Why Local Witnesses Help
Many definitions in topology quantify over every open neighborhood of a point. A basis turns that broad requirement into a more manageable task: find a suitable basis element inside a given neighborhood. The replacement is valid because basis elements are themselves open neighborhoods, and because every open neighborhood contains one. We will use this local refinement principle to test openness, closure membership, and continuity at a point.
For a topological space \((X,\mathcal{T})\), a basis \(\mathcal{B}\) is a collection of open subsets of \(X\) such that every point belongs to some member of \(\mathcal{B}\), and whenever \(x\in B_1\cap B_2\) for \(B_1,B_2\in\mathcal{B}\), there is \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\). The open sets of the topology are precisely the unions of basis elements. In particular, if \(U\) is open and \(x\in U\), there is \(B\in\mathcal{B}\) with \(x\in B\subseteq U\).
A basis is a collection that works across the space; a neighborhood base is a local collection for one point. Either can supply the smaller open set needed as a witness. The essential condition is containment: the chosen witness must contain the point and lie inside the neighborhood or set under consideration.
Openness From Basis Elements
Proof. Suppose \(A\) is open. For each \(x\in A\), the basis refinement property applied to the open neighborhood \(A\) gives \(B\in\mathcal{B}\) with \(x\in B\subseteq A\). Conversely, suppose such a basis element exists for every \(x\in A\). Then \(A\) is the union of the basis elements contained in it: each such element is a subset of \(A\), and every point of \(A\) belongs to one of them. A union of open sets is open, so \(A\) is open. If \(A=\varnothing\), the condition is vacuous and \(A\) is open as well. \(\square\)
Worked Example: An Open Set in the Real Line
Give \(\mathbb{R}\) its standard topology, with basis all open intervals \((a,b)\), where \(a<b\). Consider \(A=(0,1)\cup(2,3)\). For any \(x\in(0,1)\), let \(r=\min(x,1-x)/2\). Since \(0<x<1\), both entries in this minimum are positive, so \(r>0\). Also \(r\le x/2\) and \(r\le(1-x)/2\), which give \(x-r>0\) and \(x+r<1\). Thus \(x\in(x-r,x+r)\subseteq(0,1)\subseteq A\).
For any \(x\in(2,3)\), take \(r=\min(x-2,3-x)/2\). Again \(r>0\), and the two inequalities give \(x-r>2\) and \(x+r<3\). Therefore \(x\in(x-r,x+r)\subseteq(2,3)\subseteq A\). Every point of \(A\) has a basis element around it contained in \(A\), so the basis test proves \(A\) is open.
Check at a particular point: for \(x=5/2\), the two distances to the endpoints of \((2,3)\) are \(5/2-2=1/2\) and \(3-5/2=1/2\). Hence \(r=\min(1/2,1/2)/2=1/4\), and the witness is \((9/4,11/4)\), which lies inside \((2,3)\).
Closure Membership From a Neighborhood Base
Recall that a set is closed if its complement is open, and that \(\overline{A}\) denotes the closure of \(A\), the smallest closed set containing \(A\). The neighborhood test says that \(x\in\overline{A}\) exactly when every open neighborhood of \(x\) intersects \(A\). The next result shows why a neighborhood base is enough to perform this test.
Proof. Suppose \(x\in\overline{A}\). Every open neighborhood of \(x\) intersects \(A\), so in particular every basis element containing \(x\) intersects \(A\). Conversely, suppose every basis element containing \(x\) intersects \(A\), and let \(U\) be any open neighborhood of \(x\). By the basis property, there is \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). The assumption gives \(B\cap A\ne\varnothing\), and therefore \(U\cap A\ne\varnothing\). Since this holds for every open neighborhood \(U\), the neighborhood test gives \(x\in\overline{A}\). \(\square\)
Worked Example: Closure in a Subspace
Let \(X=[0,1]\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\), and let \(A=(0,1/2)\), viewed as a subset of \(X\). A basis for \(X\) consists of sets \((a,b)\cap[0,1]\) with \(a<b\). We claim that \(\overline{A}=[0,1/2]\), where the closure is taken in \(X\).
Every point \(x\in(0,1/2)\) belongs to \(A\), so each of its neighborhoods intersects \(A\). At \(x=0\), every basis neighborhood has the form \((a,b)\cap[0,1]\) with \(a<0<b\). It contains \(t=\min(b/2,1/4)\): this number is positive, is less than \(b\), and is at most \(1/4<1/2\). Thus \(t\in A\) and the neighborhood intersects \(A\). At \(x=1/2\), a basis neighborhood containing \(1/2\) has \(a<1/2<b\). Choose \(t=1/2-\min((1/2-a)/2,1/4)\). The minimum is positive, so \(t<1/2\); it is at most \((1/2-a)/2\), so \(t>a\); and \(t\ge1/4>0\). Hence \(t\in A\) and \(t\in(a,b)\cap[0,1]\). The closure test includes every point of \([0,1/2]\).
Now let \(x\in[0,1]\) with \(x>1/2\), and put \(r=(x-1/2)/2\). Then \(r>0\), and \(x-r=(x+1/2)/2>1/2\). The basis neighborhood \((x-r,x+r)\cap[0,1]\) is disjoint from \(A=(0,1/2)\). Thus \(x\notin\overline{A}\). Combining the two cases gives \(\overline{A}=[0,1/2]\) in the subspace \(X\).
Check: the endpoint \(1/2\) is included even though it is not in \(A\), because every basis neighborhood of it contains a point of \(A\). Points greater than \(1/2\) are excluded by the explicit disjoint basis neighborhood above.
Local Continuity From Neighborhood Bases
A function \(f:X\to Y\) is continuous at \(x\in X\) if for every open neighborhood \(V\) of \(f(x)\), there is an open neighborhood \(U\) of \(x\) such that \(f(U)\subseteq V\). Local bases can reduce either side of this condition. On the domain side, a basis or neighborhood base can provide \(U\); on the target side, a neighborhood base at \(f(x)\) can reduce the neighborhoods \(V\) that need to be tested.
Proof. If \(f\) is continuous at \(x\), then for any open neighborhood \(V\) of \(f(x)\), there is an open neighborhood \(U\) of \(x\) with \(f(U)\subseteq V\). Since \(\mathcal{N}_x\) is a neighborhood base, some \(N\in\mathcal{N}_x\) satisfies \(N\subseteq U\), so \(f(N)\subseteq f(U)\subseteq V\). Conversely, if such an \(N\) exists for every open neighborhood \(V\), then \(N\) itself is an open neighborhood of \(x\) witnessing continuity at \(x\). Now suppose the condition is checked only for members \(V_0\) of a neighborhood base at \(f(x)\). Given any open neighborhood \(V\) of \(f(x)\), choose such a \(V_0\subseteq V\). The condition supplies \(N\) with \(f(N)\subseteq V_0\subseteq V\), proving the full condition. The reverse implication follows by restricting a condition that holds for every open neighborhood to the members of the neighborhood base. \(\square\)
Worked Example: A Local Continuity Witness
Give \(X=[0,2]\) the subspace topology from \(\mathbb{R}\), give \(\mathbb{R}\) its standard topology, and define \(f:X\to\mathbb{R}\) by \(f(t)=t^2\). We verify continuity at \(x=1\) using the usual interval neighborhood base on \(\mathbb{R}\) and the basis \((a,b)\cap[0,2]\) on \(X\).
Let \(\varepsilon>0\) be arbitrary and set \(\delta=\min(1,\varepsilon/3)\). Both numbers in the minimum are positive, so \(\delta>0\), and \(3\delta\le\varepsilon\). Choose the domain basis neighborhood \(N=(1-\delta,1+\delta)\cap[0,2]\). If \(t\in N\), then \(t\in[0,2]\) and \(|t-1|<\delta\). Factoring gives \(|f(t)-f(1)|=|t^2-1|=|t-1||t+1|\). Since \(t\le2\), we have \(|t+1|=t+1\le3\), so \(|f(t)-f(1)|<3\delta\le\varepsilon\). Therefore \(f(N)\subseteq(1-\varepsilon,1+\varepsilon)\), as required.
Check with a specific tolerance: if \(\varepsilon=3/5\), then \(\delta=\min(1,(3/5)/3)=1/5\), and \(N=(4/5,6/5)\cap[0,2]\). For \(t\in N\), \(|t-1|<1/5\) and \(t+1<11/5\), so \(|t^2-1|<(1/5)(11/5)=11/25<3/5\). This verifies the estimate for that particular neighborhood as well.
Choose a Basis for the Topology You Have
A valid local witness must come from a basis for the topology actually being used. For example, give \(\mathbb{R}\) the lower limit topology, whose basis consists of intervals \([a,b)\) with \(a<b\). The set \([0,1)\) is open in this topology because it is itself a basis element. It is not valid to try to prove this using the standard basis of open intervals: every standard open interval containing \(0\) has the form \((a,b)\) with \(a<0<b\), and therefore contains negative points, so it cannot be contained in \([0,1)\). This does not show that \([0,1)\) is not open; it shows that the proposed witnesses belong to the wrong basis.
Common pitfall. In the preceding example, the incorrect argument is: “To show \([0,1)\) is open, choose a standard open interval \((a,b)\) around \(0\) contained in \([0,1)\).” Such an interval cannot exist, since \(a<0\) forces points of \((a,b)\) to be negative. The correction is to use the lower limit basis: \([0,1)\) is a basis element containing \(0\) and contained in the set. The same discipline applies in subspaces: use basis elements of the form \(U\cap X\), not ambient sets without intersecting with the subspace.
Local witnesses are useful because they make the quantifiers in topological definitions constructive. To prove a set open, give a basis element around each of its points. To prove closure membership, show every basis neighborhood intersects the set. To prove continuity at a point, find a domain neighborhood-base member whose image lies in the target neighborhood; if the target also has a neighborhood base, it is enough to handle those target witnesses. In each case, the basis reduces the number of neighborhoods to check without changing the definition.
Check Your Understanding
Use basis elements or neighborhood-base members explicitly in each response.
- In the standard topology on \(\mathbb{R}\), show that \((1,4)\) is open by giving a suitable basis element for an arbitrary \(x\in(1,4)\).
- In \(X=[0,1]\) with its subspace topology, determine whether \(1\) belongs to the closure of \(A=(0,1)\), and justify your answer using basis neighborhoods.
- State why a basis for the standard topology on \(\mathbb{R}\) cannot be used as a basis for the lower limit topology.
- Explain how a neighborhood base at \(f(x)\) reduces the target neighborhoods that need to be tested for continuity at \(x\).
- For \(g(t)=3t+1\) from \(\mathbb{R}\) to \(\mathbb{R}\), find a domain interval around \(x=2\) that maps inside \((7-\varepsilon,7+\varepsilon)\), for arbitrary \(\varepsilon>0\).