From a Subbasis to a Basis
A basis gives local building blocks for the open sets of a topology. A subbasis is a less restrictive starting collection: its members do not need to satisfy the basis intersection condition. Instead, we first intersect finitely many of them, then take arbitrary unions. The finite intersections provide the basis that makes this construction work.
Let \(X\) be a set. In this tutorial, a subbasis \(\mathcal{S}\) on \(X\) means a family of subsets of \(X\) whose union is \(X\). If a starting family does not cover \(X\), adjoining \(X\) gives a covering family without changing the topology it generates. For a finite subfamily \(\mathcal{F}\subseteq\mathcal{S}\), take its intersection; when \(\mathcal{F}\) is empty, define its intersection to be \(X\).
The empty intersection matters even when every member of \(\mathcal{S}\) is a proper subset of \(X\): it ensures that \(X\) is in \(\mathcal{B}_{\mathcal{S}}\). The condition that \(\mathcal{S}\) covers \(X\) ensures that every point belongs to some member of \(\mathcal{B}_{\mathcal{S}}\). Together these are part of why finite intersections form a basis.
Why the Finite Intersections Form a Basis
Proof. To check the basis conditions, first let \(x\in X\). Since the members of \(\mathcal{S}\) cover \(X\), some \(S\in\mathcal{S}\) contains \(x\). The set \(S\) is the intersection of the finite subfamily \(\{S\}\), so \(S\in\mathcal{B}_{\mathcal{S}}\). Thus every point of \(X\) lies in a member of \(\mathcal{B}_{\mathcal{S}}\). If \(X=\varnothing\), this condition holds vacuously.
Now take \(B_1,B_2\in\mathcal{B}_{\mathcal{S}}\) and \(x\in B_1\cap B_2\). Write \(B_1=\bigcap_{S\in\mathcal{F}_1}S\) and \(B_2=\bigcap_{S\in\mathcal{F}_2}S\) for finite subfamilies \(\mathcal{F}_1,\mathcal{F}_2\subseteq\mathcal{S}\). Then \(B_1\cap B_2=\bigcap_{S\in\mathcal{F}_1\cup\mathcal{F}_2}S\). The family \(\mathcal{F}_1\cup\mathcal{F}_2\) is finite, including when either subfamily is empty. Hence \(B_1\cap B_2\in\mathcal{B}_{\mathcal{S}}\). Taking \(B=B_1\cap B_2\) gives a basis member containing \(x\) and contained in \(B_1\cap B_2\), as required. \(\square\)
In fact, the argument gives a little more: the intersection of two members of \(\mathcal{B}_{\mathcal{S}}\) is itself a member. The basis condition only asks for a basis member around each point of that intersection, but here the whole intersection works.
Worked Example: A Subbasis on a Three-Point Set
Let \(X=\{a,b,c\}\) and \(\mathcal{S}=\bigl\{\{a,b\},\{b,c\}\bigr\}\). The subbasis covers \(X\), since its two members have union \(\{a,b,c\}\).
The empty subfamily contributes \(X\). The two one-member subfamilies contribute \(\{a,b\}\) and \(\{b,c\}\), and the intersection of the two distinct members is \(\{b\}\). Thus \(\mathcal{B}_{\mathcal{S}}=\{X,\{a,b\},\{b,c\},\{b\}\}\). There are no other finite intersections: repeating a member does not change it, and intersecting the two distinct members gives \(\{b\}\).
The generated topology consists of unions of these four sets. The empty union is \(\varnothing\); the possible nonempty unions are \(\{b\}\), \(\{a,b\}\), \(\{b,c\}\), and \(X\). Therefore the topology is \(\{\varnothing,\{b\},\{a,b\},\{b,c\},X\}\).
Check: \(\{a,b\}\cup\{b,c\}=\{a,b,c\}=X\), while \(\{a,b\}\cap\{b,c\}=\{b\}\), as listed. Also, the singleton \(\{a\}\) is not among the generated open sets.
Unions of Finite Intersections Give a Topology
Proof. Call this family \(\mathcal{T}_{\mathcal{S}}\). The empty union is \(\varnothing\), and the empty-subfamily intersection \(X\) belongs to \(\mathcal{B}_{\mathcal{S}}\), so \(X\in\mathcal{T}_{\mathcal{S}}\). A union of any collection of members of \(\mathcal{T}_{\mathcal{S}}\) is again a union of members of \(\mathcal{B}_{\mathcal{S}}\), so it belongs to \(\mathcal{T}_{\mathcal{S}}\).
For finite intersections, write \(U=\bigcup_{i\in I}B_i\) and \(V=\bigcup_{j\in J}C_j\), where each \(B_i,C_j\in\mathcal{B}_{\mathcal{S}}\). The set identity \(U\cap V=\bigcup_{(i,j)\in I\times J}(B_i\cap C_j)\) expresses \(U\cap V\) as a union of members of \(\mathcal{B}_{\mathcal{S}}\), because that family is closed under intersections. If either union is empty, the intersection is empty, which is already in \(\mathcal{T}_{\mathcal{S}}\). Thus \(\mathcal{T}_{\mathcal{S}}\) is a topology.
Each \(S\in\mathcal{S}\) belongs to \(\mathcal{B}_{\mathcal{S}}\), as the intersection of the one-member subfamily \(\{S\}\). Hence \(\mathcal{T}_{\mathcal{S}}\) contains \(\mathcal{S}\). Finally, let \(\mathcal{T}\) be any topology on \(X\) containing \(\mathcal{S}\). Its finite-intersection axiom implies that every member of \(\mathcal{B}_{\mathcal{S}}\) belongs to \(\mathcal{T}\); this includes \(X\), which belongs to every topology. Its arbitrary-union axiom then implies that every member of \(\mathcal{T}_{\mathcal{S}}\) belongs to \(\mathcal{T}\). Therefore \(\mathcal{T}_{\mathcal{S}}\subseteq\mathcal{T}\), proving that \(\mathcal{T}_{\mathcal{S}}\) is the smallest topology containing \(\mathcal{S}\). \(\square\)
Worked Example: Two Subbasic Sets on the Real Line
Let \(X=\mathbb{R}\) as a set, and take \(\mathcal{S}=\{U,V\}\), where \(U=(-1,1)\) and \(V=\mathbb{R}\setminus\{0\}\). We are defining a topology from these subsets; the topology on this \(X\) is the one generated by \(\mathcal{S}\), not necessarily the standard topology on \(\mathbb{R}\).
The subbasis covers \(X\): \(0\in U\), and every nonzero real number belongs to \(V\). The empty subfamily gives \(\mathbb{R}\). The one-member subfamilies give \(U\) and \(V\). The two-member subfamily gives \(W=U\cap V=(-1,0)\cup(0,1)\). Thus the finite-intersection family is \(\{\mathbb{R},U,V,W\}\).
Taking all unions gives exactly \(\{\varnothing,W,U,V,\mathbb{R}\}\). Indeed, \(W\subseteq U\) and \(W\subseteq V\), while \(U\cup V=\mathbb{R}\). So a union of the four basis members can only be one of the five sets displayed.
Check: \(0\in U\) and \(0\notin V,W\); \(1\in V\) and \(1\notin U,W\); and \(1/2\in W\). These membership checks agree with the displayed intersections and unions.
Worked Example: Nested Intervals as a Subbasis
Let \(X=\mathbb{R}\), and let \(\mathcal{S}=\{\mathbb{R}\}\cup\{(-1/n,1/n):n\in\mathbb{N}\}\). This family covers \(X\) because it includes \(\mathbb{R}\). For a nonempty finite collection of the intervals, let \(N\) be the largest index used. The intervals are nested, so their intersection is \((-1/N,1/N)\). Intersecting with \(\mathbb{R}\) does not change a set. Consequently, \(\mathcal{B}_{\mathcal{S}}=\{\mathbb{R}\}\cup\{(-1/N,1/N):N\in\mathbb{N}\}\).
Now take any union of members of this basis. If it includes \(\mathbb{R}\), the union is \(\mathbb{R}\). Otherwise it is a union of intervals \((-1/N,1/N)\). The empty collection gives \(\varnothing\). If the indices used form a nonempty bounded subset of \(\mathbb{N}\), they have a largest index \(N\), and the union is \((-1/N,1/N)\). If the indices are unbounded, then for every \(x\) with \(-1<x<1\), there is an index \(N\) large enough that \(1/N>|x|\), so \(x\in(-1/N,1/N)\). No interval in the union contains a point outside \((-1,1)\). Thus the union is \((-1,1)\).
The generated topology is therefore \(\{\varnothing,\mathbb{R},(-1,1)\}\cup\{(-1/N,1/N):N\in\mathbb{N}\}\). For a numerical check, the intersection of \((-1,1)\) and \((-1/2,1/2)\) is \((-1/2,1/2)\), as the formula using the larger index \(N=2\) predicts.
A Common Mistake: Using an Infinite Intersection
In the nested-interval example, the calculation \(\bigcap_{n=1}^{\infty}(-1/n,1/n)=\{0\}\) is correct as a set calculation. But this intersection uses infinitely many subbasic sets. It is not one of the finite intersections that make up \(\mathcal{B}_{\mathcal{S}}\).
It would therefore be wrong to list \(\{0\}\) as a basis member on that basis. It is also not open in the generated topology: the complete list of open sets just computed contains no \(\{0\}\). More directly, every nonempty basis member containing \(0\) is either \(\mathbb{R}\) or an interval \((-1/N,1/N)\), each of which contains nonzero points. A union of such sets containing \(0\) cannot be \(\{0\}\).
Check Your Understanding
For each question, identify the finite intersections before describing the unions.
- On \(X=\{a,b,c,d\}\), let \(\mathcal{S}=\{\{a,b\},\{b,c\},\{c,d\}\}\). List the distinct finite intersections, including the empty-subfamily intersection.
- For the subbasis in question 1, is \(\{a,b,c\}\) open in the generated topology? Give it as a union of subbasic sets or explain why it cannot be such a union of finite intersections.
- Let \(X=\{1,2,3\}\) and \(\mathcal{S}=\{\{1,2\},\{2,3\}\}\). Why would omitting the empty intersection cause the proposed basis to fail to contain \(X\)?
- In the nested-interval example, what is \((-1/3,1/3)\cap(-1/5,1/5)\)? Explain why it belongs to the finite-intersection family.
- Can an arbitrary union of basis members be an infinite intersection of subbasic sets? Distinguish the operations in your explanation.