Comparing Topologies by Refining Basis Elements
Two topologies on the same set can give different meanings to “open,” even when their basis elements look similar. A reliable way to compare them is to ask whether every basic open set in one topology can be assembled from basic open sets in the other. The direction matters: if the basis elements of one topology are open in the other, then the topology they generate is contained in the other topology.
Throughout, both topologies are on the same set \(X\). A basis \(\mathcal{B}\) for a topology on \(X\) is a collection of subsets that covers \(X\) and has the following property: if \(x\in B_1\cap B_2\), where \(B_1,B_2\in\mathcal{B}\), then some \(B_3\in\mathcal{B}\) satisfies \(x\in B_3\subseteq B_1\cap B_2\). The open sets of the topology are exactly the unions of basis elements; the empty union is \(\varnothing\).
The Basis-Refinement Criterion
Proof. Suppose first that \(\mathcal{T}_1\) is finer than \(\mathcal{T}_0\). Each \(B_0\in\mathcal{B}_0\) is open in \(\mathcal{T}_0\), so \(B_0\in\mathcal{T}_0\subseteq\mathcal{T}_1\). By the definition of a basis, every \(\mathcal{T}_1\)-open set is a union of members of \(\mathcal{B}_1\). Thus \(B_0\) is such a union. In particular, for any \(x\in B_0\), at least one of the basis elements in that union contains \(x\); that element is contained in \(B_0\).
Conversely, suppose every \(B_0\in\mathcal{B}_0\) is a union of members of \(\mathcal{B}_1\). Any \(U\in\mathcal{T}_0\) is a union of members of \(\mathcal{B}_0\). Replacing each of those basis elements by its union of \(\mathcal{B}_1\)-elements expresses \(U\) as a union of members of \(\mathcal{B}_1\). Hence \(U\in\mathcal{T}_1\), so \(\mathcal{T}_0\subseteq\mathcal{T}_1\). For the pointwise formulation, if \(x\in B_0\), membership in the union gives a \(B_1\) with \(x\in B_1\subseteq B_0\). Conversely, if such a \(B_1\) exists for every \(x\in B_0\), then \(B_0\) is the union of all the \(\mathcal{B}_1\)-elements contained in it: each is contained in \(B_0\), and every point of \(B_0\) belongs to one. If \(B_0=\varnothing\), the empty union gives the same conclusion. \(\square\)
The pointwise version is often the fastest to use. For each point in a coarser basis element, find a finer basis element that contains that point and stays inside the given set. The witness may depend on the point. One basis element need not work for every point.
Proof. If the topologies are equal, each is finer than the other, so the theorem gives both refinement conditions. If both refinement conditions hold, the theorem gives \(\mathcal{T}_0\subseteq\mathcal{T}_1\) and \(\mathcal{T}_1\subseteq\mathcal{T}_0\). The two collections of open sets are therefore equal. \(\square\)
Worked Comparisons on the Real Line
Worked Example: The Standard and Lower-Limit Topologies
Let \(\mathbb{R}\) have its standard topology, with basis all open intervals \((a,b)\), and also consider the lower-limit topology \(\mathbb{R}_\ell\), with basis intervals \([a,b)\) for \(a<b\). We determine which topology is finer.
First take any standard basis element \((a,b)\), and fix \(x\in(a,b)\). The lower-limit basis element \([x,(x+b)/2)\) contains \(x\). Because \(a<x<b\), we have \(x<(x+b)/2<b\), so every point of \([x,(x+b)/2)\) lies in \((a,b)\). Thus every point of every standard basis interval has a lower-limit basis neighborhood contained in that interval. By the criterion, the lower-limit topology is finer than the standard topology.
For example, \((0,2)\) is exactly the union of the lower-limit basis elements \([x,(x+2)/2)\) as \(x\) ranges over \((0,2)\). Each displayed interval is contained in \((0,2)\); and every \(x\in(0,2)\) belongs to its own interval \([x,(x+2)/2)\). The union therefore contains no point outside \((0,2)\) and misses none of its points.
To test the reverse direction, use the lower-limit basis element \([0,1)\). If the standard topology were finer, some standard basis interval \((c,d)\) would have to satisfy \(0\in(c,d)\subseteq[0,1)\). But \(0\in(c,d)\) implies \(c<0\), so \((c,d)\) contains points less than \(0\), which are not in \([0,1)\). No such interval exists. Hence the standard topology is not finer than the lower-limit topology.
Check: the comparison is strict: the lower-limit topology contains every standard open set, and \([0,1)\) is lower-limit open but not standard open.
Worked Example: The Cofinite and Standard Topologies
On \(\mathbb{R}\), let \(\mathcal{T}_{\mathrm{cof}}\) consist of \(\varnothing\) and all sets whose complement is finite. This is a topology: arbitrary unions of nonempty cofinite sets have finite complements (their complement is contained in the finite complement of any one of them), and a finite intersection of cofinite sets is cofinite. The cofinite sets themselves form a basis, since they cover \(\mathbb{R}\) and the intersection of two cofinite sets is cofinite.
We show that every cofinite basis element \(C=\mathbb{R}\setminus F\), with \(F\) finite, is standard open. Take \(x\in C\). If \(F=\varnothing\), any open interval around \(x\) is contained in \(C=\mathbb{R}\). If \(F\ne\varnothing\), the distances \(|x-f|\), for \(f\in F\), are finitely many positive numbers. Their minimum \(m\) is positive. The standard open interval \((x-m/2,x+m/2)\) contains \(x\) and excludes every \(f\in F\), because each such \(f\) is at distance at least \(m>m/2\) from \(x\). Thus this interval is contained in \(C\). The criterion shows that the standard topology is finer than the cofinite topology.
The reverse fails. The standard basis element \((0,1)\) cannot be a union of cofinite basis elements: if such a union contained \(0\), one of its cofinite sets would contain \(0\) and be contained in \((0,1)\). But a cofinite subset of \(\mathbb{R}\) cannot be contained in \((0,1)\), since its complement would then have to contain every real number outside \((0,1)\), an infinite set. Therefore the cofinite topology is strictly coarser than the standard topology.
Check: the argument uses finiteness precisely when it takes the minimum of finitely many positive distances from \(x\) to the excluded points.
Worked Example: The Standard and Discrete Topologies
Give \(\mathbb{R}\) the discrete topology, whose basis is the collection of all singletons \(\{x\}\). Every standard basis interval \((a,b)\) is the union of the singletons \(\{x\}\) for \(x\in(a,b)\). Each singleton in that union is contained in \((a,b)\), and each point of \((a,b)\) appears in its own singleton. The basis-refinement criterion proves that the discrete topology is finer than the standard topology.
For the reverse comparison, consider the discrete basis element \(\{0\}\). A standard basis interval containing \(0\) has the form \((c,d)\) with \(c<0<d\). It contains points other than \(0\), so it cannot be contained in \(\{0\}\). No standard basis element refines \(\{0\}\) at \(0\), and the standard topology is not finer than the discrete topology. The comparison is strict.
Check: every subset of \(\mathbb{R}\) is a union of singletons, so the topology generated by this basis is indeed the discrete topology.
Direction, Equality, and a Common Pitfall
The criterion can be organized as a short proof strategy. First identify which topology is claimed to be finer. Then start with basis elements of the proposed coarser topology. For each point in each such element, find a basis neighborhood from the proposed finer topology that stays inside it. To disprove the comparison, it is enough to find one coarser basis element and one point in it for which no such finer basis neighborhood exists.
Common pitfall: reversing the basis roles. In the lower-limit example, a mistaken argument might say: “Every standard interval is a union of lower-limit basis intervals, so the standard topology is finer.” The conclusion reverses the actual implication. The standard intervals are the basis elements of the standard topology, and expressing them as unions of lower-limit basis elements proves that they are lower-limit open. It proves \(\mathcal{T}_{\mathrm{standard}}\subseteq\mathcal{T}_{\ell}\), not the reverse. To check the reverse, start instead with a lower-limit basis element, such as \([0,1)\); at \(0\), no standard interval containing \(0\) fits inside it.
One related caution is that a basis element need not itself be a basis element of the other topology. The requirement is that it be open there, which means it is a union of that topology’s basis elements. The local containment witnesses may vary from point to point, as they did for \((a,b)\) in the lower-limit example.
| Claim to establish | Start with | What the test requires |
|---|---|---|
| \(\mathcal{T}_1\) is finer than \(\mathcal{T}_0\) | Each \(B_0\in\mathcal{B}_0\) | At each \(x\in B_0\), find \(B_1\in\mathcal{B}_1\) with \(x\in B_1\subseteq B_0\) |
| \(\mathcal{T}_0\) is finer than \(\mathcal{T}_1\) | Each \(B_1\in\mathcal{B}_1\) | At each \(x\in B_1\), find \(B_0\in\mathcal{B}_0\) with \(x\in B_0\subseteq B_1\) |
| The topologies are equal | Both collections of basis elements | Verify both directions of refinement |
Check Your Understanding
For each comparison, name the proposed finer topology before applying the pointwise test.
- In the standard topology on \(\mathbb{R}\), show directly that \((2,5)\) is a union of lower-limit basis elements.
- Why does \([2,5)\) fail to have a standard basis neighborhood contained in it at the point \(2\)?
- On \(\mathbb{R}\), compare the cofinite topology with the discrete topology using the criterion.
- Suppose two bases on \(X\) satisfy the pointwise refinement condition in one direction only. What inclusion between their topologies follows?
- Explain why knowing that each basis element of \(\mathcal{B}_0\) is itself a member of \(\mathcal{B}_1\) is sufficient, but stronger than necessary, for \(\mathcal{T}_1\) to be finer than \(\mathcal{T}_0\).