and ${\mathcal {S}}$ is a subbasis for $X,$ every cover of $X$ by elements from ${\mathcal {S}}$ has a finite subcover.

Proof

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: 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). 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 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 $x\in U.$ Since ${\mathcal {S}}$ is a subbasis generating $X$'s topology, from the definition of the topology generated by ${\mathcal {S}},$ there must exist a finite collection of subbasic open sets $S_{1},\ldots ,S_{n}\in {\mathcal {S}}$ 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. $\blacksquare$

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 $\mathbb {R}$ above, one can give a very easy proof that bounded closed intervals in $\mathbb {R}$ 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.

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, if $C_{i}\neq \varnothing$ then $C_{i}$ does 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 $\left(x_{i}\right)_{i}.$

See also

• Base (topology) – Collection of open sets used to define a topology

References

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