From Neighborhoods at a Point to Open Sets
Open sets can be recognized locally: a set is open when each of its points has an open neighborhood contained in it. A neighborhood base makes this test more efficient by recording only a collection of neighborhoods from which every other open neighborhood can be found by taking a larger set. When such data is specified at every point, it determines exactly which subsets are open.
Throughout, \((X,\mathcal{T})\) is a topological space. An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) with \(x\in U\).
The containment direction matters: a base element must fit inside the neighborhood being tested. For instance, a large open neighborhood may contain a smaller base element even though it is not itself one of the chosen elements.
The Pointwise Test for Openness
Proof. Suppose first that \(U\) is open. For every \(x\in U\), \(U\) is an open neighborhood of \(x\). By the definition of a neighborhood base, some \(B\in\mathcal{B}_x\) satisfies \(B\subseteq U\). Conversely, suppose that for every \(x\in U\), there is \(B_x\in\mathcal{B}_x\) with \(B_x\subseteq U\). Each \(B_x\) is open and contains \(x\). Therefore \(U=\bigcup_{x\in U}B_x\): the union is contained in \(U\), and every \(x\in U\) belongs to its corresponding \(B_x\). A union of open sets is open, so \(U\in\mathcal{T}\). If \(U=\varnothing\), the condition is vacuously satisfied and \(U\) is open as well. \(\square\)
This gives a precise way to recover the topology: inspect each point of a proposed set separately, and ask whether one of that point’s base neighborhoods stays inside the set. The base element may depend on the point; there is no requirement that one open set work for every point of \(U\).
Neighborhood Bases from a Global Basis
A basis \(\mathcal{B}\) for a topology is a collection of open sets such that every open set is a union of members of \(\mathcal{B}\). At a particular point, the relevant basis elements are those that contain that point.
Proof. Every member of \(\mathcal{B}_x\) is open and contains \(x\). Now let \(U\) be any open neighborhood of \(x\). Since \(\mathcal{B}\) is a basis, \(U\) is a union of basis elements. Because \(x\in U\), at least one of the basis elements in that union contains \(x\); call it \(B\). Then \(B\in\mathcal{B}_x\) and \(B\subseteq U\). This is exactly the neighborhood-base condition. \(\square\)
Worked Example: Metric Balls in the Real Line
Give a neighborhood base at \(x=2\) in \(\mathbb{R}\) with its standard topology, and use it to test \(U=(1,4)\).
For every \(r>0\), let \(B(2,r)=(2-r,2+r)\). The metric topology has open balls as a basis, so the balls centered at \(2\) form a neighborhood base at \(2\). For the proposed set \(U=(1,4)\), choose \(r=1\). Then \(B(2,1)=(1,3)\), and \(1<2<3<4\), so \(B(2,1)\subseteq(1,4)\). This supplies an open neighborhood inside \(U\) for the point \(2\).
For comparison, \(B(2,2)=(0,4)\) is not contained in \(U\), since \(0\in(0,4)\) but \(0\notin(1,4)\). A neighborhood base does not say that every base element fits inside every neighborhood; it says that at least one suitable base element can be found. More generally, if \(V\) is any open neighborhood of \(2\), the definition of the standard topology gives an \(\varepsilon>0\) with \(B(2,\varepsilon)\subseteq V\). Thus the centered balls do satisfy the required condition.
Check: The interval \(U=(1,4)\) is open in the standard topology, and the chosen ball \((1,3)\) is an open neighborhood of \(2\) contained in \(U\), as the pointwise test requires.
Worked Example: A Neighborhood Base at an Endpoint
Let \(I=[0,1]\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). Find a neighborhood base at \(0\) and apply it to \(U=[0,\tfrac12)\).
For \(r>0\), define \(N_r=(-r,r)\cap I\). Each \(N_r\) is open in the subspace topology and contains \(0\). To check the base condition, let \(V\) be any open neighborhood of \(0\) in \(I\). By the definition of the subspace topology, \(V=O\cap I\) for some open \(O\subseteq\mathbb{R}\). Since \(0\in O\), there is a \(\delta>0\) such that \((-\delta,\delta)\subseteq O\). It follows that \(N_\delta=(-\delta,\delta)\cap I\subseteq V\). Hence \(\{N_r:r>0\}\) is a neighborhood base at \(0\).
For \(U=[0,\tfrac12)\), take \(r=\tfrac14\). Then \(N_{1/4}=(-\tfrac14,\tfrac14)\cap[0,1]=[0,\tfrac14)\). Every point of \([0,\tfrac14)\) is in \([0,\tfrac12)\), so \(N_{1/4}\subseteq U\). This verifies the neighborhood-base test at the endpoint \(0\). The intersection with \(I\) is essential: an open neighborhood of \(0\) in the subspace is not required to be open as a subset of \(\mathbb{R}\).
Check: \(N_{1/4}\) contains \(0\), is open in \(I\), and is contained in \(U\); it therefore witnesses the required local condition at \(0\).
Worked Example: Half-Open Basis Elements
Let \(\mathbb{R}_\ell\) denote \(\mathbb{R}\) with the lower limit topology, whose basis consists of intervals \([a,b)\) with \(a<b\). Find a neighborhood base at \(2\), then test \(U=[2,3)\).
The basis elements that contain \(2\) are precisely the intervals \([a,b)\) with \(a\leq 2<b\). They form a neighborhood base by the proposition above. A convenient subcollection is \(\{[2,2+r):r>0\}\). Indeed, if an open neighborhood \(V\) of \(2\) is a union of basis elements, at least one interval \([a,b)\) in that union contains \(2\) and is contained in \(V\). Choose \(r=b-2>0\). Then \([2,2+r)=[2,b)\subseteq[a,b)\subseteq V\).
For \(U=[2,3)\), choose \(r=1\). The base element \([2,2+1)=[2,3)\) equals \(U\), so it is contained in \(U\). This set is open in the lower limit topology because it is itself one of the specified basis elements. It is not open in the standard topology on \(\mathbb{R}\): every standard open interval containing \(2\) also contains points less than \(2\), whereas \([2,3)\) contains none. The same half-open interval thus serves as a basic open neighborhood in one topology but not the other.
Check: The endpoint \(3\) is excluded from \([2,3)\), and the selected base element has the same excluded endpoint, so the containment is exact.
Neighborhood Bases Need Not Be Unique
Different collections can serve as neighborhood bases at the same point. The relevant question is not whether the collections have identical members, but whether each one can find sufficiently small neighborhoods inside the members of the other.
Proof. Fix \(B\in\mathcal{B}_x\). It is open and contains \(x\), so it is an open neighborhood of \(x\). Applying the neighborhood-base property for \(\mathcal{C}_x\) to this neighborhood gives \(C\in\mathcal{C}_x\) with \(C\subseteq B\). Interchanging the roles of the two bases proves the other containment. \(\square\)
Worked Example: Two Bases at the Same Point
In the standard topology on \(\mathbb{R}\), compare these two collections at \(0\): \(\mathcal{B}_0=\{(-r,r):r>0\}\) and \(\mathcal{C}_0=\{(-2r,2r):r>0\}\).
Both are neighborhood bases: each collection consists of open neighborhoods of \(0\), and any open neighborhood \(V\) of \(0\) contains some interval \((-\delta,\delta)\) with \(\delta>0\). For \(\mathcal{B}_0\), this interval is already a member. For \(\mathcal{C}_0\), the member \((-\delta/2,\delta/2)\) lies inside \((-\delta,\delta)\).
To refine a member \((-r,r)\in\mathcal{B}_0\) using \(\mathcal{C}_0\), choose \((-r/2,r/2)\in\mathcal{C}_0\), since this is the member with parameter \(r/4\): \((-2(r/4),2(r/4))=(-r/2,r/2)\subseteq(-r,r)\). Conversely, the member \((-2r,2r)\in\mathcal{C}_0\) contains \((-r,r)\in\mathcal{B}_0\). Thus each collection refines the other even though their members are not generally the same.
Check: The two collections describe the same open neighborhoods at \(0\), as their mutual refinements show.
A Common Quantifier Mistake
Common pitfall. The openness test is sometimes misread as requiring one fixed basis element \(B\) that contains every point of \(U\) and also satisfies \(B\subseteq U\). That is a much stronger condition and is not the definition.
For a concrete failure, take \(U=(-1,1)\) in the standard topology on \(\mathbb{R}\). This set is open. The correct pointwise test works: for each \(x\in(-1,1)\), choose \(r=(1-|x|)/2\), which is positive. Then every \(y\in B(x,r)\) satisfies \(|y|\leq |x|+|y-x|<|x|+r=(1+|x|)/2<1\), so \(B(x,r)\subseteq(-1,1)\).
The incorrect global requirement would ask for one open set \(B\) with \((-1,1)\subseteq B\subseteq(-1,1)\), hence \(B=(-1,1)\), that also serves as a single selected neighborhood for all points. The openness criterion makes no such demand: it asks for a possibly different neighborhood at each \(x\). The distinction is exactly between “for each point, there exists a neighborhood” and “there exists one neighborhood for all points.”
Neighborhood bases are useful because they compress the local information in a topology without losing the open sets. A global basis provides a standard way to produce one at each point, while metric balls and other concrete basic sets make the pointwise test explicit. Whenever openness is checked with local data, keep the containment direction and the order of the quantifiers in view.
Check Your Understanding
Use the neighborhood-base definition and the pointwise openness test to answer the following.
- In the standard topology on \(\mathbb{R}\), verify directly that \(\{(-r,r):r>0\}\) is a neighborhood base at \(0\).
- In the subspace \(I=[0,1]\), give a neighborhood base at \(1\) and explain why its members are open in \(I\).
- In \(\mathbb{R}_\ell\), describe the basis elements containing \(x=4\), and use one to test \(U=[4,6)\).
- State the pointwise condition that characterizes open sets when a neighborhood base is specified at every point.
- Explain why a neighborhood base at \(x\) need not contain every open neighborhood of \(x\).