From Subbasic Sets to a Topology
A finite subbasis lets us calculate a topology in two stages: first take intersections of subbasic sets, then take unions of the resulting sets. The order matters. Intersections create the basic building blocks; unions combine them into open sets. On a finite underlying set, each stage can be listed explicitly, so we can check the answer against every subset of the space.
Throughout, \(X=\{a,b,c\}\) has three distinct points. A topology on \(X\) is a collection \(\mathcal{T}\subseteq\mathcal{P}(X)\) containing \(\varnothing\) and \(X\), closed under arbitrary unions and finite intersections. A subbasis \(\mathcal{S}\) for a topology is a family of open sets whose finite intersections form a basis for that topology. In the examples below, each subbasis covers \(X\).
Repeated sets do not add anything to either list, so we list only distinct intersections and distinct unions. In a finite space, there are only finitely many subsets to check. Also, an arbitrary union of subsets of \(X\) is the union of at most eight distinct subsets, since \(\mathcal{P}(X)\) has eight members.
Why the Two-Stage Calculation Works
Proof. Write \(\mathcal{B}_{\mathcal{S}}\) for the finite intersections, including \(X\), and let \(\mathcal{T}_{\mathcal{S}}\) be their arbitrary unions. The empty union shows \(\varnothing\in\mathcal{T}_{\mathcal{S}}\), and the empty intersection \(X\) is itself a union with one member, so \(X\in\mathcal{T}_{\mathcal{S}}\). A union of unions of members of \(\mathcal{B}_{\mathcal{S}}\) is still a union of members of \(\mathcal{B}_{\mathcal{S}}\); hence \(\mathcal{T}_{\mathcal{S}}\) is closed under arbitrary unions.
Now take two members \(U,V\in\mathcal{T}_{\mathcal{S}}\). Write \(U=\bigcup_{i\in I}B_i\) and \(V=\bigcup_{j\in J}C_j\), where every \(B_i,C_j\in\mathcal{B}_{\mathcal{S}}\). Then \(U\cap V=\bigcup_{(i,j)\in I\times J}(B_i\cap C_j)\). Each \(B_i\cap C_j\) is again a finite intersection of members of \(\mathcal{S}\), so \(U\cap V\in\mathcal{T}_{\mathcal{S}}\). This identity also covers an empty index set: if \(I\) or \(J\) is empty, both the intersection and the displayed union are empty. Closure under intersections of two sets implies closure under every finite intersection, including the intersection of no sets, which is \(X\). Thus \(\mathcal{T}_{\mathcal{S}}\) is a topology.
Each member of \(\mathcal{S}\) is a finite intersection consisting of that one set, so \(\mathcal{T}_{\mathcal{S}}\) contains \(\mathcal{S}\). Finally, if \(\mathcal{T}\) is any topology containing \(\mathcal{S}\), then it contains every finite intersection of members of \(\mathcal{S}\), and it contains every union of those intersections. Therefore \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}\). This proves that \(\mathcal{T}_{\mathcal{S}}\) is the smallest topology containing \(\mathcal{S}\). \(\square\)
For a finite subbasis, the theorem gives a direct procedure: list its distinct finite intersections, then list all unions of those sets. The theorem just proved guarantees that the result is a topology and that no smaller topology containing the given sets is missed.
Worked Example: Two Overlapping Subbasic Sets
Let \(P=\{a,b\}\), \(Q=\{b,c\}\), and \(\mathcal{S}_1=\{P,Q\}\). Find the generated topology and check that the list is complete.
Stage 1: finite intersections. The empty intersection is \(X=\{a,b,c\}\). The one-set intersections are \(P=\{a,b\}\) and \(Q=\{b,c\}\). The intersection of both subbasic sets is \(P\cap Q=\{b\}\). These are all the distinct finite intersections: \(\mathcal{B}_{\mathcal{S}_1}=\{X,\{a,b\},\{b,c\},\{b\}\}\). In particular, none of these intersections is empty: the two subbasic sets overlap at \(b\).
Stage 2: unions. Unions not involving \(X\) are formed from \(\varnothing,\{a,b\},\{b,c\},\{b\}\). The union of no sets is \(\varnothing\); the union of \(\{b\}\) alone is \(\{b\}\); and the single-set unions include \(\{a,b\}\) and \(\{b,c\}\). The union \(\{a,b\}\cup\{b,c\}\) is \(X\), and any union involving \(\{b\}\) with either of those sets adds no new point. Thus \(\mathcal{T}_{\mathcal{S}_1}=\{\varnothing,\{b\},\{a,b\},\{b,c\},X\}\).
Check for missing subsets. The eight subsets of \(X\) are \(\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},X\). The list contains \(\varnothing,\{b\},\{a,b\},\{b,c\},X\); the subsets not listed are exactly \(\{a\},\{c\},\{a,c\}\). None is a union of the finite intersections: every nonempty intersection in the list contains \(b\), and a union of such sets also contains \(b\). So none of these three subsets can occur. The generated topology has exactly five sets.
Check by the generation theorem. Every listed set is a union of the four finite intersections, and the theorem ensures that these unions form the smallest topology containing \(P\) and \(Q\). The list includes both subbasic sets, as required.
Worked Example: A Disjoint Pair
Keep \(X=\{a,b,c\}\), but take \(\mathcal{S}_2=\{P,R\}\), where \(P=\{a,b\}\) and \(R=\{c\}\). Compute its topology and compare it with the preceding one.
The empty intersection is \(X\). The one-set intersections are \(P\) and \(R\), while the intersection of both is \(P\cap R=\varnothing\), since \(P\) contains \(a,b\) and \(R\) contains \(c\). Hence the complete list of distinct finite intersections is \(\{X,\{a,b\},\{c\},\varnothing\}\).
Their distinct unions are \(\varnothing\), \(\{a,b\}\), \(\{c\}\), and \(\{a,b\}\cup\{c\}=X\). Unions that include \(\varnothing\) add no points, and unions that include \(X\) equal \(X\). Therefore \(\mathcal{T}_{\mathcal{S}_2}=\{\varnothing,\{a,b\},\{c\},X\}\).
Check for missing subsets. Among the eight subsets of \(X\), the ones absent from this list are \(\{a\},\{b\},\{a,c\},\{b,c\}\). A union of \(\{a,b\}\) and \(\{c\}\) either contains both \(a,b\), contains neither, or is \(X\); it cannot contain just one of \(a,b\). This verifies that the four omitted subsets cannot be formed. Compared with the first example, the generated topology is different: it has four sets instead of five.
Check against the construction. Both subbasic sets occur in the resulting collection, and each of its four sets is a union of the displayed intersections. The generation theorem therefore confirms that this is exactly the topology generated by \(\mathcal{S}_2\).
When a Subbasic Set Is Redundant
Proof. Since \(\mathcal{S}\subseteq\mathcal{S}\cup\{V\}\), every topology containing \(\mathcal{S}\cup\{V\}\) contains \(\mathcal{S}\). In particular, \(\mathcal{T}_{\mathcal{S}\cup\{V\}}\) contains \(\mathcal{S}\), so the minimality in the generation theorem gives \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}_{\mathcal{S}\cup\{V\}}\). Conversely, \(\mathcal{T}_{\mathcal{S}}\) contains every finite intersection of members of \(\mathcal{S}\), including \(V\). It therefore contains \(\mathcal{S}\cup\{V\}\), and minimality gives \(\mathcal{T}_{\mathcal{S}\cup\{V\}}\subseteq\mathcal{T}_{\mathcal{S}}\). The two inclusions prove equality. \(\square\)
Worked Example: Adding an Intersection as a Generator
Start with \(\mathcal{S}_1=\{\{a,b\},\{b,c\}\}\) and add the set \(\{b\}\), obtaining \(\mathcal{S}_3=\{\{a,b\},\{b,c\},\{b\}\}\). Find the finite intersections and compare the generated topology directly.
The empty intersection is \(X\). The one-set intersections are \(\{a,b\},\{b,c\},\{b\}\). The two-set intersections are \(\{a,b\}\cap\{b,c\}=\{b\}\), \(\{a,b\}\cap\{b\}=\{b\}\), and \(\{b,c\}\cap\{b\}=\{b\}\). The intersection of all three sets is also \(\{b\}\). Thus the distinct finite intersections are exactly \(X,\{a,b\},\{b,c\},\{b\}\).
Their distinct unions are \(\varnothing,\{b\},\{a,b\},\{b,c\},X\): the union of \(\{a,b\}\) and \(\{b,c\}\) is \(X\), and adjoining \(\{b\}\) to either does not enlarge it. Hence \(\mathcal{T}_{\mathcal{S}_3}=\{\varnothing,\{b\},\{a,b\},\{b,c\},X\}=\mathcal{T}_{\mathcal{S}_1}\).
Check. The added set is precisely \(\{a,b\}\cap\{b,c\}=\{b\}\), a finite intersection of the original subbasic sets. The proposition predicts that it is redundant, and the direct lists give the same five open sets in both cases.
A Common Counting Pitfall
Common mistake. For \(\mathcal{S}_1=\{\{a,b\},\{b,c\}\}\), one might list \(\varnothing\) as a finite intersection simply because it is often included in a topology. That is not justified by the intersection calculation: the empty intersection is \(X\), and the only intersection using both subbasic sets is \(\{b\}\), not \(\varnothing\). In this example, \(\varnothing\) enters the topology as the empty union of finite intersections, not as a finite intersection of subbasic sets.
A second counting error is to say that the subsets missing from \(\mathcal{T}_{\mathcal{S}_1}\) are only \(\{a\}\) and \(\{c\}\). The full list of subsets also includes \(\{a,c\}\), which is absent. Checking against all eight subsets avoids both errors. The distinction between the two stages is useful: an empty union always supplies \(\varnothing\), while an empty intersection always supplies \(X\).
| Subbasis on \(X\) | Distinct finite intersections | Generated topology |
|---|---|---|
| \(\{\{a,b\},\{b,c\}\}\) | \(X,\{a,b\},\{b,c\},\{b\}\) | \(\varnothing,\{b\},\{a,b\},\{b,c\},X\) |
| \(\{\{a,b\},\{c\}\}\) | \(X,\{a,b\},\{c\},\varnothing\) | \(\varnothing,\{a,b\},\{c\},X\) |
| \(\{\{a,b\},\{b,c\},\{b\}\}\) | \(X,\{a,b\},\{b,c\},\{b\}\) | \(\varnothing,\{b\},\{a,b\},\{b,c\},X\) |
Check Your Understanding
For each question, keep the empty intersection \(X\) and empty union \(\varnothing\) distinct.
- On \(X=\{a,b,c\}\), take \(\mathcal{S}=\{\{a\},\{b,c\}\}\). List its distinct finite intersections and determine the generated topology.
- For \(\mathcal{S}=\{\{a,b\},\{a,c\}\}\), list every distinct finite intersection. Does \(\varnothing\) occur as an intersection?
- Find a subbasis on \(X\) whose generated topology has exactly \(\{\varnothing,\{a\},X\}\), and verify the list by computing intersections and unions.
- Explain why adding \(\{b,c\}\) to \(\{\{a,b\},\{b,c\}\}\) changes the generated topology, while adding \(\{b\}\) does not.
- For any finite subbasis \(\mathcal{S}\), why must the generated topology contain \(\varnothing\) even when \(\varnothing\) is not a finite intersection of members of \(\mathcal{S}\)?