Describing a Topology One Point at a Time
A basis for a topology describes open sets across the whole space: every open set is a union of basis elements. There is also a pointwise way to record the same local information. Instead of asking which basis elements occur anywhere, fix a point and keep track of the neighborhoods around it. A family that is sufficiently rich to refine every neighborhood of that point is called a neighborhood base there.
This idea applies even when the neighborhoods themselves are not open. Recall that a set is open when every one of its points lies in an open set contained in it. We use that local description to define neighborhoods first, then define a neighborhood base.
The containment requirement is the key: a neighborhood base must provide a member small enough to fit inside each neighborhood, not merely a member that intersects it. A neighborhood base need not contain every neighborhood, and different families can serve as neighborhood bases at the same point. Often it is convenient to choose open sets for its members, because open sets already express the topology directly.
Metric Balls Give Neighborhood Bases
In a metric space, the open balls centered at a point are its standard local neighborhoods. In fact, we do not need to use every possible radius: balls with radii \(1/n\), for positive integers \(n\), already refine every neighborhood. This countable family is useful because it lets us describe local behavior with a sequence of progressively smaller sets.
Proof. For every positive integer \(n\), the open ball \(B_{1/n}(x)\) is an open set containing \(x\), so it is a neighborhood of \(x\). Now let \(N\) be any neighborhood of \(x\). By definition, there is an open set \(U\) such that \(x\in U\subseteq N\). Since \(U\) is open and contains \(x\), there is an \(\varepsilon>0\) such that \(B_{\varepsilon}(x)\subseteq U\). Choose a positive integer \(n\) with \(1/n<\varepsilon\). Then $$ B_{1/n}(x)\subseteq B_{\varepsilon}(x)\subseteq U\subseteq N. $$ Thus every neighborhood of \(x\) contains a member of the proposed family, as required. \(\square\)
Worked Example: A Local Base in the Euclidean Plane
Give \(\mathbb{R}^2\) the Euclidean metric, and fix \(p=(2,-1)\). Consider the open balls \(B_{1/n}(p)\). To check directly how the refinement works, take the neighborhood $$ N=\{(s,t)\in\mathbb{R}^2:(s-2)^2+(t+1)^2<0.04\}. $$ This is the open ball \(B_{0.2}(p)\). Choose \(n=6\). Since \(1/6<0.2\), every point whose distance from \(p\) is less than \(1/6\) also has distance less than \(0.2\). Therefore \(B_{1/6}(p)\subseteq N\).
The same reasoning works for any neighborhood \(N\) of \(p\): it contains some open ball \(B_{\varepsilon}(p)\), and a sufficiently large \(n\) makes \(1/n<\varepsilon\). The family of reciprocal-radius balls is therefore a neighborhood base, even though it includes only a countable selection of all the balls centered at \(p\).
The theorem does not say that every topological space has a countable neighborhood base at every point. It says that metric spaces do. This distinction matters: having such a countable family is an additional property of a point or space, not part of the definition of a topological space.
Neighborhood Bases in a Topological Space
Worked Example: Neighborhood Bases in a Finite Space
Let \(X=\{u,v,w\}\) have topology $$ \mathcal{T}=\{\varnothing,\{u\},\{u,v\},X\}. $$ At \(u\), the open sets containing \(u\) are \(\{u\}\), \(\{u,v\}\), and \(X\). Every neighborhood of \(u\) contains an open set containing \(u\), and \(\{u\}\) is already contained in each such open set. Thus \(\{\{u\}\}\) is an open neighborhood base at \(u\).
At \(v\), the open sets containing \(v\) are \(\{u,v\}\) and \(X\), so \(\{\{u,v\}\}\) is an open neighborhood base at \(v\). At \(w\), the only open set containing \(w\) is \(X\), and hence \(\{X\}\) is an open neighborhood base at \(w\). The different local bases reflect the fact that the topology provides different amounts of local information at different points.
There is a useful connection between neighborhood bases and bases for a topology. If \(\mathcal{B}\) is a basis for a topology, then the basis elements that contain \(x\) form a neighborhood base at \(x\). Indeed, each such element is an open neighborhood; and if \(N\) is a neighborhood of \(x\), choose an open \(U\) with \(x\in U\subseteq N\). The basis property gives a basis element \(B\) with \(x\in B\subseteq U\subseteq N\).
Conversely, open neighborhood bases chosen at all points can be collected to produce a basis for the entire topology. The intersection condition in the definition of a basis follows from the local refinement property at the point in the intersection.
Proof. Every member of \(\mathcal{B}\) is open. Also, for each \(x\in X\), the family \(\mathcal{B}_x\) contains neighborhoods of \(x\), so some member of it contains \(x\). Thus the members of \(\mathcal{B}\) cover \(X\).
Now take \(B_1,B_2\in\mathcal{B}\), and let \(z\in B_1\cap B_2\). Since \(B_1\) and \(B_2\) are open, their intersection is an open neighborhood of \(z\). Because \(\mathcal{B}_z\) is a neighborhood base at \(z\), there is a \(B_3\in\mathcal{B}_z\) such that $$ z\in B_3\subseteq B_1\cap B_2. $$ Here \(z\in B_3\) because every member of \(\mathcal{B}_z\) is a neighborhood of \(z\). This verifies the basis intersection condition. Finally, if \(U\) is open and \(x\in U\), then \(U\) is a neighborhood of \(x\). The neighborhood-base property gives a \(B_x\in\mathcal{B}_x\) with \(x\in B_x\subseteq U\), so \(U\) is a union of members of \(\mathcal{B}\). The basis characterization in Bases for Topologies now shows that \(\mathcal{B}\) is a basis for the topology. \(\square\)
The theorem also explains why local bases are more than a point-by-point convenience. They can be assembled into a global basis. The converse observation above says that any global basis supplies compatible local bases simply by retaining, at each point, the basis elements that contain it.
Worked Example: Recovering a Basis from Local Bases
In the finite space above, use the local bases $$ \mathcal{B}_u=\{\{u\}\},\qquad \mathcal{B}_v=\{\{u,v\}\},\qquad \mathcal{B}_w=\{X\}. $$ Their union is \(\mathcal{B}=\{\{u\},\{u,v\},X\}\). These sets cover \(X\). The only nontrivial intersection of distinct members is $$ \{u\}\cap\{u,v\}=\{u\}, $$ which is itself in \(\mathcal{B}\); intersections with \(X\) leave the other set unchanged. Thus the union satisfies the basis conditions.
Taking unions of its members gives exactly \(\varnothing,\{u\},\{u,v\}\), and \(X\), which are the open sets of the stated topology. In particular, the local choices at the three points recover the full topology when assembled into one global basis.
Testing Continuity with Neighborhood Bases
Continuity can also be checked locally. A continuous map must pull back neighborhoods of an image point to neighborhoods of the original point. When open neighborhood bases are available in both spaces, it is enough to check a containment for each member of the target base. The source base can then refine the inverse image.
Proof. First suppose \(f\) is continuous. Fix \(x\in X\) and \(C\in\mathcal{C}_{f(x)}\). The set \(C\) is open and contains \(f(x)\), so \(f^{-1}[C]\) is open and contains \(x\). It is therefore a neighborhood of \(x\). Since \(\mathcal{B}_x\) is a neighborhood base, some \(B\in\mathcal{B}_x\) satisfies \(B\subseteq f^{-1}[C]\). This containment means exactly that \(f[B]\subseteq C\).
Conversely, suppose the stated containment condition holds. Let \(V\) be any open set in \(Y\), and take \(x\in f^{-1}[V]\). Then \(f(x)\in V\). Because \(\mathcal{C}_{f(x)}\) is a neighborhood base, there is a \(C\in\mathcal{C}_{f(x)}\) with \(C\subseteq V\). By the assumed condition, there is a \(B\in\mathcal{B}_x\) such that \(f[B]\subseteq C\subseteq V\). Hence \(x\in B\subseteq f^{-1}[V]\). Since \(B\) is open, every point of \(f^{-1}[V]\) lies in an open set contained in \(f^{-1}[V]\). The local characterization of open sets from Open Sets in Topological Spaces implies that \(f^{-1}[V]\) is open. This holds for every open \(V\), so \(f\) is continuous. \(\square\)
Worked Example: Comparing Two Topologies with Local Bases
Let \(X=\{u,v\}\). Give the domain the discrete topology \(\{\varnothing,\{u\},\{v\},X\}\) and the codomain the topology \(\{\varnothing,\{u\},X\}\). Consider the identity map from the first space to the second. Open neighborhood bases in the domain are \(\mathcal{B}_u=\{\{u\}\}\) and \(\mathcal{B}_v=\{\{v\}\}\). In the codomain, take \(\mathcal{C}_u=\{\{u\}\}\) and \(\mathcal{C}_v=\{X\}\).
For \(x=u\), the only target base member is \(\{u\}\), and the source base member \(\{u\}\) maps into it. For \(x=v\), the target base member is \(X\), and \(\{v\}\) maps into \(X\). The theorem therefore shows that the identity map is continuous. In the reverse direction, at \(v\) the target discrete base member \(\{v\}\) contains no image of an open neighborhood of \(v\) in the original topology that is contained in \(\{v\}\): the only open neighborhood of \(v\) there is \(X\). The reverse identity map is not continuous.
Why the Local View Matters
Neighborhood bases make a topology easier to use when a question is inherently local: what sets can be found arbitrarily close to a point, or what must a map do near a particular input? Metric balls answer these questions in metric spaces, while the general definition works without a distance. The continuity criterion shows how to translate local information at an image point into a required local condition in the domain.
A common pitfall is to confuse a neighborhood with an open neighborhood. A neighborhood may include points beyond an open set containing the point, so it need not itself be open. The definition of a neighborhood base accommodates this: its members need only be neighborhoods, but each must fit inside every prescribed neighborhood when the family is used to refine it. When collecting local families into a global basis or applying the continuity theorem above, we specifically use open neighborhood bases so that the basis elements and the sets used to establish openness really are open.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What condition makes a set a neighborhood of a point, and how does a neighborhood base refine it?
- Why do balls of radius \(1/n\) form a neighborhood base at a point in a metric space?
- How does a basis for a topology give a neighborhood base at each point?
- Why does the union of open neighborhood bases at all points satisfy the basis intersection condition?
- State the local-base criterion for continuity and explain what the required containment means.