Local Bases and First Countability

Why Use a Local Base?

A point can have many open neighborhoods, often too many to inspect one by one. A local base is a selected family of neighborhoods that is sufficient for this purpose: every neighborhood contains a member of the family. If the family is countable, questions involving all neighborhoods can often be reduced to a sequence of checks. This tutorial defines local bases and first countability, tests them in familiar spaces, and proves how a countable local base can produce a sequence approaching a point.

Let \((X,\mathcal{T})\) be a topological space. An open neighborhood of \(x\in X\) means an open set \(U\in\mathcal{T}\) with \(x\in U\). A family of sets is called countable if it is finite or can be listed in a sequence indexed by \(\mathbb{N}\).

Definition: A local base at \(x\) is a family \(\mathcal{B}_x\) of open neighborhoods of \(x\) such that for every open neighborhood \(U\) of \(x\), there is some \(B\in\mathcal{B}_x\) with \(B\subseteq U\). The space \(X\) is first countable if every point of \(X\) has a countable local base.

The direction of containment matters: a local-base member must fit inside each neighborhood being tested. A local base is attached to one point; it is not necessarily a basis for the topology on all of \(X\). The family of all open neighborhoods of \(x\) is always a local base, but it need not be countable.

Standard Sources of Local Bases

Worked Example: Metric Spaces

Let \((X,d)\) be a metric space, with the topology generated by its open balls. Fix \(x\in X\). Consider the family \(\mathcal{B}_x=\{B_d(x,1/n):n\in\mathbb{N}\}\), where \(B_d(x,r)=\{y\in X:d(x,y)<r\}\).

Each ball is an open neighborhood of \(x\), since \(d(x,x)=0<1/n\). To check the local-base condition, let \(U\) be any open neighborhood of \(x\). By the metric topology, there is an \(r>0\) such that \(B_d(x,r)\subseteq U\). Choose \(n\in\mathbb{N}\) with \(1/n<r\). If \(y\in B_d(x,1/n)\), then \(d(x,y)<1/n<r\), so \(y\in B_d(x,r)\subseteq U\). Thus \(B_d(x,1/n)\subseteq U\), as required. Every metric space is therefore first countable.

Check: in \(\mathbb{R}\) with its standard topology, at \(x=2\) the third ball in this family is \(B(2,1/3)=(5/3,7/3)\). Its radius is \(1/3\), and \(2-1/3=5/3\), \(2+1/3=7/3\), confirming the stated interval.

Worked Example: Discrete Spaces

Let \(X\) have the discrete topology, in which every subset is open. For any \(x\in X\), the one-element family \(\mathcal{B}_x=\{\{x\}\}\) is a local base. Indeed, every open neighborhood \(U\) of \(x\) contains \(x\), so \(\{x\}\subseteq U\). This argument applies whether \(X\) is finite or uncountable. In particular, every discrete space is first countable.

Check: take \(X=\{a,b,c\}\) with the discrete topology and \(x=b\). The neighborhood \(U=\{a,b\}\) contains \(b\), and the local-base member \(\{b\}\) satisfies \(\{b\}\subseteq\{a,b\}\). The same containment holds for every subset containing \(b\).

Local Bases in the Cofinite Topology

On a set \(X\), the cofinite topology consists of \(\varnothing\) and all subsets whose complements in \(X\) are finite. Thus every nonempty open neighborhood \(U\) has the form \(X\setminus F\) for a finite set \(F\subseteq X\). The size of \(X\) determines whether a countable local base is possible.

Theorem: In a cofinite space \(X\), a point \(x\in X\) has a countable local base if and only if \(X\setminus\{x\}\) is countable.

Proof. First suppose \(\mathcal{B}_x\) is a countable local base. For each \(B\in\mathcal{B}_x\), the set \(F_B=X\setminus B\) is finite and does not contain \(x\). Let \(y\in X\setminus\{x\}\). The set \(X\setminus\{y\}\) is an open neighborhood of \(x\), so the local-base condition gives some \(B\in\mathcal{B}_x\) with \(B\subseteq X\setminus\{y\}\). Hence \(y\notin B\), which means \(y\in F_B\). Every point of \(X\setminus\{x\}\) therefore belongs to one of the finite sets \(F_B\), so \(X\setminus\{x\}=\bigcup_{B\in\mathcal{B}_x}F_B\). A countable union of finite sets is countable.

Conversely, suppose \(X\setminus\{x\}\) is countable. List its distinct points as \(y_1,y_2,\ldots\) if it is infinite. Set \(B_n=X\setminus\{y_1,\ldots,y_n\}\). Each \(B_n\) is an open neighborhood of \(x\). Given any open neighborhood \(U=X\setminus F\) of \(x\), the finite set \(F\) does not contain \(x\), so all its points occur in the list. Choose \(n\) at least as large as every index of a point in \(F\); if \(F=\varnothing\), any \(n\) works. Then \(F\subseteq\{y_1,\ldots,y_n\}\), and consequently \(B_n\subseteq X\setminus F=U\). Thus \((B_n)\) is a countable local base. If \(X\setminus\{x\}\) is finite, use the same construction with the finite list, repeating its final set to obtain a sequence; if it is empty, use \(B_n=X\) for all \(n\). This proves both directions, including the finite cases. \(\square\)

Worked Example: A Countable Cofinite Space

Let \(X=\{p\}\cup\{y_n:n\in\mathbb{N}\}\), with all the \(y_n\) distinct, and give \(X\) the cofinite topology. At \(p\), set \(B_n=X\setminus\{y_1,\ldots,y_n\}\). To check the local-base condition, take \(U=X\setminus F\), where \(F\) is finite and \(p\notin F\). For example, if \(F=\{y_2,y_5\}\), then \(B_5=X\setminus\{y_1,y_2,y_3,y_4,y_5\}\subseteq U\). In general, choose \(n\) at least as large as each index appearing in \(F\); then every point removed in forming \(U\) is also removed in forming \(B_n\), so \(B_n\subseteq U\).

Check: \(B_5\) is open because its complement has five points, namely \(\{y_1,y_2,y_3,y_4,y_5\}\). It contains \(p\), and the example’s target neighborhood \(U=X\setminus\{y_2,y_5\}\) contains every point of \(B_5\). Thus the particular containment also follows directly from \(\{y_2,y_5\}\subseteq\{y_1,y_2,y_3,y_4,y_5\}\).

Worked Example: An Uncountable Cofinite Space

Give \(\mathbb{R}\) the cofinite topology, not its standard topology, and fix \(x\in\mathbb{R}\). Since \(\mathbb{R}\setminus\{x\}\) is uncountable, the theorem shows that \(x\) has no countable local base. This holds for every \(x\), so this cofinite space is not first countable.

Here is the obstruction directly. If \((B_n)\) were a countable local base, each complement \(F_n=\mathbb{R}\setminus B_n\) would be finite. Their union would be countable. But the neighborhood \(\mathbb{R}\setminus\{y\}\) must contain some \(B_n\) for every \(y\ne x\); that containment forces \(y\in F_n\). It would follow that \(\mathbb{R}\setminus\{x\}\subseteq\bigcup_n F_n\), contradicting uncountability.

Local Bases and Convergent Sequences

For a sequence \((x_n)\) in \(X\), we say \(x_n\to x\) if for every open neighborhood \(U\) of \(x\), there is an \(N\in\mathbb{N}\) such that \(x_n\in U\) for all \(n\ge N\). A local base lets us test this definition using only its own members.

Theorem: Let \(\mathcal{B}_x\) be a local base at \(x\). A sequence \((x_n)\) converges to \(x\) if and only if, for every \(B\in\mathcal{B}_x\), the sequence is eventually in \(B\).

Proof. Suppose \(x_n\to x\). Every \(B\in\mathcal{B}_x\) is an open neighborhood of \(x\), so the definition of convergence says that \(x_n\in B\) for all sufficiently large \(n\). Conversely, suppose the sequence is eventually in every member of \(\mathcal{B}_x\). Let \(U\) be any open neighborhood of \(x\). By the local-base property, choose \(B\in\mathcal{B}_x\) with \(B\subseteq U\). The sequence is eventually in \(B\), and hence it is eventually in \(U\). This is convergence to \(x\) by definition. \(\square\)

The countable case gives a further construction. First, any countable local base can be arranged to be nested: if \((U_n)\) is a list of its members, put \(V_n=U_1\cap\cdots\cap U_n\). Each \(V_n\) is an open neighborhood of \(x\), \(V_{n+1}\subseteq V_n\), and the family \((V_n)\) is still a local base. For the last claim, given a neighborhood \(U\), choose \(U_k\subseteq U\); then \(V_k\subseteq U_k\subseteq U\).

Theorem: Suppose \(x\) has a countable local base and \(A\subseteq X\) meets every open neighborhood of \(x\). Then there is a sequence of points of \(A\) converging to \(x\).

Proof. Replace a countable local base by the nested local base \((V_n)\) just constructed. Since each \(V_n\) is an open neighborhood of \(x\), the assumption that \(A\) meets every such neighborhood gives \(A\cap V_n\ne\varnothing\) for every \(n\). Choose \(a_n\in A\cap V_n\) for each \(n\). (This is a countable sequence of choices, as usual in standard set-theoretic foundations.) Let \(U\) be any open neighborhood of \(x\). Choose \(k\) with \(V_k\subseteq U\). For every \(n\ge k\), nesting gives \(V_n\subseteq V_k\), so \(a_n\in V_n\subseteq V_k\subseteq U\). Thus the sequence is eventually in every open neighborhood of \(x\), which means \(a_n\to x\). \(\square\)

Worked Example: Choosing a Sequence from Neighborhoods

In \(\mathbb{R}\) with its standard topology, let \(x=0\) and \(A=(0,1)\). Use the nested local base \(V_n=(-1/n,1/n)\) at \(0\). Each neighborhood \(V_n\) meets \(A\); a specific choice is \(a_n=1/(n+1)\).

For each \(n\ge1\), \(a_n>0\), and \(a_n=1/(n+1)<1/n\), so \(a_n\in(0,1/n)\subseteq V_n\) and \(a_n\in A\). To verify convergence directly, let \(U\) be any open neighborhood of \(0\). There is an \(r>0\) with \((-r,r)\subseteq U\). Choose \(N\in\mathbb{N}\) so that \(1/(N+1)<r\). For \(n\ge N\), \(0<1/(n+1)\le1/(N+1)<r\), and therefore \(a_n\in(-r,r)\subseteq U\). Hence \(a_n\to0\).

Check: at \(n=2\), \(a_2=1/(2+1)=1/3\), and \(V_2=(-1/2,1/2)\). The inequalities \(0<1/3<1/2\) verify that this selected term lies in both \(A\) and \(V_2\).

A Common Pitfall

Common mistake: In an uncountable cofinite space, one might choose a sequence of points \(y_1,y_2,\ldots\) different from \(x\) and claim that \(B_n=X\setminus\{y_1,\ldots,y_n\}\) is a local base at \(x\). Each \(B_n\) is indeed an open neighborhood of \(x\), but that alone does not make the family a local base. Choose \(z\in X\setminus\{x,y_1,y_2,\ldots\}\), which is possible because removing a countable set from an uncountable set leaves points. The open neighborhood \(X\setminus\{z\}\) contains \(x\). Yet every \(B_n\) contains \(z\), so no \(B_n\) is a subset of \(X\setminus\{z\}\). The missing refinement is exactly why the proposed family fails.

The distinction between being a neighborhood and being a local base is useful throughout topology. Metric balls and the singleton neighborhoods in a discrete space provide countable local bases, while the cofinite example shows that countability is not automatic. When a countable local base does exist, its members can be nested and used to build sequences that enter every neighborhood eventually.

Check Your Understanding

For each question, apply the local-base condition with the direction of containment stated explicitly.

In a metric space, explain why the balls \(B_d(x,1/n)\) form a local base at \(x\).
Give a countable local base at a point of an uncountable discrete space.
In a cofinite space, state the necessary and sufficient condition for \(x\) to have a countable local base.
Show that any sequence eventually in every member of a local base at \(x\) converges to \(x\).
In \(\mathbb{R}\) with the standard topology, choose a sequence in \((0,1)\) converging to \(0\) using the neighborhoods \((-1/n,1/n)\), and verify its terms lie in those neighborhoods.