From Generating Families to Open Sets
A basis or subbasis is useful because it replaces a potentially complicated list of open sets with a smaller family of building blocks. To use one correctly, we must check its defining conditions and keep track of the operations that generate the topology. Basis elements are joined by arbitrary unions; subbasic elements are first combined by finite intersections, and those intersections are then joined by arbitrary unions.
Let \(X\) be a set. A family \(\mathcal{B}\subseteq\mathcal{P}(X)\) is a basis if it covers \(X\), and whenever \(x\in B_1\cap B_2\) for \(B_1,B_2\in\mathcal{B}\), there is \(B_3\in\mathcal{B}\) such that \(x\in B_3\subseteq B_1\cap B_2\). The second condition does not require \(B_1\cap B_2\) itself to belong to \(\mathcal{B}\); it requires a basis element around each point of the intersection.
Proof. Let \(\mathcal{T}_{\mathcal{B}}\) be the family of all unions of members of \(\mathcal{B}\). The empty union shows \(\varnothing\in\mathcal{T}_{\mathcal{B}}\), and the covering condition gives \(X=\bigcup_{B\in\mathcal{B}}B\in\mathcal{T}_{\mathcal{B}}\). An arbitrary union of unions of basis elements is again a union of basis elements, so \(\mathcal{T}_{\mathcal{B}}\) is closed under arbitrary unions.
It remains to check finite intersections. Suppose \(O_1,O_2\in\mathcal{T}_{\mathcal{B}}\). If \(x\in O_1\cap O_2\), there are \(B_1,B_2\in\mathcal{B}\) with \(x\in B_1\subseteq O_1\) and \(x\in B_2\subseteq O_2\). The basis condition gives \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\subseteq O_1\cap O_2\). Thus \(O_1\cap O_2\) is the union of all basis elements contained in it: every point belongs to one by this argument, and every such basis element lies in the intersection. Hence \(O_1\cap O_2\in\mathcal{T}_{\mathcal{B}}\). This proves closure under intersections of two sets, and repeated application proves closure under every finite intersection. Therefore \(\mathcal{T}_{\mathcal{B}}\) is a topology. \(\square\)
Worked Example: Checking a Basis on a Three-Point Set
Let \(X=\{a,b,c\}\) and \(\mathcal{B}=\{X,\{a,b\},\{b,c\},\{b\}\}\). Check that \(\mathcal{B}\) is a basis, then list the topology it generates.
The family covers \(X\), since \(X\in\mathcal{B}\). To check the intersection condition, intersections involving \(X\) leave the other set unchanged, and intersections of a set with itself also cause no difficulty. The remaining distinct pairs have intersections \(\{a,b\}\cap\{b,c\}=\{b\}\), \(\{a,b\}\cap\{b\}=\{b\}\), and \(\{b,c\}\cap\{b\}=\{b\}\). In each case the intersection \(\{b\}\) is itself in \(\mathcal{B}\), so it supplies the required basis element at every point of that intersection. Thus \(\mathcal{B}\) is a basis.
Its generated topology consists of all unions of these four sets. The possible unions are \(\varnothing,\{b\},\{a,b\},\{b,c\},X\). For instance, \(\{a,b\}\cup\{b,c\}=X\), and adjoining \(\{b\}\) to either of the two larger proper sets does not change it. These five sets are therefore exactly the generated topology.
Check: The two nontrivial proper sets intersect in \(\{b\}\), which is included in the list. The intersection of any listed set with \(X\) is itself, and intersections with \(\varnothing\) are empty. This confirms directly that the list is closed under finite intersections. It is also closed under arbitrary unions: unions of the listed sets produce only the same five sets.
Subbases: Intersect First, Then Take Unions
A subbasis \(\mathcal{S}\subseteq\mathcal{P}(X)\) generates a topology by a two-stage process. First take all finite intersections of members of \(\mathcal{S}\); then take arbitrary unions of the resulting sets. We include the intersection of no sets, defined to be \(X\). This convention ensures that the finite intersections cover \(X\), including when \(\mathcal{S}\) is empty.
Proof. Write \(\mathcal{B}_{\mathcal{S}}\) for the family of finite intersections of members of \(\mathcal{S}\), with \(X\) included as the empty intersection. It covers \(X\) because \(X\in\mathcal{B}_{\mathcal{S}}\). If \(P,Q\in\mathcal{B}_{\mathcal{S}}\), then \(P\cap Q\) is also a finite intersection of members of \(\mathcal{S}\): concatenate the two finite lists of sets being intersected. This also holds if one or both lists are empty. Thus, whenever \(x\in P\cap Q\), the set \(P\cap Q\) itself is a member of \(\mathcal{B}_{\mathcal{S}}\) containing \(x\) and contained in \(P\cap Q\). So \(\mathcal{B}_{\mathcal{S}}\) is a basis. The basis theorem now shows that its arbitrary unions form a topology. This topology contains every member of \(\mathcal{S}\), since each is a one-set finite intersection. Conversely, any topology containing \(\mathcal{S}\) contains all their finite intersections and arbitrary unions. Hence the topology just constructed is the smallest topology containing \(\mathcal{S}\), as claimed. \(\square\)
Worked Example: Generating Open Sets from Rays
On \(\mathbb{R}\) with its standard topology, let \(\mathcal{S}=\{(-\infty,1),(-\infty,3),(0,\infty),(2,\infty)\}\). Identify the possible finite intersections and construct two open sets in the generated topology.
Consider first an intersection involving only left rays. If it uses no left rays, the intersection contributes no restriction and is \(\mathbb{R}\). If it uses one or more left rays, write their endpoints as \(a_1,\ldots,a_m\). Membership requires \(x<a_i\) for every \(i\), which is equivalent to \(x<\min(a_1,\ldots,a_m)\). The intersection is therefore the left ray with the smallest endpoint. For this \(\mathcal{S}\), the possible nonempty left-ray results are \((-\infty,1)\) and \((-\infty,3)\).
For right rays, no selected right rays again means no restriction, giving \(\mathbb{R}\). If the selected right-ray endpoints are \(b_1,\ldots,b_n\), membership requires \(x>b_i\) for every \(i\), or equivalently \(x>\max(b_1,\ldots,b_n)\). The possible right-ray results are \((0,\infty)\) and \((2,\infty)\). In particular, the intersection of no rays is \(\mathbb{R}\), not \(\varnothing\).
If both types of ray occur, let \(a\) be the smallest selected left endpoint and \(b\) the largest selected right endpoint. The intersection is \((b,a)\) when \(b<a\), and is empty when \(b\ge a\). The possible nonempty bounded intervals here are \((0,1)\), \((0,3)\), and \((2,3)\): these come from endpoint pairs \((b,a)=(0,1),(0,3),(2,3)\), respectively. The remaining pair \((2,1)\) gives the empty set because \(2\ge1\).
Each finite intersection is a basis element for the generated topology, so arbitrary unions of these intersections are open. For example, \((0,1)=(-\infty,1)\cap(0,\infty)\) and \((2,3)=(-\infty,3)\cap(2,\infty)\). Consequently \(U=(0,1)\cup(2,3)\) is open in the generated topology. Also \((-\infty,1)\cup(2,\infty)\) is open, since it is a union of two subbasic sets.
Check: The endpoints are excluded in every ray and interval. Thus \(1\notin(0,1)\) and \(2\notin(2,3)\), while \(1/2\in(0,1)\) because \(0<1/2<1\), and \(5/2\in(2,3)\) because \(2<5/2<3\). These tests confirm that the constructed union has the stated membership conditions.
Coordinate Strips and Open Rectangles
Worked Example: Building Open Sets in the Plane
Give \(\mathbb{R}^2\) the topology generated by the subbasis \(\mathcal{S}=\{U\times\mathbb{R},\mathbb{R}\times V: U,V\text{ are open in }\mathbb{R}\}\), where \(\mathbb{R}\) has its standard topology. Show how to construct a particular open set from subbasic sets.
A horizontal-coordinate restriction \(U\times\mathbb{R}\) and a vertical-coordinate restriction \(\mathbb{R}\times V\) have intersection \[ (U\times\mathbb{R})\cap(\mathbb{R}\times V)=U\times V. \] Indeed, \((x,y)\) is in the left side exactly when \(x\in U\) and \(y\in V\), which is precisely membership in \(U\times V\). Thus every open rectangle \(U\times V\) is a finite intersection of subbasic sets.
For explicit intervals, take \(U_1=(-1,2)\), \(V_1=(0,3)\), \(U_2=(4,5)\), and \(V_2=(-2,1)\). Each is open in the standard topology of \(\mathbb{R}\). The set \[ W=\bigl((-1,2)\times(0,3)\bigr)\cup\bigl((4,5)\times(-2,1)\bigr) \] is open in the generated topology, because its first rectangle is the intersection of \((-1,2)\times\mathbb{R}\) and \(\mathbb{R}\times(0,3)\), its second is the intersection of \((4,5)\times\mathbb{R}\) and \(\mathbb{R}\times(-2,1)\), and \(W\) is the union of those two finite intersections.
Check: The point \((0,1)\) belongs to the first rectangle because \(-1<0<2\) and \(0<1<3\). The point \((4.5,0)\) belongs to the second because \(4<4.5<5\) and \(-2<0<1\). The point \((2,1)\) belongs to neither rectangle: its first coordinate is not in \((-1,2)\), and it is not in \((4,5)\). These membership checks agree with the strict endpoint conditions defining the intervals.
A Common Construction Error
A subbasis is not generally itself the topology it generates. For a concrete check, take \(X=\{a,b,c\}\) and \(\mathcal{S}=\{\{a,b\},\{b,c\}\}\). The intersection of the two subbasic sets is \(\{b\}\). Therefore the generated topology includes \(\{b\}\), as well as the empty intersection \(X\), the two subbasic sets, and their unions. The full topology is \(\{\varnothing,\{b\},\{a,b\},\{b,c\},X\}\).
Common pitfall. A wrong calculation is to list only \(\varnothing,X,\{a,b\},\{b,c\}\), on the grounds that these are the subbasic sets together with the empty set and \(X\). This list is not a topology: it contains \(\{a,b\}\) and \(\{b,c\}\) but omits their intersection \(\{b\}\), so it is not closed under finite intersections. The correction is to calculate finite intersections first, then take arbitrary unions. This example shows why the two stages in the subbasis construction cannot be skipped.
When working with a proposed generating family, keep the relevant operations visible. For a basis, test covering and the local refinement condition, then take unions. For a subbasis, include the empty intersection \(X\), calculate finite intersections, and then take unions. In either case, a set is open in the generated topology only after it has been expressed by the permitted operations.
Check Your Understanding
For each question, state which generating operation is being used and check any endpoint conditions.
- On \(X=\{p,q,r\}\), decide whether \(\{X,\{p,q\},\{q,r\}\}\) is a basis. If so, list its generated topology.
- Let \(\mathcal{S}=\{(-\infty,0),(1,\infty)\}\) on the standard real line. List all finite intersections, including the empty intersection.
- For the coordinate-strip subbasis on \(\mathbb{R}^2\), express \((1,4)\times(-3,2)\) as an intersection of subbasic sets.
- Explain why the family \(\{\varnothing,X,\{a,b\},\{b,c\}\}\) on \(\{a,b,c\}\) is not a topology.
- State the two conditions a family must satisfy to be a basis, and explain why its intersection condition is local to points of the intersection.