Topology/Bases

Motivation

Often it is difficult to list every open set of a topology directly. A common strategy is to specify a smaller family of sets and then declare the open sets to be the unions of members of that family. This page develops that idea carefully for bases and subbases (sometimes called semibases).

Definition: Base for a Given Topology

Let $(X,𝒯)$ be a topological space. A collection $ℬ\subseteq 𝒯$ is called a base (basis) for $𝒯$ if every open set $U\in 𝒯$ can be written as a union of elements of $ℬ$.

Remarks.

• Trivially, $𝒯$ is a base for itself.
• If $ℬ$ is a base and $U\in 𝒯$, then $U=\bigcup \{B\in ℬ\mid B\subseteq U\}$. (No choice is needed: for each $x\in U$, there is some $B\in ℬ$ with $x\in B\subseteq U$, hence $x$ lies in the displayed union.)

Pointwise Characterization

In a topological space $(X,𝒯)$, a collection $ℬ\subseteq 𝒯$ is a base for $𝒯$ if and only if for every open set $U\in 𝒯$ and every point $x\in U$, there exists $B\in ℬ$ such that $x\in B\subseteq U$.

Proof. The “only if” direction is immediate because $U$ is a union of elements of $ℬ$. For the converse, set $U^{\prime }=\bigcup \{B\in ℬ\mid B\subseteq U\}$. The hypothesis implies every $x\in U$ lies in some such $B$, hence $x\in U^{\prime }$; therefore $U\subseteq U^{\prime }$. The reverse inclusion is obvious, so $U=U^{\prime }$ is a union of basis elements. ■

Constructing a Topology from a Base

Let $X$ be any set and $ℬ$ a collection of subsets of $X$. We say that $ℬ$ is a base on $X$ if the following two conditions hold:

1. (B1) For each $x\in X$, there exists $B\in ℬ$ with $x\in B$.
2. (B2) If $x\in B_1\cap B_2$ for $B_1,B_2\in ℬ$, then there exists $B\in ℬ$ with $x\in B\subseteq B_1\cap B_2$.

Given such $ℬ$, define $𝒯(ℬ)$ to be the collection of all $U\subseteq X$ such that for every $x\in U$ there exists $B\in ℬ$ with $x\in B\subseteq U$. We call $𝒯(ℬ)$ the topology generated by $ℬ$.

Theorem. If $ℬ$ satisfies (B1)–(B2), then $𝒯(ℬ)$ is a topology on $X$. Moreover, every element of $ℬ$ lies in $𝒯(ℬ)$, and every set in $𝒯(ℬ)$ is a union of elements of $ℬ$.

Proof. The empty set is vacuously in $𝒯(ℬ)$; by (B1) the whole set $X$ is in $𝒯(ℬ)$. Arbitrary unions are handled pointwise using the definition. For finite intersections, if $x\in U_1\cap U_2$ with $U_i\in 𝒯(ℬ)$, pick $B_i\in ℬ$ such that $x\in B_i\subseteq U_i$. By (B2) there is $B\in ℬ$ with $x\in B\subseteq B_1\cap B_2\subseteq U_1\cap U_2$, showing $U_1\cap U_2\in 𝒯(ℬ)$. The remaining assertions follow directly from the definition. ■

Lemma (Unions of basis elements). If $ℬ$ satisfies (B1)–(B2), then $𝒯(ℬ)=\{\bigcup 𝒰|𝒰\subseteq ℬ\}$.

Minimality. $𝒯(ℬ)$ is the intersection of all topologies on $X$ that contain $ℬ$.

Extracting a Base from a Given Topology

Let $(X,𝒯)$ be a topological space. If $𝒞\subseteq 𝒯$ has the property that for every open set $U\in 𝒯$ and every $x\in U$, there exists $C\in 𝒞$ with $x\in C\subseteq U$, then $𝒞$ is a base for $𝒯$ (and $𝒯=𝒯(𝒞)$).

Proof. The pointwise property is precisely the criterion from the first section. ■

Comparing Topologies via Bases

Suppose $ℬ$ and ${ℬ}^{\prime }$ are bases for topologies $𝒯$ and ${𝒯}^{\prime }$ on the same set $X$. Then the following are equivalent:

• ${𝒯}^{\prime }$ is finer than $𝒯$ (that is, $𝒯\subseteq {𝒯}^{\prime }$).
• For each $x\in X$ and each $B\in ℬ$ with $x\in B$, there exists $B^{\prime }\in {ℬ}^{\prime }$ with $x\in B^{\prime }\subseteq B$.

(2 ⇒ 1). Let $U\in 𝒯$. Fix $x\in U$. Because $ℬ$ is a base for $𝒯$, there exists $B\in ℬ$ with $x\in B\subseteq U$. By (2), there is $B^{\prime }\in {ℬ}^{\prime }$ such that $x\in B^{\prime }\subseteq B\subseteq U$. Since this argument works for each $x\in U$, the pointwise (basis) definition of openness for ${ℬ}^{\prime }$ shows $U\in {𝒯}^{\prime }$.

• (Note we did not need to choose all $B^{\prime }$ simultaneously; the definition of openness is pointwise.)*

(1 ⇒ 2). Suppose ${𝒯}^{\prime }$ is finer than $𝒯$. Let $x\in X$ and $B\in ℬ$ with $x\in B$. Then $B\in 𝒯\subseteq {𝒯}^{\prime }$. Because ${ℬ}^{\prime }$ is a base for ${𝒯}^{\prime }$, there exists $B^{\prime }\in {ℬ}^{\prime }$ with $x\in B^{\prime }\subseteq B$. ∎

Subbases (a.k.a. Semibases)

A collection $𝒮$ of subsets of $X$ is a subbasis (many texts say semibase) if $\bigcup 𝒮=X$. The topology generated by $𝒮$ is defined to be

$𝒯(𝒮)=\{\bigcup 𝒰|𝒰\subseteq ℬ\}$,

where $ℬ$ is the collection of all finite intersections of elements of $𝒮$ (allowing the empty intersection, which is $X$). In particular, $ℬ$ is a base and $𝒯(𝒮)=𝒯(ℬ)$.

Proof that $ℬ$ is a base. (B1) holds because each $x\in X$ lies in some $S\in 𝒮$, hence in a finite intersection containing $S$. For (B2), the intersection of two finite intersections of elements of $𝒮$ is again a finite intersection of elements of $𝒮$.

Minimality. $𝒯(𝒮)$ is the intersection of all topologies on $X$ that contain $𝒮$.

Comparison via subbases

Let $\mathrm{\Sigma }$ and ${\mathrm{\Sigma }}^{\prime }$ be subbases generating $𝒯$ and ${𝒯}^{\prime }$, respectively, on $X$. Then ${𝒯}^{\prime }$ is finer than $𝒯$ iff for each $x\in X$ and each $S\in \mathrm{\Sigma }$ with $x\in S$, there exist $S{\text{'}}_1,\dots ,S{\text{'}}_n\in {\mathrm{\Sigma }}^{\prime }$ (with $n\ge 1$) such that

$x\in \bigcap _{i=1}^{n}S{\text{'}}_i\subseteq S.$

Proof. The base generated by ${\mathrm{\Sigma }}^{\prime }$ consists of all finite intersections of members of ${\mathrm{\Sigma }}^{\prime }$. Apply the basis comparison lemma above with $ℬ=$ finite intersections from $\mathrm{\Sigma }$ (in practice each $S\in \mathrm{\Sigma }$ is itself a basis element) and ${ℬ}^{\prime }=$ finite intersections from ${\mathrm{\Sigma }}^{\prime }$. ∎

Examples

1. (Standard topology on $ℝ$.) The collection of all open intervals $(a,b)$ is a base.
2. (Lower limit topology.) The half-open intervals $[a,b)$ (with $a<b$) form a base for the lower limit (or Sorgenfrey) topology on $ℝ$, often denoted ${ℝ}_{\ell }$.
3. ($K$-topology.) Let $K=\{1/n\mid n\in \mathbb{N}\}$. The family consisting of all open intervals $(a,b)$ together with all sets of the form $(a,b)\setminus K$ is a base for a topology on $ℝ$, called the $K$-topology, denoted ${ℝ}_K$.
4. (Discrete topology.) For any set $X$, the singletons $\{x\}$ ($x\in X$) form a base for the discrete topology.
5. (The plane.) In ${ℝ}^2$, the sets $\{(x,y)\mid (x-a{)}^2+(y-b{)}^2<r^2\}$ (open disks) form a base; so do the axis-parallel open rectangles. Each base generates the usual Euclidean topology, and each refines the other at every point (small rectangles fit inside disks and vice versa).

Further Terminology

• A local base (or neighborhood base) at a point $x\in X$ is a collection ${ℬ}_x$ of neighborhoods of $x$ such that every neighborhood of $x$ contains some member of ${ℬ}_x$.
• A space is second countable if it has a countable base. For example, the intervals $(a,b)$ with rational endpoints form a countable base for the standard topology on $ℝ$.

Exercises

1. Show that $ℬ=\{(a,b)\mid a,b\in ℝ,a<b\}$ is a base for a topology on $ℝ$ and that this topology is the usual one.
2. Show that $𝒞=\{[a,b]\mid a<b\}$ is not a base for any topology on $ℝ$ (identify which base axiom fails).
3. Show that $ℒ=\{(a,b]\mid a<b\}$ is a base for a topology on $ℝ$. Compare this topology to the lower limit topology.
4. Let $𝒮=\{[a,b)\mid a<b\}$. Prove that $𝒮$ is a subbasis on $ℝ$ and describe explicitly the associated base of finite intersections.
5. (Comparing topologies.) Let $𝒯$ be the standard topology on $ℝ$, ${𝒯}_{\ell }$ the lower limit topology, and ${𝒯}_K$ the $K$-topology. Show that ${𝒯}_{\ell }$ and ${𝒯}_K$ are each strictly finer than $𝒯$, but are not comparable to each other.
6. (Countable base.) Prove that $ℬ=\{(a,b)\mid a<b,a,b\in \mathbb{Q}\}$ is a base that generates the standard topology on $ℝ$.
7. (Minimality.) Let $𝒜$ be any family of subsets of $X$. Show that the topology generated by $𝒜$ equals the intersection of all topologies containing $𝒜$. Do this both when $𝒜$ is assumed to be a base and when it is assumed to be a subbasis.
8. (Neighborhood bases.) In a metric space, prove that for each $x$ the sets $B(x,1/n)$ ($n\in \mathbb{N}$) form a countable neighborhood base at $x$.