Families Viewed Near Each Point
Local finiteness and discreteness describe how a family of sets is distributed around each point of a space. Both conditions ask what a sufficiently small neighborhood can meet, but they impose different limits: local finiteness allows finitely many members, while discreteness allows at most one. That distinction matters for bases, which need not themselves be discrete even when they can be divided into discrete layers.
Throughout, \((X,\mathcal{T})\) is a topological space. A family \(\mathcal{F}\) is a collection of subsets of \(X\); its members need not be open unless it is specified to be a family of base elements. A neighborhood in this tutorial means an open neighborhood.
The neighborhood in either definition may depend on \(x\). A neighborhood meeting no members satisfies both conditions. In particular, the empty family is both locally finite and discrete. Discreteness concerns the number of members a neighborhood intersects, not the number of points in those members.
Here “countable union” allows finitely many layers as well: unused layers can be empty. The members of each layer are still open, since they are members of the base. The layers themselves need not cover \(X\) individually; together they must contain all the base elements.
Why Discreteness Implies Local Finiteness
Proof. Assume that \(\mathcal{F}\) is discrete, and let \(x\in X\). By the definition of a discrete family, there is an open neighborhood \(U\) of \(x\) that intersects at most one member of \(\mathcal{F}\). Since “at most one” is finite, \(U\) intersects only finitely many members. This holds for every \(x\), so \(\mathcal{F}\) is locally finite. \(\square\)
The proof uses the same neighborhood for both tests; it is the permitted number of members that changes. Consequently, each discrete layer in a sigma-discrete base is locally finite.
Proof. Suppose \(\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{B}_n\), where each \(\mathcal{B}_n\) is discrete. The theorem shows that every \(\mathcal{B}_n\) is locally finite. The same countable collection of layers therefore expresses \(\mathcal{B}\) as a countable union of locally finite families. \(\square\)
Worked Examples
Worked Example: The Integer Singletons in the Real Line
Give \(\mathbb{R}\) its standard topology and consider the family \(\mathcal{F}=\{\{n\}:n\in\mathbb{Z}\}\). We check directly that it is discrete.
If \(x=m\) for some \(m\in\mathbb{Z}\), the open interval \((m-\tfrac12,m+\tfrac12)\) contains \(m\) and no other integer. It therefore intersects exactly one member of \(\mathcal{F}\), namely \(\{m\}\).
If \(x\notin\mathbb{Z}\), let \(m\) be the integer with \(m<x<m+1\), and set \(\delta=\min(x-m,m+1-x)\). Both numbers in the minimum are positive, so \(\delta>0\). The interval \((x-\delta/2,x+\delta/2)\) lies strictly between \(m\) and \(m+1\), and therefore contains no integer. It intersects no member of \(\mathcal{F}\).
These neighborhoods verify discreteness at every real point. They also verify local finiteness: each chosen neighborhood intersects either one member or none. Check: at \(x=3\), the interval \((2.5,3.5)\) contains the integer \(3\) and no other integer, so it meets exactly \(\{3\}\).
Worked Example: Locally Finite but Not Discrete
In \(\mathbb{R}\) with its standard topology, take \(\mathcal{G}=\{(-1,1),(0,2)\}\). This family is locally finite: for every \(x\in\mathbb{R}\), every neighborhood of \(x\) intersects at most the two members of \(\mathcal{G}\). Two is finite, so the local-finiteness condition holds at every point.
It is not discrete. The point \(x=\tfrac12\) belongs to both members. Every open neighborhood \(U\) of \(\tfrac12\) intersects each member, because \(\tfrac12\in U\cap(-1,1)\) and \(\tfrac12\in U\cap(0,2)\). Thus no neighborhood of \(\tfrac12\) meets at most one member.
Check: the particular neighborhood \((\tfrac14,\tfrac34)\) intersects both intervals, and the same argument applies to every neighborhood of \(\tfrac12\). This is a direct failure of discreteness, not merely a failure of one proposed choice of neighborhood.
Worked Example: A Sigma-Discrete Base That Is Not Discrete
Let \(X=\mathbb{N}\) have the discrete topology, so every subset of \(X\) is open. Consider the base \(\mathcal{B}=\{\{n\}:n\in\mathbb{N}\}\cup\{X\}\).
To check the base condition, let \(U\subseteq X\) be open and \(x\in U\). The member \(\{x\}\in\mathcal{B}\) satisfies \(x\in\{x\}\subseteq U\). Thus \(\mathcal{B}\) is a base.
Write \(\mathcal{B}_1=\{\{n\}:n\in\mathbb{N}\}\) and \(\mathcal{B}_2=\{X\}\). For any \(x\in X\), the open neighborhood \(\{x\}\) intersects exactly one member of \(\mathcal{B}_1\), namely \(\{x\}\). Hence \(\mathcal{B}_1\) is discrete. The family \(\mathcal{B}_2\) has only one member, so every neighborhood intersects at most one member; it is discrete as well. Therefore \(\mathcal{B}\) is sigma-discrete.
The whole base \(\mathcal{B}\), however, is not discrete. For any \(n\in\mathbb{N}\), every open neighborhood \(U\) of \(n\) contains \(n\). It intersects both \(\{n\}\) and \(X\), which are distinct members of \(\mathcal{B}\). So no such neighborhood meets at most one member of \(\mathcal{B}\). By the corollary, this sigma-discrete base is also sigma-locally finite.
Check: at \(n=1\), the neighborhood \(\{1\}\) intersects the two base members \(\{1\}\) and \(X\). It meets only two members, which is consistent with local finiteness, but it does not meet at most one.
Comparing the Base Conditions
The implication proved above compares the two conditions precisely: a sigma-discrete base is necessarily sigma-locally finite. The reason is layer-by-layer: every discrete layer is locally finite. The definitions do not make the reverse implication automatic. A locally finite family is allowed to have a neighborhood intersecting several members, as the interval example demonstrates; the argument proving the corollary cannot be reversed.
There is an important distinction between a base and a particular family of sets. The interval example shows that local finiteness of a family does not imply discreteness of that family. It does not, by itself, show that a space with some sigma-locally-finite base cannot have some other sigma-discrete base. The base conditions ask whether there exists a suitable decomposition of a base, so an example about one family cannot settle that separate question.
When reading a decomposition, check the quantifiers in each layer. For each point, a discrete layer needs a neighborhood meeting at most one of its members; a locally finite layer needs a neighborhood meeting finitely many. The neighborhood may vary from point to point and from layer to layer. No single neighborhood is required to work throughout the space or for every layer at once.
Check Your Understanding
For each question, apply the neighborhood definitions at every point that matters.
- In \(\mathbb{R}\) with the standard topology, is the family \(\{(3n,3n+1):n\in\mathbb{Z}\}\) discrete? Explain how neighborhoods near interval endpoints behave.
- Give a locally finite family that is not discrete, and identify a point at which every neighborhood meets at least two members.
- Why does a discrete family automatically satisfy the local-finiteness condition, including when the family is empty?
- In the discrete topology on \(\mathbb{N}\), verify that the family of all singleton sets is a base and is discrete.
- Does the interval example alone prove that a sigma-locally-finite base cannot be sigma-discrete? Explain the difference between a claim about one family and a claim about the existence of a base decomposition.