What a Topology Specifies
A topology begins with a set \(X\), whose elements are the points under discussion, and selects certain subsets of \(X\) to call open. The selection cannot be arbitrary: it must be stable under specified unions and intersections. Those rules are the definition of a topology. Once a topology is chosen, it determines which sets count as open, but the axioms do not require distances, coordinates, or any other extra structure.
For example, if \(X=\{a,b,c\}\), a possible collection of subsets is \(\{\varnothing,X,\{a\},\{a,b\}\}\). The question is not whether these subsets look geometrically open; it is whether the collection obeys the four axioms. The same test applies when \(X\) is infinite, although the requirement about unions then covers families of any size.
- \(\varnothing\in\mathcal{T}\) and \(X\in\mathcal{T}\).
- If \((U_i)_{i\in J}\) is any family of members of \(\mathcal{T}\), then \(\bigcup_{i\in J}U_i\in\mathcal{T}\).
- If \(U_1,\dots,U_n\in\mathcal{T}\) for a positive integer \(n\), then \(\bigcap_{k=1}^{n}U_k\in\mathcal{T}\).
In the union axiom, \(J\) is an index set and may be finite or infinite. Different indices may refer to the same set; this does not affect the union. In the finite-intersection axiom, the number of sets is finite and at least one. We state \(\varnothing\) and \(X\) separately, so there is no need to adopt a convention for an intersection of zero sets.
There is a useful convention for the union axiom: the union of a family with no members is \(\varnothing\). Thus, if “any family” includes the empty family, the union axiom itself implies \(\varnothing\in\mathcal{T}\). The definition still lists the empty-set axiom explicitly because it is one of the basic requirements and because some presentations phrase the union condition only for nonempty families.
Reading the Axioms Precisely
The first axiom supplies two particular subsets. It does not say that every subset of \(X\) is open. The second axiom applies to a whole family at once: even an infinite union of open sets must be open. The third axiom applies only to a finite list. These quantifiers matter. A collection can be stable under finite unions and still fail to contain the union of an infinite family; it can be stable under arbitrary unions and still fail to contain an intersection of two open sets.
The union and intersection conditions concern sets that are already members of \(\mathcal{T}\). They do not assert that arbitrary subsets of \(X\) belong to \(\mathcal{T}\). To check a proposed collection, first confirm that every listed object is a subset of \(X\), then test whether the required unions and intersections remain in the collection.
Worked Example: Checking a Collection on Three Points
Let \(X=\{a,b,c\}\) and let \(\mathcal{T}=\{\varnothing,X,\{a\},\{a,b\}\}\). We check all four requirements.
Empty set and whole set. Both \(\varnothing\) and \(X=\{a,b,c\}\) are explicitly in \(\mathcal{T}\).
Arbitrary unions. There are only four possible members of \(\mathcal{T}\). A union of a family of these sets is \(\varnothing\) if every set in the family is empty, is \(\{a\}\) if the union includes \(\{a\}\) but neither \(\{a,b\}\) nor \(X\), is \(\{a,b\}\) if the union includes \(\{a,b\}\) but not \(X\), and is \(X\) if the union includes \(X\). Each possible union belongs to \(\mathcal{T}\). This accounts for finite and infinite families, since repeating any of the four available sets cannot create another union.
Finite intersections. An intersection involving \(\varnothing\) is \(\varnothing\). An intersection involving \(X\) and any other set equals that other set, while an intersection of copies of the same set equals that set. The only distinct nonempty proper sets in the collection are \(\{a\}\) and \(\{a,b\}\), and \(\{a\}\cap\{a,b\}=\{a\}\). Thus every finite nonempty intersection is in \(\mathcal{T}\).
Conclusion. All four axioms hold, so \(\mathcal{T}\) is a topology on \(X\). The argument checked unions of arbitrary size, not just unions of two listed sets.
Examples That Fail an Axiom
A collection may contain \(\varnothing\) and \(X\) but still not be a topology. The following examples isolate specific ways a requirement can fail. A failed axiom is enough to rule out the collection, even if all the other listed sets appear reasonable.
Worked Example: A Missing Whole-Space Set
Let \(X=\{a,b\}\) and \(\mathcal{C}=\{\varnothing,\{a\}\}\). Every member is a subset of \(X\), and \(\varnothing\in\mathcal{C}\), but \(X=\{a,b\}\notin\mathcal{C}\). Therefore the first axiom fails and \(\mathcal{C}\) is not a topology on \(X\).
This example also satisfies the arbitrary-union condition: a union of members of \(\mathcal{C}\) is either \(\varnothing\) or \(\{a\}\). It satisfies the finite-intersection condition for the same reason: an intersection of one or more members is either \(\varnothing\) or \(\{a\}\). The missing whole space is the obstruction.
Worked Example: A Missing Union
Let \(X=\{a,b,c\}\) and \(\mathcal{C}=\{\varnothing,X,\{a\},\{b\}\}\). The empty set and whole space belong to \(\mathcal{C}\). Intersections of members also belong: the intersection of \(\{a\}\) and \(\{b\}\) is \(\varnothing\), intersection with \(X\) leaves the other set unchanged, and intersection with \(\varnothing\) gives \(\varnothing\).
However, \(\{a\}\) and \(\{b\}\) are both members while their union is \(\{a,b\}\), which is not in \(\mathcal{C}\). The arbitrary-union axiom fails; in fact, it already fails for a family of just two sets. Hence \(\mathcal{C}\) is not a topology.
Worked Example: A Missing Finite Intersection
Let \(X=\{a,b,c\}\) and \(\mathcal{D}=\{\varnothing,X,\{a,b\},\{b,c\}\}\). The union of any family of members belongs to \(\mathcal{D}\): if the family contains \(X\), the union is \(X\); if it contains both distinct nonempty proper sets, their union is \(X\); and otherwise its union is one of the listed sets or \(\varnothing\). Thus the arbitrary-union condition holds.
But \(\{a,b\}\cap\{b,c\}=\{b\}\), and \(\{b\}\notin\mathcal{D}\). The finite-intersection axiom fails, so \(\mathcal{D}\) is not a topology. Notice that checking only unions would not have detected this failure.
Two Useful Checks
Proof. Take an index set \(J=\varnothing\), so the family has no members. By the convention for an empty union, \(\bigcup_{i\in\varnothing}U_i=\varnothing\). The arbitrary-union axiom says that this union belongs to the collection. Therefore \(\varnothing\) belongs to the collection. \(\square\)
This explains why a presentation may either list \(\varnothing\) as its own axiom or build it into the union condition. If the union condition is restricted to nonempty index sets, this argument does not apply; then the empty-set requirement must be imposed separately.
Proof. If \(\mathcal{C}\) is closed under arbitrary unions, it is closed under unions of two members by taking a two-element index set. Conversely, suppose the union of any two members of \(\mathcal{C}\) belongs to \(\mathcal{C}\). Consider any family of members of \(\mathcal{C}\). Its distinct values form a subcollection of \(\mathcal{C}\), which is finite because \(\mathcal{C}\) is finite. If there are no distinct values, the union is \(\varnothing\), which belongs to \(\mathcal{C}\). Otherwise, list the distinct values as \(V_1,\dots,V_m\). Repeated use of closure under unions of two shows that \(V_1\cup\cdots\cup V_m\in\mathcal{C}\): the claim holds for \(m=1\), and if it holds for \(m-1\), then \((V_1\cup\cdots\cup V_{m-1})\cup V_m\in\mathcal{C}\). This is the union of the original family, so arbitrary-union closure follows. \(\square\)
The finiteness of \(\mathcal{C}\) is essential to that test: it ensures that a family of members has only finitely many distinct values. For an infinite collection, checking unions of two members alone does not by itself verify unions of every possible family.
A Common Mistake
Pitfall: Checking only that the union of two open sets is open does not establish the arbitrary-union axiom for a general collection. For example, take \(X=\{1,2,3,\dots\}\) and let \(\mathcal{C}\) consist of \(\varnothing\) and all finite subsets of \(X\). The union of two members is finite, so it remains in \(\mathcal{C}\). But the family of singletons \(\{\{n\}:n\in X\}\) consists entirely of members of \(\mathcal{C}\), while its union is \(X\), which is infinite and is not in \(\mathcal{C}\). Thus \(\mathcal{C}\) fails arbitrary-union closure. This calculation also shows why the finite-collection proposition cannot be applied here: \(\mathcal{C}\) is not finite.
When testing a proposed topology, keep the scope of each quantifier in view: unions range over arbitrary families, while intersections range over finite nonempty lists. A successful check must address each requirement, not just the ones that are easiest to verify.
Check Your Understanding
For each question, identify the relevant axiom and state the set calculation that decides it.
- On \(X=\{p,q\}\), does \(\{\varnothing,X,\{p\}\}\) satisfy all four topology axioms?
- On \(X=\{p,q,r\}\), what union of two members fails for \(\{\varnothing,X,\{p\},\{q\}\}\)?
- Why must the union of an empty indexed family be specified when interpreting the arbitrary-union axiom?
- For a finite collection containing \(\varnothing\), why can unions of two members suffice to test arbitrary-union closure?
- Give a collection that contains \(\varnothing\) and \(X\), is closed under unions of two members, but fails closure under some finite intersection.