Why Two Local Conditions Characterize a Basis
A basis is a collection of sets that can be combined by unions to describe the open sets of a topology. Two requirements make this work: the collection must reach every point of the space, and wherever two of its sets overlap, there must be a basis set around each point of overlap that stays inside the intersection. The second requirement is local: it does not demand that the intersection itself belong to the collection.
Throughout, \(X\) is a set and \(\mathcal{B}\subseteq\mathcal{P}(X)\) is a collection of subsets. If \(\mathcal{B}\) is to be a basis for a particular topology \(\mathcal{T}\), every member of \(\mathcal{B}\) must be open in \(\mathcal{T}\). The criterion below first asks when \(\mathcal{B}\) is a basis for some topology on \(X\), the topology whose open sets are unions of members of \(\mathcal{B}\).
- \(\bigcup_{B\in\mathcal{B}}B=X\); and
- whenever \(B_1,B_2\in\mathcal{B}\) and \(x\in B_1\cap B_2\), there is a \(B_3\in\mathcal{B}\) such that \(x\in B_3\subseteq B_1\cap B_2\).
The first condition says that each point lies in at least one basis set. In the second, the quantifiers matter: for every pair \(B_1,B_2\), and for each point \(x\) in their intersection, a suitable \(B_3\) must exist. The set \(B_3\) may depend on \(x\). It need not equal \(B_1\cap B_2\).
Why a Basis Must Satisfy the Conditions
Suppose first that \(\mathcal{B}\) is a basis for a topology \(\mathcal{T}\). Since \(X\) is open, it is a union of members of \(\mathcal{B}\). That union must contain every point of \(X\), so the basis covers \(X\).
Now take \(B_1,B_2\in\mathcal{B}\) and \(x\in B_1\cap B_2\). Both basis sets are open, so their intersection is open. Because \(\mathcal{B}\) is a basis, \(B_1\cap B_2\) is a union of members of \(\mathcal{B}\). Since \(x\) belongs to that union, at least one of those members contains \(x\). Call it \(B_3\). As one of the sets in the union representing \(B_1\cap B_2\), it satisfies \(B_3\subseteq B_1\cap B_2\). This proves the pointwise condition. The next proof shows that these necessary conditions are also sufficient.
Why the Two Conditions Are Sufficient
The empty union is \(\varnothing\), so \(\varnothing\in\mathcal{T}_{\mathcal{B}}\). The covering condition gives \(X=\bigcup_{B\in\mathcal{B}}B\), so \(X\in\mathcal{T}_{\mathcal{B}}\).
A union of unions of members of \(\mathcal{B}\) is itself a union of members of \(\mathcal{B}\). Therefore an arbitrary union of sets in \(\mathcal{T}_{\mathcal{B}}\) belongs to \(\mathcal{T}_{\mathcal{B}}\).
Let \(U,V\in\mathcal{T}_{\mathcal{B}}\). Write \(U=\bigcup_{i\in I}B_i\) and \(V=\bigcup_{j\in J}C_j\), where \(B_i,C_j\in\mathcal{B}\). If \(x\in U\cap V\), there are indices \(i\in I\) and \(j\in J\) with \(x\in B_i\cap C_j\). By the pointwise condition, there is \(D\in\mathcal{B}\) with \(x\in D\subseteq B_i\cap C_j\subseteq U\cap V\). Thus every point of \(U\cap V\) belongs to a basis set contained in \(U\cap V\). Conversely, each basis set contained in \(U\cap V\) is a subset of \(U\cap V\). Hence \(U\cap V\) is the union of all basis sets contained in it, and so belongs to \(\mathcal{T}_{\mathcal{B}}\). This argument also covers \(U\cap V=\varnothing\), which is already in \(\mathcal{T}_{\mathcal{B}}\).
These statements are precisely the topology axioms, so \(\mathcal{T}_{\mathcal{B}}\) is a topology. Each \(B\in\mathcal{B}\) belongs to \(\mathcal{T}_{\mathcal{B}}\), since it is the union consisting of that one member. By definition, every set in \(\mathcal{T}_{\mathcal{B}}\) is a union of members of \(\mathcal{B}\). Therefore \(\mathcal{B}\) is a basis for this topology. If \(X=\varnothing\), the same argument applies: the covering condition holds, and the resulting topology is \(\{\varnothing\}\). \(\square\)
Using the Criterion for a Specified Topology
The two conditions characterize when a collection is a basis for the topology it determines. If a topology \(\mathcal{T}\) has already been specified, those two conditions alone do not guarantee that \(\mathcal{B}\) is a basis for \(\mathcal{T}\). In that setting, one must also ensure that the members of \(\mathcal{B}\) are open in \(\mathcal{T}\), and that they can describe every open set of \(\mathcal{T}\).
Proof. If \(\mathcal{B}\) is a basis for \(\mathcal{T}\), then \(X\) and every \(U\in\mathcal{T}\) are unions of members of \(\mathcal{B}\). Thus the members cover \(X\), and each \(x\in U\) belongs to one of the basis members in a union representing \(U\); that member is contained in \(U\). Conversely, suppose the stated conditions hold. For each \(U\in\mathcal{T}\), take the union of all \(B\in\mathcal{B}\) with \(B\subseteq U\). Every such \(B\) lies in \(U\), so this union is contained in \(U\). Every \(x\in U\) lies in at least one such \(B\) by the pointwise condition, so the union contains \(U\). Therefore \(U\) is a union of members of \(\mathcal{B}\). Since \(\mathcal{B}\subseteq\mathcal{T}\), these are open sets, and \(\mathcal{B}\) is a basis for \(\mathcal{T}\). \(\square\)
Worked Examples of the Conditions
Worked Example: A Basis from a Partition
Let \(X=\{a,b,c\}\) and \(\mathcal{B}=\{\{a,b\},\{c\}\}\). Check the criterion and describe the topology whose basis is \(\mathcal{B}\).
The union of the two members is \(\{a,b\}\cup\{c\}=X\), so the covering condition holds. The possible intersections of members are \(\{a,b\}\cap\{a,b\}=\{a,b\}\), \(\{c\}\cap\{c\}=\{c\}\), and \(\{a,b\}\cap\{c\}=\varnothing\). For a point in either nonempty intersection, the corresponding basis member itself contains the point and is contained in that intersection. The empty intersection has no points to check. Thus the pointwise condition holds.
The unions of subcollections are \(\varnothing\), \(\{a,b\}\), \(\{c\}\), and \(\{a,b\}\cup\{c\}=X\). Hence the resulting topology is \(\{\varnothing,\{a,b\},\{c\},X\}\). As a check, the intersection of the two nontrivial proper open sets is \(\{a,b\}\cap\{c\}=\varnothing\), which is in the list; their union is \(X\), also in the list.
Worked Example: An Overlap with a Smaller Basis Set
Let \(X=\{a,b,c\}\) and \(\mathcal{B}=\{X,\{a,b\},\{b,c\},\{b\}\}\). Here \(\{a,b\}\cap\{b,c\}=\{b\}\), and the intersection itself is a basis member. Check the other requirements and find all unions.
The collection covers \(X\), because \(X\in\mathcal{B}\). For the pair \(\{a,b\},\{b,c\}\), the only point of the intersection is \(b\), and the member \(\{b\}\) contains \(b\) and lies in the intersection. For any pair involving \(X\), the intersection is the other member, so that member works for each of its points. Intersections of a member with itself are that member, and \(\{b\}\) is contained in each of the other members that contains \(b\). These cases check every pair, so the criterion holds.
The distinct unions are \(\varnothing\), \(\{b\}\), \(\{a,b\}\), \(\{b,c\}\), and \(X\). For example, \(\{a,b\}\cup\{b,c\}=X\), while \(\{a,b\}\cup\{b\}=\{a,b\}\). The resulting topology is therefore \(\{\varnothing,\{b\},\{a,b\},\{b,c\},X\}\). In particular, the condition does not require every pairwise intersection to be a listed basis member in general; it requires a suitable member at each point.
Worked Example: Nonempty Intersections Are Not Enough
Let \(X=\{a,b,c\}\) and \(\mathcal{B}=\{X,\{a,b\},\{b,c\}\}\). The collection covers \(X\). Does it satisfy the basis criterion?
The intersection \(\{a,b\}\cap\{b,c\}\) is \(\{b\}\). At \(b\), the criterion asks for some \(D\in\mathcal{B}\) such that \(b\in D\subseteq\{b\}\). The only subset of \(\{b\}\) containing \(b\) is \(\{b\}\), but \(\{b\}\notin\mathcal{B}\). Thus the pointwise condition fails.
One can also see the obstruction directly. The distinct unions of members of \(\mathcal{B}\) are \(\varnothing,\{a,b\},\{b,c\},X\). The intersection of the two sets \(\{a,b\}\) and \(\{b,c\}\) is \(\{b\}\), which is absent from this collection of unions. That collection is not a topology, since a topology must be closed under intersections of two open sets.
Worked Example: Why Coverage Is Necessary
Let \(X=\{a,b\}\) and \(\mathcal{B}=\{\{a\}\}\). The intersection condition holds: the only pair of basis members has intersection \(\{a\}\), and \(\{a\}\) itself works at its only point. But \(\bigcup_{B\in\mathcal{B}}B=\{a\}\ne X\), so the covering condition fails.
The only unions of members are \(\varnothing\) and \(\{a\}\). In particular, \(X\) is not a union of members of \(\mathcal{B}\), so \(\mathcal{B}\) cannot be a basis for a topology on \(X\). This example shows why the intersection condition cannot replace coverage.
A Common Pitfall and a Practical Checklist
Common pitfall. A tempting but incorrect test is: “Whenever two basis sets overlap, their intersection must be nonempty.” That condition is not enough. In the third example, the intersection is nonempty, but there is no basis member containing \(b\) and lying inside it. The correction is to check the required inclusion point by point: for each \(x\in B_1\cap B_2\), find \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\). Merely knowing that \(B_1\cap B_2\ne\varnothing\), or that some basis set meets the intersection, does not establish this.
When applying the criterion, keep the two checks separate. First ask whether every point of \(X\) is covered. Then take an arbitrary pair of members and examine each point of their intersection. If an intersection is empty, there are no points for the second check; if it is nonempty, the required smaller member must exist around every point in it. These quantifiers are what make the criterion both precise and useful.
Check Your Understanding
For each question, distinguish the covering requirement from the pointwise intersection requirement.
- State the basis criterion for a collection of subsets of \(X\).
- Why does a basis for a topology necessarily cover \(X\)?
- For \(X=\{a,b,c\}\) and \(\mathcal{B}=\{\{a,b\},\{b,c\},\{b\}\}\), does \(\mathcal{B}\) satisfy the two conditions? Explain.
- Can the intersection condition hold while the collection fails to be a basis? Give an example and identify the failed condition.
- What additional pointwise requirement characterizes a basis for an already specified topology \(\mathcal{T}\)?