Sigma-Locally-Finite Bases

From Local Finiteness to a Base

A base can describe all the open sets of a space even when the base itself is uncountable. In metrization arguments, a useful additional feature is that the base can be separated into countably many families, each locally finite. Local finiteness controls one family at a time: near any point, only finitely many members of that family occur. The countably many families together can still provide base elements small enough to fit inside any prescribed open neighborhood.

Throughout, $(X,\mathcal{T})$ is a topological space. A family here is a collection of subsets of $X$; its members need not be disjoint.

Definition: A family $\mathcal{F}$ of subsets of $X$ is locally finite if for every $x\in X$, there is an open neighborhood $U$ of $x$ that intersects only finitely many members of $\mathcal{F}$. A family $\mathcal{B}$ of open subsets of $X$ is a base if, whenever $x\in O\in\mathcal{T}$, there is $B\in\mathcal{B}$ with $x\in B\subseteq O$. A base is sigma-locally-finite if it can be written as $\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{B}_n$, where every family $\mathcal{B}_n$ is locally finite.

The index $n$ identifies the layer; a layer can contain many base elements. Local finiteness is required separately for each layer. It does not say that the entire union of the layers is locally finite.

How the Layers Generate the Topology

Theorem: If $\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{B}_n$ is a base for $(X,\mathcal{T})$, then every open set $O\in\mathcal{T}$ is the union of the base elements from these layers that are contained in $O$: $$ O=\bigcup\{B\in\mathcal{B}:B\subseteq O\}. $$

Proof. Every set in the union on the right is contained in $O$, so that union is a subset of $O$. For the reverse inclusion, take any $x\in O$. The base property gives $B\in\mathcal{B}$ with $x\in B\subseteq O$. Since $\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{B}_n$, this $B$ belongs to at least one layer and occurs in the union on the right. Thus $x$ belongs to that union. Both inclusions give the equality. If $X=\varnothing$, the equality holds as well: both sides are empty for $O=\varnothing$. $\square$

This statement explains the role of the layers: their union is the base, and the base elements lying inside any particular open set recover that open set by union. It does not assert that the topology is a countable collection of open sets. A topology can contain unions of arbitrarily many base elements.

Countable Bases and Sigma-Local Finiteness

Proposition: Every countable base is sigma-locally-finite. A sigma-locally-finite base need not be countable.

Proof. Let $\mathcal{B}$ be a countable base. If it is finite, list its elements and use singleton families containing one listed element each; after the list ends, take the remaining layers to be empty. If it is countably infinite, enumerate its elements as $B_1,B_2,\ldots$ and set $\mathcal{B}_n=\{B_n\}$. Each such family is locally finite: any open neighborhood of any point intersects at most one member of a family with only one member. These layers have union $\mathcal{B}$, so the base is sigma-locally-finite.

For the failure of the converse, take an uncountable set $X$ with the discrete topology $\mathcal{T}=\mathcal{P}(X)$. The family $\mathcal{B}=\{\{x\}:x\in X\}$ is a base: for $x\in O\subseteq X$, the member $\{x\}$ satisfies $x\in\{x\}\subseteq O$. It is locally finite, since the open neighborhood $\{x\}$ of $x$ intersects only one member of $\mathcal{B}$, namely $\{x\}$. Therefore it is itself a sigma-locally-finite base, by taking $\mathcal{B}_1=\mathcal{B}$ and all later layers empty.

This space has no countable base. Indeed, if $\mathcal{C}$ were a base and $x\in X$, apply the base property to the open set $\{x\}$. It forces some $C\in\mathcal{C}$ to satisfy $x\in C\subseteq\{x\}$, hence $C=\{x\}$. Thus a base must contain a distinct singleton for every $x\in X$; no countable family can do so when $X$ is uncountable. $\square$

The two properties should therefore not be identified. A countable base is one sufficient way to obtain a sigma-locally-finite base, but layers may be uncountable even when each is locally finite.

Locally Finite Metric-Ball Covers

Here is a useful condition that turns a sequence of covers into a sigma-locally-finite base. In a metric space $(X,d)$, the diameter of a nonempty set $V\subseteq X$ is $\operatorname{diam}(V)=\sup\{d(y,z):y,z\in V\}$.

Theorem: Suppose that for each $n\in\mathbb{N}$, $\mathcal{U}_n$ is a locally finite open cover of a metric space $(X,d)$, every member of $\mathcal{U}_n$ is a metric ball, and every member has diameter less than $1/n$. Then $\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{U}_n$ is a sigma-locally-finite base for the metric topology on $X$.

Proof. By hypothesis, each $\mathcal{U}_n$ is locally finite. It remains to prove the base property. Let $x\in O$, where $O$ is open in the metric topology. By the definition of that topology, there is $\varepsilon>0$ such that $B_d(x,\varepsilon)\subseteq O$. Choose $n\in\mathbb{N}$ with $1/n<\varepsilon$. Since $\mathcal{U}_n$ covers $X$, there is $V\in\mathcal{U}_n$ with $x\in V$. For every $y\in V$, the definition of diameter gives $d(x,y)\leq\operatorname{diam}(V)<1/n<\varepsilon$. Hence $x\in V\subseteq B_d(x,\varepsilon)\subseteq O$. This is the base property, and the specified locally finite layers show that $\mathcal{B}$ is sigma-locally-finite. If $X=\varnothing$, all the cover and base conditions hold vacuously. $\square$

The shrinking diameters do the base work: a cover member containing $x$ is eventually small enough to fit inside a metric ball around $x$. Local finiteness is a separate requirement on each cover. Neither condition alone is being used as a substitute for the other.

Worked Example: A Finite Discrete Metric Space

Let $X=\{a,b,c\}$ and define $d(x,y)=0$ when $x=y$, and $d(x,y)=1$ when $x\ne y$. Give $X$ the metric topology. For every $n\in\mathbb{N}$, let $\mathcal{U}_n=\{B_d(a,1/2),B_d(b,1/2),B_d(c,1/2)\}$.

For any $x\in X$, the ball $B_d(x,1/2)$ consists of the points $y$ with $d(x,y)<1/2$. The distance is $0$ for $y=x$ and $1$ for either other point, so $B_d(x,1/2)=\{x\}$. Consequently, each $\mathcal{U}_n$ is exactly $\{\{a\},\{b\},\{c\}\}$ and covers $X$. Every family on this finite space is locally finite: $X$ itself is an open neighborhood of each point and meets at most the three members of any $\mathcal{U}_n$.

Each singleton has diameter $0$, and $0<1/n$ for every $n\in\mathbb{N}$. The metric-ball theorem therefore applies. Directly, every subset $O\subseteq X$ is the union of the singletons $\{x\}$ for $x\in O$, so these layers do form a base for the discrete topology. The calculation of the balls also checks that no other points accidentally enter a layer member.

Worked Example: A Locally Finite Ball Base on the Real Line

Give $\mathbb{R}$ its standard metric $d(x,y)=|x-y|$. For $n\in\mathbb{N}$, define $$ \mathcal{U}_n=\{B(k/n,1/n):k\in\mathbb{Z}\}. $$

First, each $\mathcal{U}_n$ covers $\mathbb{R}$. For any $x\in\mathbb{R}$, choose an integer $k$ nearest to $nx$, so that $|nx-k|\leq 1/2$. Dividing by $n>0$ gives $|x-k/n|\leq1/(2n)<1/n$, and hence $x\in B(k/n,1/n)$.

Next, each layer is locally finite. Fix $x\in\mathbb{R}$ and use the open neighborhood $B(x,1/n)$. If this neighborhood intersects $B(k/n,1/n)$, there is $y$ in their intersection. The triangle inequality gives $$ |k/n-x|\leq |k/n-y|+|y-x|<1/n+1/n=2/n. $$ Thus $|k-nx|<2$, which permits only finitely many integer values of $k$. So the chosen neighborhood intersects only finitely many members of $\mathcal{U}_n$.

Every member of the layer is an interval of radius $1/n$, so its diameter is $2/n$, not $1/n$. To apply the theorem with the stated diameter bound, instead use the covers $$ \mathcal{V}_n=\{B(k/n,1/(2n)):k\in\mathbb{Z}\}. $$

We check this adjusted choice carefully. It does not cover $\mathbb{R}$: a point halfway between consecutive centers is at distance $1/(2n)$ from each, and the balls are open. Thus this tempting radius adjustment fails the cover condition. A working choice is $$ \mathcal{W}_n=\{B(k/n,2/n):k\in\mathbb{Z}\}. $$

These balls cover because the nearest-center distance is at most $1/(2n)<2/n$. The same local-finiteness argument works with a neighborhood $B(x,2/n)$: an intersecting center satisfies $|k/n-x|<4/n$, leaving finitely many integers $k$. Each ball has diameter $4/n$, so these layers also do not meet the theorem's $1/n$ bound. We need finer center spacing. Set $$ \mathcal{U}_n=\{B(k/n^2,1/n):k\in\mathbb{Z}\}. $$

For this final choice, every $x$ is within $1/(2n^2)$ of a center, and $1/(2n^2)<1/n$, so it covers. If $B(x,1/n)$ intersects a member, its center is within $2/n$ of $x$; only finitely many grid centers $k/n^2$ lie in that bounded interval. Each ball has diameter $2/n$, again not strictly less than $1/n$. The metric-ball theorem above can instead be applied after reindexing scales: use radius $1/(3n)$ and center spacing $1/(3n^2)$. The nearest-center distance is at most $1/(6n^2)<1/(3n)$, and each ball has diameter $2/(3n)$. These diameters tend to zero, so for any $\varepsilon>0$ choose $n$ with $2/(3n)<\varepsilon$. As in the theorem's proof, the layer containing $x$ is then contained in $B(x,\varepsilon)$. Hence the union is a base and is sigma-locally-finite.

For a concrete check, take $x=3/10$ and $n=2$. The centers in this last construction are $k/12$; $k=4$ gives center $1/3$, and $$ |3/10-1/3|=|9/30-10/30|=1/30<1/6. $$ So $x\in B(1/3,1/6)$, as required by the cover condition. The lesson in this computation is to check cover, local finiteness, and diameter separately; a small diameter alone does not guarantee coverage.

A Common Mistake and Its Correction

Common pitfall. One might reason: “There are countably many locally finite layers, so the base is countable.” That conclusion is wrong: countably many families can each be uncountable. In the uncountable discrete example, the single layer of all singleton sets is locally finite and is an uncountable base. What is countable is the set of layer indices, not necessarily the number of members in each layer.

The converse implication also needs care. It is correct that a countable base is sigma-locally-finite: put one base element in each singleton layer. It is incorrect to reverse that implication; the uncountable discrete space verifies the failure directly. Keeping these distinctions clear is useful when a metrization criterion asks for a sigma-locally-finite base rather than a countable one.

Check Your Understanding

Use the definitions to decide what each hypothesis contributes.

In the uncountable discrete space, explain why the family of all singleton sets is locally finite and why no countable base exists.
In the metric-ball theorem, identify where the cover condition is used and where the diameter bound is used in the base-property proof.
For the standard metric on $\mathbb{R}$, show that the family $\{B(k,1):k\in\mathbb{Z}\}$ is not a cover by testing $x=1/2$.
Show that a finite base is sigma-locally-finite by specifying its layers, including the case of an empty base.
Suppose each $\mathcal{U}_n$ is a locally finite open cover of a metric space and all its members have diameter less than $1/n$, but the members are not required to be balls. Does the proof that their union is a base still work? Justify your answer.

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