Comparing Topologies by Checking Generators
A topology generated by a subbasis can contain many sets, because it includes arbitrary unions of finite intersections of the subbasic sets. Comparing two such topologies by listing all their open sets may therefore be inefficient. The central test is simpler: to check whether one generated topology is contained in another, check whether each generator of the first is open in the second.
Throughout, \(X\) is a set and \(\mathcal{S}\) and \(\mathcal{R}\) are families of subsets of \(X\). We use the convention that a subbasis may be any family of subsets; it need not cover \(X\). The empty intersection is \(X\), so the construction below always includes \(X\). If a convention requires a subbasis to cover \(X\), adjoining \(X\) to either family makes it cover \(X\) without changing its generated topology.
This description includes all the operations needed to form the generated topology: first take finite intersections of generators, then take arbitrary unions. The empty intersection supplies \(X\), and the empty union supplies \(\varnothing\).
Proof. The empty union and the empty intersection show that \(\varnothing,X\in\mathcal{T}_{\mathcal{S}}\). A union of any collection of members of \(\mathcal{T}_{\mathcal{S}}\) is again a union of finite intersections of members of \(\mathcal{S}\), so it belongs to \(\mathcal{T}_{\mathcal{S}}\). To check closure under finite intersections, write two members as \(U=\bigcup_{i\in I}B_i\) and \(V=\bigcup_{j\in J}C_j\), where every \(B_i\) and \(C_j\) is a finite intersection of members of \(\mathcal{S}\). Then \[ U\cap V=\bigcup_{(i,j)\in I\times J}(B_i\cap C_j). \] Each \(B_i\cap C_j\) is itself a finite intersection of members of \(\mathcal{S}\). If either union is empty, the intersection is \(\varnothing\), which is already in \(\mathcal{T}_{\mathcal{S}}\). Thus \(\mathcal{T}_{\mathcal{S}}\) is a topology. Each \(S\in\mathcal{S}\) belongs to it, since \(S\) is a one-set finite intersection. Finally, any topology containing every member of \(\mathcal{S}\) contains their finite intersections and arbitrary unions, so it contains every member of \(\mathcal{T}_{\mathcal{S}}\). This proves the asserted minimality. \(\square\)
The Subbasis Comparison Criterion
Proof. Suppose first that \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\). Each \(S\in\mathcal{S}\) belongs to \(\mathcal{T}_{\mathcal{S}}\), because it is a finite intersection with one factor. The assumed inclusion therefore gives \(S\in\mathcal{T}_{\mathcal{R}}\).
Conversely, suppose every \(S\in\mathcal{S}\) belongs to \(\mathcal{T}_{\mathcal{R}}\). Since \(\mathcal{T}_{\mathcal{R}}\) is a topology, it contains every finite intersection of members of \(\mathcal{S}\), and it contains every union of those intersections. By the definition of \(\mathcal{T}_{\mathcal{S}}\), every member of \(\mathcal{T}_{\mathcal{S}}\) is such a union. Hence \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\). Both implications hold, proving the equivalence. \(\square\)
To prove equality, apply the criterion in both directions: check that every \(\mathcal{S}\)-generator is open in \(\mathcal{T}_{\mathcal{R}}\), and every \(\mathcal{R}\)-generator is open in \(\mathcal{T}_{\mathcal{S}}\). To prove strict inclusion, one direction suffices for containment, followed by one open set in the larger topology that is absent from the smaller one.
Worked Example: Two Topologies on a Three-Point Set
Let \(X=\{a,b,c\}\), with the topologies understood to be generated by the indicated families. Take \(\mathcal{S}=\{\{a\},\{b,c\}\}\) and \(\mathcal{R}=\{\{a\},\{b,c\},\{a,b\}\}\). Compute both topologies and compare them.
For \(\mathcal{S}\), the empty intersection is \(X\), the one-set intersections are \(\{a\}\) and \(\{b,c\}\), and the intersection of the two distinct generators is \(\varnothing\). Taking unions of these sets gives \[ \mathcal{T}_{\mathcal{S}}=\{\varnothing,\{a\},\{b,c\},X\}. \]
For \(\mathcal{R}\), the finite intersections include \(X\), the three generators, and \(\{a\}\cap\{a,b\}=\{a\}\), \(\{a\}\cap\{b,c\}=\varnothing\), and \(\{a,b\}\cap\{b,c\}=\{b\}\). Unions of these intersections give \[ \mathcal{T}_{\mathcal{R}} =\{\varnothing,\{a\},\{b\},\{a,b\},\{b,c\},X\}. \] For example, \(\{a\}\cup\{b\}=\{a,b\}\), while \(\{a\}\cup\{b,c\}=X\). The displayed list includes every possible union of the available sets.
Each member of \(\mathcal{S}\) occurs in the displayed list for \(\mathcal{T}_{\mathcal{R}}\), so the theorem gives \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\). The inclusion is strict: \(\{b\}\in\mathcal{T}_{\mathcal{R}}\) but \(\{b\}\notin\mathcal{T}_{\mathcal{S}}\). Thus \(\mathcal{T}_{\mathcal{S}}\subsetneq\mathcal{T}_{\mathcal{R}}\).
Check: the generators \(\{a\}\) and \(\{b,c\}\) are open in \(\mathcal{T}_{\mathcal{R}}\), as the criterion requires. The set \(\{b\}\), which witnesses strictness, cannot be formed as a union of \(\varnothing,\{a\},\{b,c\},X\): any union containing \(b\) must include \(\{b,c\}\), and hence contains \(c\) as well.
Equality from Different Subbases on the Real Line
For a second comparison, let \(\mathbb{R}\) have its standard topology. Consider the subbasis of open rays \[ \mathcal{S}=\{(-\infty,a):a\in\mathbb{R}\}\cup\{(b,\infty):b\in\mathbb{R}\}, \] and let \(\mathcal{R}\) be the family of all bounded open intervals \((u,v)\) with \(u<v\). We compare their generated topologies directly by checking the generators in each direction.
Worked Example: Open Rays and Bounded Intervals
First take any generator \((-\infty,a)\in\mathcal{S}\). It is the union of bounded open intervals \[ (-\infty,a)=\bigcup_{\substack{n\in\mathbb{N}\\-n<a}}(-n,a). \] Every interval on the right is a member of \(\mathcal{R}\). To verify the equality, each such interval is contained in \((-\infty,a)\). Conversely, if \(x<a\), choose \(n\in\mathbb{N}\) with \(n>-x\); then \(-n<x<a\), so \(x\in(-n,a)\). Thus \((-\infty,a)\in\mathcal{T}_{\mathcal{R}}\).
For a generator \((b,\infty)\in\mathcal{S}\), use \[ (b,\infty)=\bigcup_{\substack{n\in\mathbb{N}\\b<n}}(b,n). \] Every interval on the right belongs to \(\mathcal{R}\). Each is contained in \((b,\infty)\); and if \(x>b\), choose \(n\in\mathbb{N}\) with \(n>x\), giving \(b<x<n\). Hence \((b,\infty)\in\mathcal{T}_{\mathcal{R}}\). The criterion now gives \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\).
For the reverse containment, take any generator \((u,v)\in\mathcal{R}\). It satisfies \[ (u,v)=(u,\infty)\cap(-\infty,v). \] Both rays on the right belong to \(\mathcal{S}\), so their finite intersection belongs to \(\mathcal{T}_{\mathcal{S}}\). The criterion gives \(\mathcal{T}_{\mathcal{R}}\subseteq\mathcal{T}_{\mathcal{S}}\). Therefore \(\mathcal{T}_{\mathcal{S}}=\mathcal{T}_{\mathcal{R}}\).
Check: the first containment followed from checking both types of ray, not just one of them. The reverse containment used the interval identity for every pair \(u<v\), so both inclusions required by the equality claim have been verified.
When a Generator Fails the Test
Worked Example: A Containment That Does Not Hold
Let \(X=\{a,b,c\}\), \(\mathcal{S}=\{\{a,b\}\}\), and \(\mathcal{R}=\{\{a\}\}\). Neither family is required to cover \(X\) under the convention stated above. For \(\mathcal{S}\), the finite intersections are \(X,\{a,b\}\); their unions give \[ \mathcal{T}_{\mathcal{S}}=\{\varnothing,\{a,b\},X\}. \] For \(\mathcal{R}\), the finite intersections are \(X,\{a\}\); their unions give \[ \mathcal{T}_{\mathcal{R}}=\{\varnothing,\{a\},X\}. \]
The sole \(\mathcal{S}\)-generator \(\{a,b\}\) is not in \(\mathcal{T}_{\mathcal{R}}\). The criterion therefore rules out \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\). Indeed, the explicit lists show that \(\{a,b\}\in\mathcal{T}_{\mathcal{S}}\) and \(\{a,b\}\notin\mathcal{T}_{\mathcal{R}}\). This confirms the failed containment directly.
Check: adjoining \(X\) to either family would not change either displayed topology, because \(X\) is already the empty intersection. Thus the same comparison works even if subbases are required to cover \(X\).
A Common Mistake and a Practical Test
Common pitfall. If every generator in one family is also a generator in another, the first generated topology is contained in the second; but this does not imply that the topologies are equal. In the first worked example, \(\mathcal{S}\subseteq\mathcal{R}\), so the criterion gives \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\). A mistaken conclusion would be \(\mathcal{T}_{\mathcal{S}}=\mathcal{T}_{\mathcal{R}}\). The computed set \(\{b\}\) corrects it: it is open in \(\mathcal{T}_{\mathcal{R}}\) but not in \(\mathcal{T}_{\mathcal{S}}\). For equality, the reverse generator test is also necessary.
To prove \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{R}}\), test the members of \(\mathcal{S}\) for openness in \(\mathcal{T}_{\mathcal{R}}\).
Express each tested set as a union of finite intersections of members of \(\mathcal{R}\).
The theorem gives the desired containment. For equality, repeat the test with \(\mathcal{R}\) and \(\mathcal{T}_{\mathcal{S}}\); for strictness, exhibit an open set in the larger topology missing from the smaller one.
The criterion matters because a subbasis is designed to generate open sets through repeated unions and finite intersections. The theorem lets us compare the resulting topologies without enumerating every generated open set. In examples with infinitely many generators, that can turn a large global comparison into a direct check of the generators.
Check Your Understanding
For each question, identify which family supplies the generators that must be tested, and state the direction of the proposed containment.
- On \(X=\{p,q,r\}\), let \(\mathcal{S}=\{\{p\}\}\) and \(\mathcal{R}=\{\{p\},\{q\}\}\). Compute both generated topologies and determine whether the inclusion is strict.
- On the standard real line, show that every bounded open interval is open in the topology generated by the open rays.
- If every member of \(\mathcal{R}\) is open in \(\mathcal{T}_{\mathcal{S}}\), which inclusion follows from the criterion?
- In the failed-containment example, which specific generator fails the openness test?
- State the two generator checks needed to prove \(\mathcal{T}_{\mathcal{S}}=\mathcal{T}_{\mathcal{R}}\).