From Local Building Blocks to a Topology
A basis is a collection of subsets designed to build open sets locally: points are covered by basis elements, and overlapping basis elements can be refined around each point of their overlap. The natural construction is to declare every union of basis elements to be open. We will check that this collection really is a topology, including the finite-intersection axiom, where the basis intersection condition does the essential work.
Let \(X\) be a set. A family \(\mathcal{B}\subseteq\mathcal{P}(X)\) is a basis for \(X\) if (1) every point of \(X\) belongs to at least one member of \(\mathcal{B}\), and (2) whenever \(B_1,B_2\in\mathcal{B}\) and \(x\in B_1\cap B_2\), there is \(B_3\in\mathcal{B}\) such that \(x\in B_3\subseteq B_1\cap B_2\). These conditions allow \(X=\varnothing\), in which case the empty family is a basis.
Equivalently, \(U\in\mathcal{T}_{\mathcal{B}}\) means there is a subfamily \(\mathcal{C}\subseteq\mathcal{B}\) with \(U=\bigcup_{B\in\mathcal{C}}B\). The subfamily may be empty, finite, or infinite. In particular, the definition does not require a basis element itself to be a union of just one other, distinct basis element.
Why the Unions Satisfy the Topology Axioms
Proof. We verify the topology axioms. First, \(\varnothing\in\mathcal{T}_{\mathcal{B}}\), since it is the union of the empty subfamily. The basis covers \(X\), so \(X=\bigcup_{B\in\mathcal{B}}B\) and \(X\in\mathcal{T}_{\mathcal{B}}\).
Next let \((U_\lambda)_{\lambda\in\Lambda}\) be any family of members of \(\mathcal{T}_{\mathcal{B}}\), where \(\Lambda\) may be empty. For each \(\lambda\), choose a subfamily \(\mathcal{C}_\lambda\subseteq\mathcal{B}\) such that \(U_\lambda=\bigcup_{B\in\mathcal{C}_\lambda}B\). Then \(\bigcup_{\lambda\in\Lambda}U_\lambda=\bigcup_{\lambda\in\Lambda}\bigcup_{B\in\mathcal{C}_\lambda}B\), which is a union of basis elements. If \(\Lambda\) is empty, this is the empty union, already shown to belong to \(\mathcal{T}_{\mathcal{B}}\). Thus arbitrary unions belong to \(\mathcal{T}_{\mathcal{B}}\).
For finite intersections it suffices first to consider two sets. Write \(U=\bigcup_{B\in\mathcal{C}}B\) and \(V=\bigcup_{D\in\mathcal{D}}D\), where \(\mathcal{C},\mathcal{D}\subseteq\mathcal{B}\). A point \(x\) lies in \(U\cap V\) exactly when there are \(B\in\mathcal{C}\) and \(D\in\mathcal{D}\) with \(x\in B\cap D\). By the basis intersection condition, for that point there is a basis element \(E\) with \(x\in E\subseteq B\cap D\). Consequently, \(U\cap V=\bigcup\{E\in\mathcal{B}: E\subseteq B\cap D\text{ for some }B\in\mathcal{C},D\in\mathcal{D}\}\). Indeed, each such \(E\) is contained in \(U\cap V\), and the basis condition supplies one of these \(E\)'s for every \(x\in U\cap V\). Hence \(U\cap V\in\mathcal{T}_{\mathcal{B}}\). Repeated application gives closure under every nonempty finite intersection; the intersection of no sets is \(X\), which is already in \(\mathcal{T}_{\mathcal{B}}\). All topology axioms hold. \(\square\)
The proof identifies the role of the basis condition precisely. Intersections of unions reduce to intersections of pairs of basis elements. Those pairwise intersections need not themselves be basis elements, but every point in each intersection must lie in a basis element contained in it. That is enough to express the whole intersection as a union of basis elements.
Worked Examples
Worked Example: Open Intervals on the Real Line
Let \(X=\mathbb{R}\) with its standard topology, and let \(\mathcal{B}\) be the family of all open intervals \((a,b)\) with \(a<b\). We verify the basis conditions and identify the generated topology.
For any \(x\in\mathbb{R}\), the interval \((x-1,x+1)\) contains \(x\), so the family covers \(\mathbb{R}\). Now suppose \(x\in(a,b)\cap(c,d)\). Then \(x\) is strictly greater than both \(a,c\) and strictly less than both \(b,d\). Thus \(\max(a,c)<x<\min(b,d)\). Choose \(\varepsilon=\frac12\min(x-\max(a,c),\min(b,d)-x)\). Both quantities in the minimum are positive, so \(\varepsilon>0\); the interval \((x-\varepsilon,x+\varepsilon)\) contains \(x\) and is contained in \((a,b)\cap(c,d)\). This verifies the intersection condition.
It follows that the unions of open intervals form a topology. In this example those unions are exactly the open sets of the standard topology on \(\mathbb{R}\): a union of open intervals is open, and each standard open set is, by the definition of openness on \(\mathbb{R}\), a union of open intervals. Thus the construction recovers the standard topology.
Check: if two intervals are disjoint, their intersection is empty and contributes no points to the union of refining basis elements. If they overlap, the calculation above supplies a basis interval around each point of the overlap. Both cases are covered by the intersection argument.
Worked Example: A Basis That Is Not Closed Under Intersections
Let \(X=\{1,2,3,4\}\) and \(\mathcal{B}=\{X,\{1,2,3\},\{2,3,4\},\{2\},\{3\}\}\). We check the basis conditions and list every set in the generated topology.
The sets \(\{1,2,3\}\) and \(\{2,3,4\}\) together cover \(X\); hence the family covers \(X\). Their intersection is \(\{2,3\}\). At the point \(2\), the basis element \(\{2\}\) is contained in that intersection; at the point \(3\), the basis element \(\{3\}\) is contained in it. All other intersections of basis elements are either one of the sets already listed, empty, or obtained by intersecting with \(X\). For each point in those intersections, the intersection itself is a basis element, or one of \(\{2\},\{3\}\) lies inside it. So the basis intersection condition holds.
Every union of basis elements is one of \(\varnothing,\{2\},\{3\},\{2,3\},\{1,2,3\},\{2,3,4\},X\). To check that the list is complete, a union containing \(\{1,2,3\}\) but not \(\{2,3,4\}\) is either \(\{1,2,3\}\) or \(X\); a union containing \(\{2,3,4\}\) but not \(\{1,2,3\}\) is either \(\{2,3,4\}\) or \(X\). If it contains neither, it is one of \(\varnothing,\{2\},\{3\},\{2,3\}\). If it contains both, it is \(X\).
Check: the listed collection contains \(\varnothing\) and \(X\). Unions of its members remain in the list: the only two listed proper sets whose union introduces both \(1\) and \(4\) are \(\{1,2,3\}\) and \(\{2,3,4\}\), and their union is \(X\); all other unions either stay among the listed sets or also equal \(X\). Intersections of the two larger proper sets equal \(\{2,3\}\), and their intersections with the smaller listed sets are among \(\varnothing,\{2\},\{3\}\). The other intersections remain listed as well. This is consistent with the theorem.
In particular, \(\{1,2,3\}\cap\{2,3,4\}=\{2,3\}\) is not a member of \(\mathcal{B}\), even though it is a union of basis elements. A basis need not be closed under intersections.
Worked Example: The Discrete and Indiscrete Topologies
First take a set \(X\) and the basis \(\mathcal{B}=\{\{x\}:x\in X\}\). If \(X\ne\varnothing\), every point belongs to its singleton, and two singleton basis elements that share a point must be the same singleton. The intersection condition therefore holds. Every subset \(A\subseteq X\) is the union of the singletons \(\{x\}\) for \(x\in A\); when \(A=\varnothing\), it is the empty union. Thus \(\mathcal{T}_{\mathcal{B}}=\mathcal{P}(X)\), the discrete topology. If \(X=\varnothing\), the basis is empty and its only union is \(\varnothing=X\), so the same conclusion holds.
Now suppose \(X\ne\varnothing\) and take \(\mathcal{B}=\{X\}\). It covers \(X\), and the intersection of its only basis element with itself is \(X\), satisfying the basis condition. Its unions are just \(\varnothing\) and \(X\), so it generates the indiscrete topology. The condition \(X\ne\varnothing\) matters: if \(X=\varnothing\), then \(\{X\}=\{\varnothing\}\) also generates \(\{\varnothing\}=\{\varnothing,X\}\), the topology on the empty set.
The Smallest Topology Containing the Basis
Proof. Each \(B\in\mathcal{B}\) belongs to \(\mathcal{T}_{\mathcal{B}}\), since \(B\) is the union of the one-element subfamily \(\{B\}\). Now let \(\mathcal{T}\) be any topology on \(X\) with \(\mathcal{B}\subseteq\mathcal{T}\). Every member of \(\mathcal{T}_{\mathcal{B}}\) is a union of members of \(\mathcal{B}\), and therefore is a union of members of \(\mathcal{T}\). Since a topology is closed under arbitrary unions, every such union belongs to \(\mathcal{T}\). Hence \(\mathcal{T}_{\mathcal{B}}\subseteq\mathcal{T}\), as required. \(\square\)
This minimality result explains the word “generated”: the basis elements are required to be open, and the topology axioms force their arbitrary unions to be open. The construction adds exactly those unions, with no extra sets. The basis intersection condition ensures that this collection is already closed under finite intersections, so no further sets need to be added to meet the topology axioms.
Common pitfall. A tempting but incorrect argument is that the topology axioms hold because intersections of basis elements are basis elements. The finite example above disproves that claim: \(\{1,2,3\}\cap\{2,3,4\}=\{2,3\}\notin\mathcal{B}\). The correct argument is pointwise: \(\{2\}\) refines the intersection at \(2\), and \(\{3\}\) refines it at \(3\), so the intersection is the union \(\{2\}\cup\{3\}\) of basis elements. The basis criterion requires refinement around every point of an overlap, not closure of the basis under intersections.
Check Your Understanding
Use the definition of a basis and the union construction in each question.
- Why is \(\varnothing\) in every topology generated by a basis, even when the basis contains no empty set?
- In the finite example, express \(\{2,3\}\) as a union of basis elements and explain why it is not itself a basis element.
- For \(X=\{p,q\}\), determine the topology generated by \(\{\{p\},X\}\). Verify the basis intersection condition.
- Which step in the proof that \(\mathcal{T}_{\mathcal{B}}\) is a topology uses the basis intersection condition?
- If \(\mathcal{B}\) is a basis and \(\mathcal{T}\) is a topology containing \(\mathcal{B}\), explain why every union of basis elements must belong to \(\mathcal{T}\).