Countable Neighborhood Control
At a point in a topological space, neighborhoods can be numerous and differently shaped. A countable local base gives a manageable collection of neighborhoods that is sufficient for testing every neighborhood of that point. The condition is imposed separately at each point; it does not require a single countable collection to work throughout the space.
Let \((X,\mathcal{T})\) be a topological space. An open neighborhood of \(x\in X\) is an open set containing \(x\). A family of open neighborhoods of \(x\) is a local base at \(x\) if every open neighborhood of \(x\) contains at least one member of the family. A local base may be finite; “countable” here includes finite families.
The defining containment matters: members of a local base need not equal every neighborhood, but one of them must fit inside each neighborhood. A pointwise countable local base is therefore a way to refine neighborhoods, not a claim that the entire topology is countable.
Why Metric Spaces Are First Countable
For a metric space \((X,d)\), the open ball with center \(x\) and radius \(r>0\) is \(B_d(x,r)=\{y\in X:d(x,y)<r\}\). The metric topology consists of the sets \(U\) such that for every \(x\in U\), some ball \(B_d(x,r)\) is contained in \(U\). Balls with smaller radii fit inside balls with larger radii at the same center, so it is natural to use radii that decrease to zero.
Proof. Fix \(x\in X\). For every \(n\in\mathbb{N}\), the set \(B_d(x,1/n)\) is an open neighborhood of \(x\), since \(d(x,x)=0<1/n\). Let \(U\) be any open neighborhood of \(x\). By the definition of the metric topology, there is \(r>0\) such that \(B_d(x,r)\subseteq U\). Choose \(n\in\mathbb{N}\) with \(1/n<r\), which is possible because the positive integers are unbounded. If \(y\in B_d(x,1/n)\), then \(d(y,x)<1/n<r\), so \(y\in B_d(x,r)\subseteq U\). Thus \(B_d(x,1/n)\subseteq U\), proving that the displayed family is a local base. If \(X\) is empty, the assertion about each \(x\) is vacuous, as required. \(\square\)
The choice of \(n\) may change with the neighborhood \(U\). That is exactly what the local-base condition allows: each neighborhood needs some sufficiently small ball, not one ball that fits inside all neighborhoods.
Worked Example: A Local Base at 2 in the Real Line
Use the standard topology on \(\mathbb{R}\). At \(x=2\), the family \(B(2,1/n)\), for \(n\in\mathbb{N}\), is a local base. For instance, the neighborhood \(U=(3/2,5/2)\) contains \(B(2,1/2)\), since \(B(2,1/2)=(2-1/2,2+1/2)=(3/2,5/2)=U\).
For an arbitrary open neighborhood \(V\) of \(2\), openness gives some \(r>0\) with \(B(2,r)\subseteq V\). Choose \(n\) with \(1/n<r\). For every \(y\in B(2,1/n)\), the inequalities \(2-1/n<y<2+1/n\) imply \(2-r<y<2+r\), so \(y\in B(2,r)\subseteq V\). This verifies the local-base condition for every neighborhood, not just the particular interval \(U\).
Check: with \(n=2\), the radius is \(1/2\), and the endpoints are \(2-1/2=3/2\) and \(2+1/2=5/2\), as stated.
Subspaces and First Countability
The metric-space theorem also gives examples with the subspace topology. More generally, first countability passes from a space to each of its subspaces. Recall that if \(A\subseteq X\), its subspace topology consists of sets \(A\cap V\), where \(V\) is open in \(X\).
Proof. Fix \(a\in A\), and let \(\{U_n:n\in\mathbb{N}\}\) be a countable local base at \(a\) in \(X\). Each \(A\cap U_n\) is open in the subspace topology and contains \(a\). Let \(W\) be any open neighborhood of \(a\) in \(A\). By the definition of the subspace topology, \(W=A\cap V\) for some open \(V\subseteq X\); since \(a\in W\), \(V\) is a neighborhood of \(a\) in \(X\). Choose \(n\) with \(U_n\subseteq V\). Then \(A\cap U_n\subseteq A\cap V=W\). Thus \(\{A\cap U_n:n\in\mathbb{N}\}\) is a countable local base at \(a\) in \(A\). This holds for every \(a\in A\), so \(A\) is first countable. \(\square\)
Worked Example: The Subspace [0,1] at 0
Give \(A=[0,1]\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). At \(0\), intersect the balls centered at \(0\) with \(A\): \(A\cap B(0,1/n)=[0,1/n)\) for each \(n\in\mathbb{N}\). The right endpoint is excluded because \(B(0,1/n)\) consists of points whose distance from \(0\) is strictly less than \(1/n\).
To check this family directly, let \(W\) be any open neighborhood of \(0\) in \(A\). Write \(W=A\cap V\) with \(V\) open in \(\mathbb{R}\). There is an \(r>0\) such that \((-r,r)\subseteq V\). Choose \(n\) with \(1/n<r\). Then \([0,1/n)\subseteq A\cap(-r,r)\subseteq W\). So these intersections form a countable local base at \(0\). The proposition ensures first countability at every other point of \(A\) as well.
Check: for \(n=4\), \(A\cap B(0,1/4)=[0,1/4)\); \(0\) is included, and \(1/4\) is excluded because its distance from \(0\) equals, rather than is less than, the radius.
Uncountably Many Points, Countable Local Bases
First countability is a condition on neighborhoods around each individual point. It does not limit the number of points in the space. In particular, an uncountable discrete space is first countable.
Worked Example: An Uncountable Discrete Space
Take \(X=\mathbb{R}\) as a set, but give it the discrete topology \(\mathcal{T}=\mathcal{P}(\mathbb{R})\), in which every subset is open. The space has uncountably many points. Fix any \(x\in X\). Its singleton \(\{x\}\) is open, and every open neighborhood \(U\) of \(x\) contains \(x\), so \(\{x\}\subseteq U\). Therefore the one-element family \(\{\{x\}\}\) is a local base at \(x\). This works at every point, so \(X\) is first countable despite being uncountable.
The local base in this example depends on the point: at \(x\) it is \(\{\{x\}\}\), while at a different point \(y\) it is \(\{\{y\}\}\). The example illustrates why the definition is pointwise.
Check: \(\{x\}\) contains \(x\), is open in the discrete topology, and is contained in every set \(U\) that contains \(x\). Thus it meets both requirements for a local base.
First Countable Does Not Mean Metrizable
Every metric space is first countable, but the definition of first countability itself refers only to open neighborhoods. It does not require a metric. A small finite example makes this distinction explicit.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). At \(a\), the family \(\{\{a\}\}\) is a local base: the open neighborhoods of \(a\) are \(\{a\}\) and \(X\), and the singleton is contained in both. At \(b\), the only open neighborhood is \(X\), so \(\{X\}\) is a local base. Hence this space is first countable.
This space is not metrizable. Indeed, in any metric space, every singleton \(\{x\}\) is closed: for each \(y\ne x\), the ball \(B(y,d(x,y)/2)\) contains \(y\) and excludes \(x\), so the complement of \(\{x\}\) is open. In the displayed topology, \(\{a\}\) is not closed, because its complement \(\{b\}\) is not open. Therefore this topology cannot come from a metric.
Check: the open sets are exactly \(\varnothing,\{a\},\{a,b\}\). In particular, \(\{b\}\) is not open, confirming that \(\{a\}\) is not closed.
A Common Pitfall: Choosing a Ball That Is Too Large
In the metric-space proof, the radius must be small enough that the ball lies inside the given neighborhood. Reversing the inequality can invalidate the argument.
Wrong calculation. In \(\mathbb{R}\), take \(x=0\) and \(U=(-1/10,1/10)\). Choosing \(n=5\) gives radius \(1/5\), and \(B(0,1/5)=(-1/5,1/5)\). Since \(1/5>1/10\), this ball is not contained in \(U\).
Correction. Choose \(n=11\). Then \(1/11<1/10\), because \(10<11\). Hence every \(y\in B(0,1/11)\) satisfies \(-1/10<y<1/10\), so \(B(0,1/11)\subseteq U\). In general, if \(B(x,r)\subseteq U\), choose \(n\) so that \(1/n<r\), not so that \(1/n\) is larger than \(r\).
Countable local bases are useful because they replace the full collection of neighborhoods at a point with a countable refining family. The metric-ball theorem supplies such a family in every metric space; the subspace proposition shows that the property is inherited by subspaces. The discrete example also warns against confusing a countable local base at each point with a restriction on the total number of points.
Check Your Understanding
For each question, use the definition of a local base and state the topology involved.
- In a metric space, why does \(1/n<r\) ensure \(B_d(x,1/n)\subseteq B_d(x,r)\)?
- Give a countable local base at \(1\) in the subspace \([0,1]\) of the standard real line.
- Explain why the singleton \(\{x\}\) is a local base at \(x\) in a discrete space.
- In the Sierpinski space above, list the open neighborhoods of \(a\) and of \(b\), and give a local base at each.
- For \(U=(-1/10,1/10)\) in the standard real line, does \(B(0,1/8)\subseteq U\)? Justify your answer by comparing the radii.