Why These Errors Matter
A basis is a concise way to describe a topology, but a short description is useful only if its conditions are checked correctly. Two separate requirements govern a basis: its members must cover the space, and intersections must have enough basis members to refine them around each point. A subbasis works differently: its finite intersections form the basis, and arbitrary unions of those intersections form the topology. Confusing these steps can produce a collection that does not describe the topology claimed.
Throughout, \((X,\mathcal{T})\) denotes a topological space. A basis for \(\mathcal{T}\) is a collection \(\mathcal{B}\) of open sets such that every open set is a union of members of \(\mathcal{B}\). A subbasis is a collection \(\mathcal{S}\) of open sets whose finite intersections form a basis. The empty union is \(\varnothing\), and the intersection of no sets is \(X\).
Error 1: Forgetting That Basis Members Must Cover
It is not enough for a collection to have a well-behaved intersection pattern. Every point of \(X\) must belong to at least one basis member. In the usual basis criterion, this covering requirement is written \(\bigcup_{B\in\mathcal{B}}B=X\).
Worked Example: A Collection That Misses a Point
Take \(X=\{a,b\}\) and \(\mathcal{B}=\{\{a\}\}\). The only basis member is \(\{a\}\), so the union of all members is \(\{a\}\), not \(X\). Thus \(\mathcal{B}\) does not cover \(X\) and cannot be a basis for a topology on \(X\).
The intersection-refinement condition alone does not catch the problem: the only pair to test is \(\{a\}\) with itself. Their intersection is \(\{a\}\), and at its point \(a\), the member \(\{a\}\) is a suitable refinement. But the collection of all unions of members of \(\mathcal{B}\) is just \(\{\varnothing,\{a\}\}\). It lacks \(X\), so it is not a topology on \(X\).
Check: The missing point is \(b\), and no member of \(\mathcal{B}\) contains \(b\). This directly contradicts the covering requirement.
For \(X=\varnothing\), the empty collection does cover \(X\), because its union is \(\varnothing\); it can be a basis for the unique topology on the empty space. Thus the covering condition also handles the empty-space case without requiring a basis member to exist.
Error 2: Checking Intersections Without Refining at Each Point
The second condition is more specific than saying that two basis members have a nonempty intersection. Whenever \(x\in B_1\cap B_2\), there must be some \(B_3\in\mathcal{B}\) with \(x\in B_3\subseteq B_1\cap B_2\). The refinement may depend on the point \(x\). Merely observing that \(B_1\cap B_2\ne\varnothing\) does not supply such a member.
Worked Example: A Nonempty Intersection Is Not Enough
On \(X=\mathbb{R}\) with its standard topology, let \(\mathcal{B}=\{\mathbb{R},(-1,1),(0,2)\}\). These sets cover \(\mathbb{R}\), because \(\mathbb{R}\in\mathcal{B}\). The two bounded members intersect in \((0,1)\), which is nonempty.
A mistaken argument might say that the intersection condition is satisfied because \((0,1)\ne\varnothing\). Test instead the point \(x=\tfrac12\). The members of \(\mathcal{B}\) containing \(x\) are \(\mathbb{R}\), \((-1,1)\), and \((0,2)\). None is contained in \((0,1)\): the first contains points outside \((0,1)\), the second contains negative points, and the third contains points greater than \(1\). There is no required refinement at \(x\), so \(\mathcal{B}\) is not a basis.
Indeed, unions of members of \(\mathcal{B}\) cannot produce \((0,1)\): any union containing \(\mathbb{R}\) is \(\mathbb{R}\), and unions using only the two bounded members are \(\varnothing\), one of those members, or their union \((-1,2)\). The intersection of the open sets \((-1,1)\) and \((0,2)\) is \((0,1)\), so this collection of unions is not a topology.
Check: The failure is pointwise: at \(x=\tfrac12\), no member of \(\mathcal{B}\) contains \(x\) and lies inside the intersection.
Proof. First suppose the two conditions hold, and let \(\mathcal{T}_{\mathcal{B}}\) be the collection of all unions of members of \(\mathcal{B}\). It contains \(\varnothing\) as the empty union and \(X\) by the covering condition. An arbitrary union of members of \(\mathcal{T}_{\mathcal{B}}\) is still a union of members of \(\mathcal{B}\). For finite intersections, write two such unions as \(U=\bigcup_{i\in I}B_i\) and \(V=\bigcup_{j\in J}C_j\). A point \(x\in U\cap V\) lies in some \(B_i\cap C_j\). The refinement condition gives a basis member containing \(x\) and contained in \(B_i\cap C_j\), hence contained in \(U\cap V\). Taking all such refinements shows that \(U\cap V\) is a union of basis members. Thus \(\mathcal{T}_{\mathcal{B}}\) is a topology containing \(\mathcal{B}\). Any topology containing \(\mathcal{B}\) contains every union of its members, so \(\mathcal{T}_{\mathcal{B}}\) is the smallest such topology. Conversely, if \(\mathcal{B}\) is a basis, \(X\) is a union of basis members. If \(x\in B_1\cap B_2\), that intersection is open, so its expression as a union of basis members includes one containing \(x\); that member is the required refinement. \(\square\)
Error 3: Treating Every Open Set as One Basis Member
A basis member is open, but an open set need not itself be a basis member. The definition permits an open set to be a union of many basis members. The basis criterion’s refinement condition is local: it guarantees a member around each point of an intersection, not that the entire intersection is one member.
Worked Example: Rational Intervals on the Real Line
Let \(X=\mathbb{R}\) with its standard topology and let \(\mathcal{B}=\{(p,q):p,q\in\mathbb{Q},\ p<q\}\). Consider the open set \(U=(0,\sqrt{2})\). It is not a member of \(\mathcal{B}\), because its right endpoint \(\sqrt{2}\) is irrational.
Nevertheless, \(U\) is a union of members of \(\mathcal{B}\): take all rational-endpoint intervals \((p,q)\) with \(0<p<q<\sqrt{2}\). Each such interval lies in \(U\). Conversely, for any \(x\in U\), density of \(\mathbb{Q}\) in \(\mathbb{R}\) gives rationals \(p,q\) with \(0<p<x<q<\sqrt{2}\). Then \(x\in(p,q)\subseteq U\). Therefore the union of those basis members is exactly \(U\).
Check: Every point of \(U\) lies in one of the stated intervals, and no such interval contains a point outside \(U\). Thus \(U\) is open even though it is not a single basis member.
The same distinction matters when using the basis criterion: for an intersection \(B_1\cap B_2\), it is enough to find a suitable basis member around each point in the intersection. Demanding that the whole intersection itself belong to \(\mathcal{B}\) is unnecessarily strong and is not part of the criterion.
Error 4: Omitting Finite Intersections of Subbasic Sets
A subbasis is not generally a basis as it stands. To use a subbasis \(\mathcal{S}\), first form all finite intersections of its members. This collection includes \(X\), the intersection of no subbasic sets. The topology generated by \(\mathcal{S}\) consists of unions of these finite intersections.
Proof. The collection \(\mathcal{B}_{\mathcal{S}}\) covers \(X\), since \(X\) itself is one of its members. The intersection of two finite intersections is again a finite intersection of members of \(\mathcal{S}\), so it belongs to \(\mathcal{B}_{\mathcal{S}}\). In particular, for every point of that intersection, this same member is a refinement. The basis criterion therefore shows that unions of members of \(\mathcal{B}_{\mathcal{S}}\) form a topology. This topology contains every \(S\in\mathcal{S}\), since \(S\) is a finite intersection with one factor. Any topology containing \(\mathcal{S}\) contains its finite intersections and their unions, so the constructed topology is the smallest one containing \(\mathcal{S}\). \(\square\)
Worked Example: Two Open Rays Produce an Open Interval
In \(\mathbb{R}\) with its standard topology, take the subbasis \(\mathcal{S}=\{(-\infty,1),(0,\infty)\}\). Its finite intersections include \(X=\mathbb{R}\), each of the two rays, and their intersection \((-\infty,1)\cap(0,\infty)=(0,1)\). The empty set is also a union, namely the empty union.
Consequently \((0,1)\) belongs to the topology generated by \(\mathcal{S}\), even though it is not itself one of the two subbasic sets. A mistaken calculation that lists only the members of \(\mathcal{S}\) misses this open set. The correction is to form finite intersections first and then take arbitrary unions.
Check: A real number lies in the intersection exactly when it is greater than \(0\) and less than \(1\), so the intersection is precisely \((0,1)\), with both endpoints excluded.
A Reliable Checklist
When checking a basis or subbasis argument, write down the construction before drawing a conclusion. For a proposed basis, verify coverage and test intersections at an arbitrary point \(x\) in the intersection. For a proposed subbasis, list the finite intersections before taking unions. These checks prevent a nonempty intersection from being mistaken for a refinement, and prevent a generating family from being mistaken for the full collection of open sets.
Common pitfall. For the collection \(\{\mathbb{R},(-1,1),(0,2)\}\), the claim “the two bounded members intersect, so the collection is a basis” is wrong: at \(x=\tfrac12\), no member lies inside their intersection \((0,1)\) while containing \(x\). The corrected test checks the required containment point by point. In a subbasis argument, the analogous correction is to include intersections such as \((-\infty,1)\cap(0,\infty)\), not just the original rays.
Check Your Understanding
For each question, identify which condition or construction is relevant before deciding.
- On \(X=\{a,b\}\), why does \(\{\{a\}\}\) fail to be a basis, even though its only member intersects itself?
- For \(\mathcal{B}=\{\mathbb{R},(-1,1),(0,2)\}\), identify a point where the refinement condition fails and state the intersection that must be refined.
- With rational open intervals as a basis for \(\mathbb{R}\), can an open interval with an irrational endpoint be open without being a basis member? Explain using unions.
- For the subbasis \(\{(-\infty,1),(0,\infty)\}\), what finite intersection gives \((0,1)\)?
- In the subbasis construction, why must the intersection of no subbasic sets be included?