Ordering Topologies by Their Open Sets
A topology determines which subsets of a set count as open. On a fixed underlying set, we can therefore compare two topologies by asking whether every open set of one is also open in the other. This comparison is useful because it records how much open-set structure each topology has: a finer topology has at least as many open sets, while a coarser topology has no more.
Throughout, when comparing \(\mathcal{T}_1\) and \(\mathcal{T}_2\), both are topologies on the same set \(X\). In particular, each is a collection of subsets of \(X\), so inclusion between them is meaningful. A topology is a collection of open sets containing \(\varnothing\) and \(X\), closed under arbitrary unions and finite intersections.
The direction is worth keeping in view: the finer topology is the larger collection. For example, if \(\mathcal{T}_1\subseteq\mathcal{T}_2\), every set open in \((X,\mathcal{T}_1)\) is open in \((X,\mathcal{T}_2)\), but the reverse need not hold. If neither collection contains the other, the topologies are incomparable. Thus two topologies on the same set need not be related by refinement in either direction.
Refinement Is a Partial Order
Proof. Let \(\mathcal{T}_1,\mathcal{T}_2,\mathcal{T}_3\) be topologies on \(X\). Reflexivity follows from \(\mathcal{T}_1\subseteq\mathcal{T}_1\). If \(\mathcal{T}_1\subseteq\mathcal{T}_2\) and \(\mathcal{T}_2\subseteq\mathcal{T}_3\), then every member of \(\mathcal{T}_1\) belongs to \(\mathcal{T}_2\), and hence to \(\mathcal{T}_3\); so \(\mathcal{T}_1\subseteq\mathcal{T}_3\). For antisymmetry, if \(\mathcal{T}_1\subseteq\mathcal{T}_2\) and \(\mathcal{T}_2\subseteq\mathcal{T}_1\), each collection has exactly the same members as the other, so \(\mathcal{T}_1=\mathcal{T}_2\). \(\square\)
In particular, two topologies are equal exactly when each is finer than the other. The proposition does not say that every pair is comparable: inclusion is a partial order, not necessarily a total order. The examples below show both comparable and incomparable pairs.
Worked Example: Discrete and Indiscrete Topologies
Let \(X\) be a set. The discrete topology is \(\mathcal{P}(X)\), the collection of all subsets of \(X\). The indiscrete topology is \(\{\varnothing,X\}\). Both are topologies: \(\mathcal{P}(X)\) contains every possible union and intersection of subsets of \(X\), and unions or finite intersections of sets chosen from \(\{\varnothing,X\}\) remain in \(\{\varnothing,X\}\).
Since every member of \(\{\varnothing,X\}\) is a subset of \(X\), it is a member of \(\mathcal{P}(X)\). Therefore the indiscrete topology is coarser than the discrete topology. If \(X\) has at least two points, choose \(x\in X\). Then \(\{x\}\) is a subset of \(X\), so \(\{x\}\in\mathcal{P}(X)\), but \(\{x\}\ne\varnothing\) and \(\{x\}\ne X\). Thus \(\{x\}\notin\{\varnothing,X\}\), and the inclusion is strict.
If \(X=\varnothing\), then \(\mathcal{P}(X)=\{\varnothing\}\) and \(\{\varnothing,X\}=\{\varnothing\}\), so the two topologies are equal. If \(X\) has exactly one point, its only subsets are \(\varnothing\) and \(X\), so they are again equal. This checks the degenerate cases as well as the strict case.
Check: the only open sets of the indiscrete topology are open in the discrete topology, while the singleton \(\{x\}\), when \(X\) has at least two points, is an open set of the discrete topology that is not open in the indiscrete topology. This confirms both the direction and strictness of the comparison.
A Comparison on the Real Line
Worked Example: Cofinite and Standard Topologies on \(\mathbb{R}\)
Give \(\mathbb{R}\) its standard topology in one comparison, and its cofinite topology in the other. The cofinite topology consists of \(\varnothing\) and all subsets \(U\subseteq\mathbb{R}\) for which \(\mathbb{R}\setminus U\) is finite. We compare these topologies by checking both inclusions.
Every cofinite-open set is standard-open. The empty set is standard-open. If \(U\) is cofinite-open, write \(\mathbb{R}\setminus U=F\), where \(F\) is finite. A finite subset \(F\) of the standard real line is closed: each singleton \(\{x\}\) is closed, and a finite union of closed sets is closed. Hence \(\mathbb{R}\setminus F=U\) is standard-open. This proves that the cofinite topology is contained in the standard topology.
The inclusion is strict. The interval \((0,1)\) is standard-open. Its complement is not finite: for every \(n\in\mathbb{N}\), the distinct number \(-n\) lies in \(\mathbb{R}\setminus(0,1)\). Consequently \((0,1)\) is not cofinite-open. The standard topology is therefore strictly finer than the cofinite topology.
Check: the argument has established that every cofinite-open set is standard-open, and has exhibited one standard-open set that is not cofinite-open. Both parts are needed to conclude strict inclusion, rather than merely observing that the topologies differ.
Comparable and Incomparable Finite Examples
Worked Example: A Strict Inclusion on a Three-Point Set
Let \(X=\{a,b,c\}\), and consider \(\mathcal{T}_1=\{\varnothing,\{a\},X\}\) and \(\mathcal{T}_2=\{\varnothing,\{a\},\{a,b\},X\}\).
Both collections are topologies. Each contains \(\varnothing\) and \(X\). For \(\mathcal{T}_1\), a union of members is one of \(\varnothing,\{a\},X\), and a finite intersection of members is also one of those sets. For \(\mathcal{T}_2\), its nonempty proper members are nested, \(\{a\}\subseteq\{a,b\}\). A union of any selection of its members is the largest selected member, and a finite intersection of nonempty selected members is the smallest selected member; if \(\varnothing\) is selected for an intersection, the result is \(\varnothing\). Thus both collections satisfy the topology axioms.
Every member of \(\mathcal{T}_1\) appears in \(\mathcal{T}_2\), so \(\mathcal{T}_1\subseteq\mathcal{T}_2\). The set \(\{a,b\}\) belongs to \(\mathcal{T}_2\) but not \(\mathcal{T}_1\), so the inclusion is strict. Hence \(\mathcal{T}_2\) is strictly finer. The extra open set separates \(b\) from \(c\) in a way unavailable in \(\mathcal{T}_1\), which has no proper open set containing \(b\).
Check: listing the members gives three open sets in \(\mathcal{T}_1\) and four in \(\mathcal{T}_2\); more importantly, every member of the smaller list occurs in the larger one, and the extra member witnesses strictness.
Worked Example: Two Incomparable Topologies
On \(X=\{a,b,c\}\), let \(\mathcal{S}_1=\{\varnothing,\{a\},X\}\) and \(\mathcal{S}_2=\{\varnothing,\{b\},X\}\). Each collection is a topology: it contains \(\varnothing\) and \(X\), and unions and finite intersections of its members remain in that collection.
The set \(\{a\}\) belongs to \(\mathcal{S}_1\) but not to \(\mathcal{S}_2\), so \(\mathcal{S}_1\not\subseteq\mathcal{S}_2\). The set \(\{b\}\) belongs to \(\mathcal{S}_2\) but not to \(\mathcal{S}_1\), so \(\mathcal{S}_2\not\subseteq\mathcal{S}_1\). Neither topology is finer than the other; they are incomparable.
Check: the shared open sets are exactly \(\varnothing\) and \(X\). Each topology also has an open set missing from the other, so neither possible inclusion holds.
Common Pitfall: Reversing the Inclusion
A common error is to say that a topology is finer because it has fewer open sets. That reverses the definition. For example, on a set \(X\) with at least two points, the indiscrete topology has only \(\varnothing\) and \(X\), while the discrete topology contains all subsets. The incorrect calculation \(\mathcal{P}(X)\subseteq\{\varnothing,X\}\) would claim that the discrete topology is coarser. But for any \(x\in X\), the singleton \(\{x\}\) belongs to \(\mathcal{P}(X)\) and does not belong to \(\{\varnothing,X\}\), so that inclusion is false. The correct inclusion is \(\{\varnothing,X\}\subsetneq\mathcal{P}(X)\): the discrete topology is finer.
Another useful check is to read an inclusion aloud. If \(\mathcal{T}_1\subseteq\mathcal{T}_2\), say: “Every \(\mathcal{T}_1\)-open set is also \(\mathcal{T}_2\)-open.” If you cannot verify that sentence, the inclusion has not been established. And if you want strict refinement, also identify a set open in \(\mathcal{T}_2\) but not in \(\mathcal{T}_1\).
Check Your Understanding
For each comparison, state which topology is finer, whether the inclusion is strict, or whether the topologies are incomparable.
- On a set \(X\) with exactly one point, compare the discrete and indiscrete topologies. Are they distinct?
- On a set \(X\) with at least two points, identify an open set that proves the discrete topology is strictly finer than the indiscrete topology.
- On \(\mathbb{R}\), why is \(\mathbb{R}\setminus\{0,1\}\) open in the cofinite topology and in the standard topology?
- For the three-point topologies \(\mathcal{T}_1\) and \(\mathcal{T}_2\) in the worked example, which set witnesses that the inclusion is strict?
- For \(\mathcal{S}_1\) and \(\mathcal{S}_2\), give one open set in each topology that is not open in the other.