From Subbasic Sets to a Topology
A subbasis describes open sets indirectly: its members need not themselves be enough to build every open neighborhood, but finite intersections of them provide the local building blocks. Arbitrary unions of those building blocks then produce a topology. The key test is pointwise: a set is open in the generated topology precisely when each of its points lies in one of these finite intersections that stays inside the set.
Let \(X\) be a set and let \(\mathcal{S}\subseteq\mathcal{P}(X)\). A finite intersection of members of \(\mathcal{S}\) means a set \(S_1\cap\cdots\cap S_n\), where \(n\) is a nonnegative integer and each \(S_i\in\mathcal{S}\). When \(n=0\), the intersection is defined to be \(X\), called the empty intersection. Thus \(X\) is included whether or not it belongs to \(\mathcal{S}\).
The covering condition matters: every point must lie in some basic set. If \(\mathcal{S}\) does not cover \(X\), it is not a subbasis under this definition. One can still generate a topology by adjoining \(X\) to the collection first. If \(X=\varnothing\), the empty collection covers \(X\), and its empty intersection is also \(\varnothing\).
The Subbasis Criterion
Proof. We check the basis conditions. First, the members of \(\mathcal{B}\) cover \(X\): each \(x\in X\) belongs to some \(S\in\mathcal{S}\), and \(S\) is itself a finite intersection, with one factor. If \(X=\varnothing\), coverage holds vacuously.
Next, take \(B_1,B_2\in\mathcal{B}\). Each is a finite intersection of members of \(\mathcal{S}\), so \(B_1\cap B_2\) is also a finite intersection of members of \(\mathcal{S}\). If either intersection is empty as a list of factors, it contributes \(X\) and does not change the result. Therefore \(B_1\cap B_2\in\mathcal{B}\). In particular, for every \(x\in B_1\cap B_2\), the basic set \(B_1\cap B_2\) contains \(x\) and is contained in \(B_1\cap B_2\), as required by the basis intersection condition. The basis criterion therefore gives a topology consisting of unions of members of \(\mathcal{B}\).
Every \(S\in\mathcal{S}\) belongs to \(\mathcal{B}\), so it is open in this topology. Now let \(\mathcal{T}\) be any topology on \(X\) containing every member of \(\mathcal{S}\). A topology contains \(X\), is closed under finite intersections, and is closed under arbitrary unions. Hence it contains every finite intersection in \(\mathcal{B}\), and then every union of members of \(\mathcal{B}\). The topology constructed from \(\mathcal{B}\) is therefore contained in every topology containing \(\mathcal{S}\). This proves it is the smallest such topology. \(\square\)
Proof. If \(U\) is open, the theorem says \(U\) is a union of members of \(\mathcal{B}\). For any \(x\in U\), at least one of those basic sets contains \(x\); that set is contained in \(U\). Conversely, suppose the stated pointwise condition holds. For every \(x\in U\), choose a basic set \(B_x\) with \(x\in B_x\subseteq U\). Then \(U=\bigcup_{x\in U}B_x\): every point of \(U\) lies in its chosen set, and every chosen set lies inside \(U\). This is a union of basic sets, hence open in the generated topology. If \(U=\varnothing\), it is the empty union and is open as well. \(\square\)
The test is a local refinement condition: each point of a proposed open set must have a finite intersection of subbasic sets around it that does not leave the proposed set. The number of factors can depend on the point, and the chosen finite intersection can differ from point to point.
Worked Computations
Worked Example: Two Subsets of a Three-Point Set
Let \(X=\{a,b,c\}\) and \(\mathcal{S}=\{\{a,b\},\{b,c\}\}\). Since the union of these two sets is \(X\), they cover \(X\). Find the finite intersections and the generated topology.
The empty intersection is \(X=\{a,b,c\}\). Intersecting either subbasic set with itself leaves it unchanged, and the intersection of the two distinct subbasic sets is \(\{a,b\}\cap\{b,c\}=\{b\}\). Thus the distinct finite intersections are \(X,\{a,b\},\{b,c\},\{b\}\). There is no finite intersection equal to \(\varnothing\): every nonempty list of factors has intersection \(\{a,b\}\), \(\{b,c\}\), or \(\{b\}\), and the empty intersection is \(X\).
The generated topology consists of unions of those four basic sets. The empty union contributes \(\varnothing\). The nonempty unions give \(\{b\}\), \(\{a,b\}\), \(\{b,c\}\), or \(X\): in particular, \(\{a,b\}\cup\{b,c\}=X\). Therefore the generated topology is \(\{\varnothing,\{b\},\{a,b\},\{b,c\},X\}\).
Check: the collection contains \(\varnothing\) and \(X\). The union of \(\{a,b\}\) and \(\{b,c\}\) is \(X\), and unions involving \(\{b\}\) add no new set. Intersections of the two larger proper sets give \(\{b\}\), and the other pairwise intersections stay in the collection. These checks agree with the topology construction.
Worked Example: Intersections Create Singletons
Let \(X=\{1,2,3\}\) and \(\mathcal{S}=\{\{1,2\},\{2,3\},\{1,3\}\}\). The three subbasic sets cover \(X\). Determine the basic sets and test whether \(\{1\}\) is open in the generated topology.
The empty intersection is \(X\). The three intersections of distinct pairs of subbasic sets are \(\{1,2\}\cap\{2,3\}=\{2\}\), \(\{2,3\}\cap\{1,3\}=\{3\}\), and \(\{1,2\}\cap\{1,3\}=\{1\}\). The intersection of all three sets is \(\varnothing\), since \(\{1,2\}\cap\{2,3\}=\{2\}\) and \(\{2\}\cap\{1,3\}=\varnothing\). Intersecting a set with itself makes no change. Thus the basic sets are \(X\), the three two-point sets, the three singletons, and \(\varnothing\).
For \(U=\{1\}\), the finite intersection \(B=\{1,2\}\cap\{1,3\}=\{1\}\) satisfies \(1\in B\subseteq U\). There are no other points in \(U\) to test. The pointwise criterion gives that \(\{1\}\) is open. In fact, every subset of \(X\) is a union of the singleton basic sets, so the generated topology is the discrete topology \(\mathcal{P}(X)\).
Check: each of the eight subsets of \(X\) is a union of basic singletons: \(\varnothing\) is the empty union, and any nonempty subset is the union of its singleton members. There are exactly \(2^3=8\) subsets, so this accounts for the entire power set.
Worked Example: The Standard Topology on the Real Line
Let \(X=\mathbb{R}\) with its standard topology, and let \(\mathcal{S}\) be the collection of all open intervals \((a,b)\), where \(a,b\in\mathbb{R}\) and \(a<b\). Show that the topology generated by \(\mathcal{S}\) is the standard topology.
The intervals cover \(\mathbb{R}\): for any \(x\in\mathbb{R}\), the interval \((x-1,x+1)\) contains \(x\). A finite intersection of open intervals is either empty or an open interval. Indeed, for a nonempty finite list \((a_i,b_i)\), its intersection has lower endpoint \(\max_i a_i\) and upper endpoint \(\min_i b_i\); it is empty if \(\max_i a_i\geq\min_i b_i\), and otherwise it is \((\max_i a_i,\min_i b_i)\). The empty intersection is \(\mathbb{R}\).
Every generated open set is a union of finite intersections of open intervals, hence is standard-open. Conversely, if \(U\) is standard-open and \(x\in U\), the definition of the standard topology gives an \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq U\). This interval is a member of \(\mathcal{S}\), so the pointwise refinement test holds at \(x\). It holds for every \(x\in U\), and therefore \(U\) is generated by \(\mathcal{S}\). The two topologies are equal.
Check at a particular point: take \(U=(-2,3)\) and \(x=1\). The interval \((0,2)\) belongs to \(\mathcal{S}\), and \(1\in(0,2)\subseteq(-2,3)\). Thus the criterion supplies the required local basic set at this point. The general argument above applies to every point of every standard-open set.
A Common Mistake: Empty Intersection Is Not Empty Set
The phrase “finite intersection” includes the intersection of zero sets. By convention, that empty intersection is \(X\), not \(\varnothing\). The empty set may arise as an intersection of one or more subbasic sets when those sets have empty common intersection, but that is a separate question. By contrast, the empty union is always \(\varnothing\), and it belongs to every topology constructed as all unions of basic sets.
Common pitfall. For \(\mathcal{S}=\{\{a,b\},\{b,c\}\}\) on \(X=\{a,b,c\}\), a wrong calculation might list \(\varnothing\) among the finite intersections simply because it is the empty intersection. This confuses the two conventions. The actual empty intersection is \(X\); the intersection of the two distinct subbasic sets is \(\{b\}\), not \(\varnothing\). In this example \(\varnothing\) enters the generated topology as the empty union of basic sets, as the first worked example verifies.
The pointwise test also prevents a different error: it is not enough that \(U\) be a union of some subbasic sets in an informal sense. The test requires that every \(x\in U\) lie in an actual finite intersection of subbasic sets contained in \(U\). Establishing this containment is what certifies openness in the generated topology.
Check Your Understanding
Use the finite-intersection construction and the pointwise refinement test to justify your answers.
- If \(\mathcal{S}\) covers \(X\), why must the empty intersection be included among its finite intersections?
- For \(X=\{p,q,r\}\) and \(\mathcal{S}=\{\{p,q\},\{q,r\}\}\), list the distinct finite intersections and the generated topology.
- Give a subbasis on a three-point set for which the intersection of two distinct subbasic sets is empty. Explain how this differs from the empty intersection.
- In the standard topology on \(\mathbb{R}\), apply the pointwise test to \(U=(0,4)\) at \(x=3\) by choosing an open interval contained in \(U\).
- Prove directly from the pointwise test that the empty set is open.