From Basis Elements to a Topology
A basis gives a local description of open sets: each open set is a union of basis elements. This description also provides a direct way to construct the topology itself. Starting with a basis \(\mathcal{B}\) on a set \(X\), take all possible unions of its members, including the empty union. The resulting collection is the topology generated by \(\mathcal{B}\).
The basis conditions matter here. They ensure that the collection of unions really is a topology, as established in Open Sets in Topological Spaces. Once that construction is available, a further question is useful: among all topologies on \(X\) that contain the basis elements, where does the generated topology sit? It is the smallest such topology.
Because \(\mathcal{B}\) is a basis, the construction theorem from Open Sets in Topological Spaces guarantees that \(\mathcal{T}(\mathcal{B})\) is a topology. Also, every basis element belongs to this topology: it is the union of the one-member subcollection containing itself. The basis does not generally equal the topology; the topology additionally contains all unions of basis elements.
Proof. Let \(\mathcal{T}\) be a topology on \(X\) containing \(\mathcal{B}\). Take any \(U\in\mathcal{T}(\mathcal{B})\). By definition, \(U\) is a union of members of \(\mathcal{B}\), so there is a subcollection \(\mathcal{C}\subseteq\mathcal{B}\) such that $$ U=\bigcup_{C\in\mathcal{C}}C. $$ Each \(C\in\mathcal{C}\) belongs to \(\mathcal{T}\), since \(\mathcal{B}\subseteq\mathcal{T}\). A topology is closed under arbitrary unions, so \(U\in\mathcal{T}\). This proves \(\mathcal{T}(\mathcal{B})\subseteq\mathcal{T}\), including when \(\mathcal{C}\) is empty, because every topology contains \(\varnothing\). Since \(\mathcal{T}(\mathcal{B})\) is itself a topology containing \(\mathcal{B}\), it is the smallest one with that property. \(\square\)
This minimality result is a useful way to understand “generated.” The phrase does not mean that the basis elements are the only open sets. It means that we include exactly the unions required to make a topology, and no more. If a topology is already specified and \(\mathcal{B}\) is one of its bases, the theorem that every open set is a union of basis elements shows that the generated topology is exactly the specified topology.
Worked Example: Generating a Topology on a Finite Set
Let \(X=\{a,b,c\}\) and take $$ \mathcal{B}=\big\{\{a,b\},\{b,c\},\{b\}\big\}. $$ These sets cover \(X\). To check the basis intersection condition, the only intersection of distinct nonempty members that is not one of those members is $$ \{a,b\}\cap\{b,c\}=\{b\}, $$ which is itself in \(\mathcal{B}\). Intersections involving \(\{b\}\) are either \(\{b\}\) or already covered by the same condition. Thus \(\mathcal{B}\) is a basis.
Its arbitrary unions are $$ \varnothing,\quad \{b\},\quad \{a,b\},\quad \{b,c\},\quad X. $$ For example, \(\{a,b\}\cup\{b,c\}=X\), while adjoining \(\{b\}\) to either of those sets does not change it. Therefore $$ \mathcal{T}(\mathcal{B})=\big\{\varnothing,\{b\},\{a,b\},\{b,c\},X\big\}. $$ The topology has five open sets, although the basis has only three elements. In particular, \(\varnothing\) and \(X\) are open because they arise as the empty union and as the union of the two sets \(\{a,b\}\) and \(\{b,c\}\), respectively.
A Basis Can Generate More Than One Way
A topology may have several different bases. This does not create ambiguity about the topology: each basis for the same topology generates that topology when all its unions are taken. The collection of basis elements can change while the open sets remain unchanged.
For example, on a space with a given topology, one basis might use relatively large open sets and another might use smaller ones. Each open set is still a union of members of either basis. The pointwise refinement criterion from Bases for Topologies can be used to compare two proposed bases: locally, each basis element must be refinable by elements of the other when they are to describe the same topology.
Worked Example: Half-Open Intervals Generate a Topology on \(\mathbb{R}\)
Consider the collection $$ \mathcal{B}=\{[a,b):a,b\in\mathbb{R},\ a<b\}. $$ It covers \(\mathbb{R}\), since \(x\in[x,x+1)\) for every \(x\in\mathbb{R}\). If two members intersect, then $$ [a,b)\cap[c,d)=[\max(a,c),\min(b,d)) $$ whenever \(\max(a,c)<\min(b,d)\); if this inequality fails, the intersection is empty. Thus every point in the intersection belongs to a half-open interval contained in the intersection (indeed, the intersection itself is a basis element). These sets form a basis, and their arbitrary unions make up the generated topology.
For instance, \([1,3)\) is open in the generated topology because it is a basis element. The set \((-\infty,0)\) is also open: it is the union $$ (-\infty,0)=\bigcup_{n=1}^{\infty}[-n,0). $$ Each negative real number belongs to one of these intervals, and none contains \(0\) or a positive number. This example shows why the generated topology is not just the basis collection: unions can produce open sets that are not bounded half-open intervals.
The interval \([1,3)\) is not open in the usual topology on \(\mathbb{R}\), since no usual open interval around \(1\) is contained in \([1,3)\). Thus the topology generated by this basis is different from the usual topology. The description by unions specifies precisely which sets are open without requiring them all to be listed individually.
When the Starting Collection Is Not a Basis
It is tempting to start with any collection \(\mathcal{S}\) of subsets of \(X\) and declare its arbitrary unions to be open. This can fail: arbitrary unions of the chosen sets need not be closed under finite intersections. The basis conditions are what prevent this problem. When a collection fails those conditions, a standard repair is to first take all finite intersections of its members.
This construction turns an arbitrary family into a basis. It then allows us to form a topology by taking arbitrary unions, using the basis construction theorem from Open Sets in Topological Spaces. In the language of general topology, \(\mathcal{S}\) is a subbasis for the topology obtained this way.
Proof. The collection \(\mathcal{C}\) contains \(X\), so its members cover \(X\). If \(C_1,C_2\in\mathcal{C}\), each is a finite intersection of members of \(\mathcal{S}\), where \(X\) may represent the empty intersection. Then \(C_1\cap C_2\) is again a finite intersection of members of \(\mathcal{S}\), so \(C_1\cap C_2\in\mathcal{C}\). In particular, for any \(x\in C_1\cap C_2\), the member \(C_1\cap C_2\) itself is a basis element containing \(x\) and contained in \(C_1\cap C_2\). This verifies the basis intersection condition; hence \(\mathcal{C}\) is a basis.
Let \(\mathcal{T}(\mathcal{C})\) be the topology generated by this basis. Every \(S\in\mathcal{S}\) belongs to \(\mathcal{C}\), as a one-member intersection, so \(S\in\mathcal{T}(\mathcal{C})\). Now let \(\mathcal{T}\) be any topology containing \(\mathcal{S}\). Closure under finite intersections implies that \(\mathcal{T}\) contains every member of \(\mathcal{C}\); it contains \(X\) as well. Closure under arbitrary unions then implies that \(\mathcal{T}\) contains every member of \(\mathcal{T}(\mathcal{C})\). Therefore \(\mathcal{T}(\mathcal{C})\subseteq\mathcal{T}\), proving that the constructed topology is the smallest topology containing \(\mathcal{S}\). \(\square\)
Worked Example: Repairing a Collection That Is Not a Basis
Let \(X=\{p,q,r,s\}\) and let $$ \mathcal{S}=\big\{\{p,q\},\{q,r\}\big\}. $$ This collection is not a basis for \(X\): it does not cover \(s\), and the intersection condition also fails. In particular, $$ \{p,q\}\cap\{q,r\}=\{q\}, $$ but neither member of \(\mathcal{S}\) is contained in \(\{q\}\).
Taking finite intersections and including \(X\) gives $$ \mathcal{C}=\big\{X,\{p,q\},\{q,r\},\{q\}\big\}. $$ These sets form a basis. Their arbitrary unions are $$ \varnothing,\quad \{q\},\quad \{p,q\},\quad \{q,r\},\quad \{p,q,r\},\quad X. $$ For example, \(\{p,q,r\}=\{p,q\}\cup\{q,r\}\). The resulting topology contains the original sets in \(\mathcal{S}\), and the theorem shows it is the smallest topology on \(X\) that does so.
The extra set \(\{q\}\) is not an arbitrary addition: it is forced by closure under finite intersections. The set \(X\) is included so the finite intersections cover the entire underlying set, even though the original family did not contain \(s\). Both steps are necessary in this construction.
What the Construction Does—and Does Not—Guarantee
There are two related procedures. If the starting collection is already a basis, take all its arbitrary unions. If it is an arbitrary family, first take finite intersections, including \(X\), and then take arbitrary unions. The first procedure is shorter because a basis already has the local intersection property needed for its unions to form a topology.
A common pitfall is to skip that check and assume that arbitrary unions alone repair any collection. For \(\mathcal{S}=\{\{p,q\},\{q,r\}\}\), both chosen sets are unions of members of \(\mathcal{S}\), but their intersection \(\{q\}\) is not such a union. Thus the collection of arbitrary unions of \(\mathcal{S}\) is not a topology. Adding finite intersections first resolves the issue.
The minimality statements also explain why these procedures are canonical. Once the starting basis, or the starting family, is fixed, the construction adds exactly the unions and intersections required by the topology axioms. Any topology that contains the starting sets must contain all the sets produced by the construction.
Check Your Understanding
Use the constructions and results in this tutorial to answer the following questions.
- How is the topology generated by a basis defined, and what is the empty union?
- Why is the topology generated by a basis contained in every topology that contains the basis?
- What basis condition ensures that intersections of two basis elements can be handled locally?
- How do finite intersections help when an initial family is not a basis?
- Why must \(X\) be included among the finite intersections generated by an arbitrary family?