Why Local Finiteness Matters
Some arguments about a family of sets work because, near each point, only finitely many members need to be considered. Local finiteness formalizes this condition. It is stronger than requiring each point to belong to only finitely many members: a set may come arbitrarily close to a point without containing it. The distinction matters when studying open covers and unions of closed sets.
Throughout, \((X,\mathcal{T})\) is a topological space. A family is indexed as \(\{A_i:i\in I\}\), so “finitely many members” means finitely many indices. This convention also counts two equal sets at different indices as different family members.
In the point-finite condition, the test is whether \(x\in A_i\). In the locally finite condition, the test is whether a neighborhood of \(x\) intersects \(A_i\). Thus local finiteness implies point-finiteness: if a neighborhood \(U\) meets only finitely many members, then every member containing \(x\) meets \(U\), since \(x\in U\).
Refinement does not by itself mean local finiteness. It says each new set lies inside an old one; local finiteness is an additional condition about how many new sets can meet a neighborhood at once.
Basic Operations on Locally Finite Families
Proof. For a subfamily, use the neighborhood that works for the original family: a subfamily cannot have more members meeting that neighborhood than the original family does. For finitely many locally finite families, write them as \(\{A_i:i\in I\}\), \(\{B_j:j\in J\}\), and so on. At \(x\), choose an open neighborhood for each family that meets only finitely many of its members. The intersection of these finitely many neighborhoods is open and contains \(x\). It can meet only members from the finite lists for the separate families, so it meets only finitely many members of their combined family.
For the closure assertion, fix \(x\in X\) and choose an open neighborhood \(U\) of \(x\) meeting only finitely many \(A_i\). If \(U\cap\overline{A_i}\ne\varnothing\), choose \(y\in U\cap\overline{A_i}\). Since \(U\) is open and contains \(y\), it is a neighborhood of \(y\). Every neighborhood of a point in \(\overline{A_i}\) intersects \(A_i\), so \(U\cap A_i\ne\varnothing\). Consequently, every index for which \(U\) meets \(\overline{A_i}\) is among the finitely many indices for which \(U\) meets \(A_i\). This proves local finiteness of the closure family. \(\square\)
Proof. Since \(A_i\subseteq\bigcup_{i\in I}A_i\), every closed set containing \(\bigcup_{i\in I}A_i\) also contains \(\overline{A_i}\). Equivalently, \(\overline{A_i}\subseteq\overline{\bigcup_{i\in I}A_i}\) for every \(i\), giving one inclusion.
For the reverse inclusion, fix \(x\notin\bigcup_{i\in I}\overline{A_i}\). Choose an open neighborhood \(U\) of \(x\) meeting only finitely many \(A_i\), with those indices \(i_1,\ldots,i_k\). For each \(j\in\{1,\ldots,k\}\), \(x\notin\overline{A_{i_j}}\); by the neighborhood test for closure, there is an open neighborhood \(W_j\) of \(x\) disjoint from \(A_{i_j}\). If \(k=0\), take \(W=U\). If \(k\ge 1\), take \(W=U\cap W_1\cap\cdots\cap W_k\). In either case \(W\) is an open neighborhood of \(x\). It misses every \(A_i\): it misses the listed sets by construction, and any unlisted set does not meet \(U\). Thus \(W\) misses \(\bigcup_{i\in I}A_i\), so \(x\notin\overline{\bigcup_{i\in I}A_i}\). This proves the reverse inclusion. If every \(A_i\) is closed, the displayed equality says that their union equals its closure, hence is closed. \(\square\)
The proof uses local finiteness to reduce the question near \(x\) to finitely many sets. The same reasoning would not work for an arbitrary infinite family: intersecting infinitely many neighborhoods need not produce an open neighborhood.
Examples: Refinements and the Two Finiteness Conditions
Worked Example: A Locally Finite Open Refinement of a Cover of the Real Line
Give a locally finite open refinement of the cover \(\mathcal{U}=\{(-m,m):m\in\mathbb{N}\}\) of \(\mathbb{R}\), with its standard topology.
For each integer \(k\), set \(V_k=(k-1,k+1)\), and let \(\mathcal{V}=\{V_k:k\in\mathbb{Z}\}\). These sets are open. Every real number \(x\) lies between some consecutive integers \(k\) and \(k+1\), and then \(k-1<x<k+1\), so \(x\in V_k\). Thus \(\mathcal{V}\) covers \(\mathbb{R}\).
For each \(k\), choose a positive integer \(m>|k|+1\). If \(y\in V_k\), then \(k-1<y<k+1\), which implies \(-m<y<m\). Hence \(V_k\subseteq(-m,m)\), so \(\mathcal{V}\) refines \(\mathcal{U}\).
To check local finiteness, fix \(x\in\mathbb{R}\) and take \(W=(x-1,x+1)\). The intervals \(W\) and \(V_k\) can intersect only if \(x-2<k<x+2\): intersection requires \(k+1>x-1\) and \(k-1<x+1\). There are only finitely many integers in this bounded interval. Therefore \(W\) meets only finitely many \(V_k\), as required.
Worked Example: Point-Finite but Not Locally Finite
In \(\mathbb{R}\) with its standard topology, consider \(A_n=(0,1/n)\) for \(n\in\mathbb{N}\). This family is point-finite. If \(x\le 0\), then \(x\) belongs to none of its members. If \(x>0\), membership requires \(0<x<1/n\), or equivalently \(n<1/x\); only finitely many positive integers satisfy this bound.
But the family is not locally finite at \(0\). Let \(W\) be any open neighborhood of \(0\). It contains \((-\delta,\delta)\) for some \(\delta>0\). For each \(n\), the number \(t=\min(\delta/2,1/(2n))\) satisfies \(0<t<\delta\) and \(t<1/n\). Thus \(t\in W\cap A_n\). The neighborhood \(W\) meets every member, so no neighborhood of \(0\) meets only finitely many members.
Worked Example: Closure Commutes with a Locally Finite Union
In \(\mathbb{R}\), let \(A_n=(n,n+1)\) for \(n\in\mathbb{Z}\). The family is locally finite: given \(x\in\mathbb{R}\), the neighborhood \((x-1,x+1)\) can meet \(A_n\) only when \(x-2<n<x+1\), which holds for only finitely many integers \(n\).
The union is \(\bigcup_{n\in\mathbb{Z}}A_n=\mathbb{R}\setminus\mathbb{Z}\), and the closure of each interval is \(\overline{A_n}=[n,n+1]\). Every real number is in one of these closed intervals: if it is an integer \(n\), it belongs to \([n,n+1]\); if it is not an integer, it lies between consecutive integers. Hence \(\bigcup_{n\in\mathbb{Z}}\overline{A_n}=\mathbb{R}\). The theorem gives \(\overline{\bigcup_{n\in\mathbb{Z}}A_n}=\mathbb{R}\), in agreement with this computation.
A Common Pitfall: Point-Finite Does Not Mean Locally Finite
Consider the family \(F_n=\{1/n\}\) in \(\mathbb{R}\). Its members are closed, and each point belongs to at most one member, so it is point-finite. A tempting but incorrect calculation is to apply the locally finite closed-union result and conclude that \(\bigcup_{n\in\mathbb{N}}F_n\) is closed. That result requires local finiteness, not merely point-finiteness.
Indeed, every open neighborhood of \(0\) contains \((-\delta,\delta)\) for some \(\delta>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\delta\). Then \(0<1/n<\delta\), so the neighborhood meets \(F_n\); in fact, it meets infinitely many such singletons by taking arbitrarily large \(n\). Thus this family is not locally finite at \(0\). Its union is \(\{1/n:n\in\mathbb{N}\}\), which omits \(0\), while every neighborhood of \(0\) meets the union. Therefore \(0\) lies in its closure, and the union is not closed.
This example also shows why the closure-of-a-union formula cannot be used without its hypothesis. Here \(\bigcup_n\overline{F_n}=\{1/n:n\in\mathbb{N}\}\), but \(\overline{\bigcup_n F_n}=\{1/n:n\in\mathbb{N}\}\cup\{0\}\).
| Condition | What is tested at \(x\)? | Relationship |
|---|---|---|
| Point-finite | Whether \(x\) belongs to a member | Finitely many members contain \(x\) |
| Locally finite | Whether a neighborhood of \(x\) intersects a member | Implies point-finite; the converse can fail |
| Refinement | Whether each new set lies inside an old cover member | Does not, on its own, assert local finiteness |
Check Your Understanding
For each question, distinguish carefully between meeting a neighborhood and containing the point itself.
- Prove that every subfamily of a locally finite family is locally finite.
- In \(\mathbb{R}\), is the family \(\{(n,n+1):n\in\mathbb{Z}\}\) locally finite? Justify your answer using neighborhoods.
- For \(A_n=[n,n+1/2]\), \(n\in\mathbb{Z}\), explain why the union is closed.
- Give a point-finite family in \(\mathbb{R}\) that is not locally finite, and identify the point where local finiteness fails.
- Which step in the closure-of-a-union proof uses local finiteness?