Compare the Open Sets, Not Just the Topology Names
Two topologies on the same set may have different collections of open sets. To determine whether one is finer than the other, compare those collections directly. The key test is set inclusion: every open set in the coarser topology must also be open in the finer one. A single missing open set disproves that inclusion.
Throughout, let \(X\) be a set and let \(\mathcal{T}\) and \(\mathcal{S}\) be topologies on \(X\). A topology is a collection of subsets of \(X\) that contains \(\varnothing\) and \(X\), is closed under arbitrary unions, and is closed under finite intersections. In particular, \(\mathcal{T}\) and \(\mathcal{S}\) are collections of subsets of the same underlying set, so it makes sense to ask whether one collection is contained in the other.
Thus, to test whether \(\mathcal{S}\) is finer than \(\mathcal{T}\), inspect every member of \(\mathcal{T}\) and check that it belongs to \(\mathcal{S}\). To prove strictness, also exhibit a member of \(\mathcal{S}\) that is not in \(\mathcal{T}\). To prove incomparability, find one set in \(\mathcal{T}\setminus\mathcal{S}\) and another in \(\mathcal{S}\setminus\mathcal{T}\).
The Open-Set and Closed-Set Tests
Proof. By definition, a set is \(\mathcal{T}\)-open precisely when it is a member of the collection \(\mathcal{T}\), and it is \(\mathcal{S}\)-open precisely when it is a member of \(\mathcal{S}\). Therefore every \(\mathcal{T}\)-open set is \(\mathcal{S}\)-open exactly when every member of \(\mathcal{T}\) belongs to \(\mathcal{S}\), which is \(\mathcal{T}\subseteq\mathcal{S}\).
For the closed-set statement, first suppose \(\mathcal{T}\subseteq\mathcal{S}\) and let \(C\) be closed in \(\mathcal{T}\). By the definition of closed, \(X\setminus C\in\mathcal{T}\). The inclusion gives \(X\setminus C\in\mathcal{S}\), so \(C\) is closed in \(\mathcal{S}\). Conversely, suppose every \(\mathcal{T}\)-closed set is \(\mathcal{S}\)-closed. Let \(U\in\mathcal{T}\). Then \(X\setminus U\) is \(\mathcal{T}\)-closed and hence \(\mathcal{S}\)-closed. Its complement \(U\) is therefore \(\mathcal{S}\)-open, so \(U\in\mathcal{S}\). This proves \(\mathcal{T}\subseteq\mathcal{S}\). \(\square\)
The direction in the closed-set version matters: the closed sets of the coarser topology must be closed in the finer topology. This follows by taking complements of the open sets in the inclusion. It does not say that every closed set of the finer topology is closed in the coarser one.
Worked Example: A Strict Inclusion on Three Points
Let \(X=\{a,b,c\}\), and set \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\) and \(\mathcal{S}=\{\varnothing,\{a\},X\}\). Determine which topology is finer.
Both listed collections are topologies. In \(\mathcal{T}\), the sets are nested in the order \(\varnothing\subseteq\{a\}\subseteq\{a,b\}\subseteq X\). The union or intersection of any selection of these sets is one of the listed sets. In \(\mathcal{S}\), the same check applies to the nested collection \(\varnothing\subseteq\{a\}\subseteq X\). Each collection contains \(\varnothing\) and \(X\), so both satisfy the topology axioms.
Now test inclusion by checking the members of \(\mathcal{S}\): \(\varnothing\in\mathcal{T}\), \(\{a\}\in\mathcal{T}\), and \(X\in\mathcal{T}\). Thus \(\mathcal{S}\subseteq\mathcal{T}\). This inclusion is strict because \(\{a,b\}\in\mathcal{T}\) but \(\{a,b\}\notin\mathcal{S}\). Therefore \(\mathcal{T}\) is finer than \(\mathcal{S}\). The test depends on all the open sets, not on which collection is written first.
Check using closed sets: The \(\mathcal{S}\)-closed sets are the complements of its open sets: \(X\), \(\{b,c\}\), and \(\varnothing\). They are all \(\mathcal{T}\)-closed: these are respectively the complements of \(\varnothing\), \(\{a\}\), and \(X\), all of which are \(\mathcal{T}\)-open. But \(\{c\}=X\setminus\{a,b\}\) is \(\mathcal{T}\)-closed and is not \(\mathcal{S}\)-closed. This agrees with the theorem: the \(\mathcal{S}\)-closed sets are included among the \(\mathcal{T}\)-closed sets because \(\mathcal{S}\subseteq\mathcal{T}\).
Incomparable Topologies Have Witnesses in Both Directions
Proof. Suppose first that \(\mathcal{T}\) and \(\mathcal{S}\) are incomparable. Since \(\mathcal{T}\nsubseteq\mathcal{S}\), the definition of subset gives a set \(A\in\mathcal{T}\) such that \(A\notin\mathcal{S}\). Since \(\mathcal{S}\nsubseteq\mathcal{T}\), there is a set \(B\in\mathcal{S}\) such that \(B\notin\mathcal{T}\). Conversely, if both such witnesses exist, then \(A\in\mathcal{T}\setminus\mathcal{S}\) prevents \(\mathcal{T}\subseteq\mathcal{S}\), and \(B\in\mathcal{S}\setminus\mathcal{T}\) prevents \(\mathcal{S}\subseteq\mathcal{T}\). Hence the topologies are incomparable. \(\square\)
Worked Example: Two Incomparable Topologies
On \(X=\{a,b,c\}\), let \(\mathcal{T}=\{\varnothing,\{a\},X\}\) and \(\mathcal{S}=\{\varnothing,\{b\},X\}\). Both are topologies: each contains \(\varnothing\) and \(X\), and in each collection the sets are nested, so unions and finite intersections remain in the collection.
The set \(\{a\}\) belongs to \(\mathcal{T}\) but not to \(\mathcal{S}\), so \(\mathcal{T}\nsubseteq\mathcal{S}\). The set \(\{b\}\) belongs to \(\mathcal{S}\) but not to \(\mathcal{T}\), so \(\mathcal{S}\nsubseteq\mathcal{T}\). These two witnesses prove that the topologies are incomparable. Neither is finer than the other.
The fact that both collections have three open sets does not establish incomparability in general; the witnesses do. Inclusion must be checked by membership, not inferred from a count.
Worked Example: More Open Sets Does Not Mean Finer
Let \(X=\{a,b,c\}\), and define \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\) and \(\mathcal{S}=\{\varnothing,\{a\},\{c\},\{a,c\},X\}\). Decide whether either topology is finer.
For \(\mathcal{T}\), the sets form a nested chain, so arbitrary unions and finite intersections of its members belong to \(\mathcal{T}\). For \(\mathcal{S}\), unions and intersections of \(\{a\}\) and \(\{c\}\) are respectively \(\{a,c\}\) and \(\varnothing\); unions or intersections involving \(\{a,c\}\), \(\varnothing\), or \(X\) also remain in the list. Both collections contain \(\varnothing\) and \(X\), so each is a topology.
The set \(\{a,b\}\) belongs to \(\mathcal{T}\) but not \(\mathcal{S}\), so \(\mathcal{T}\nsubseteq\mathcal{S}\). The set \(\{c\}\) belongs to \(\mathcal{S}\) but not \(\mathcal{T}\), so \(\mathcal{S}\nsubseteq\mathcal{T}\). Thus they are incomparable, even though \(\mathcal{S}\) has five open sets and \(\mathcal{T}\) has four. A larger number of open sets alone does not imply finer: the additional sets may fail to include the open sets of the other topology.
Closed-set check: The complements of the \(\mathcal{T}\)-open sets give \(\mathcal{T}\)-closed sets \(X,\{b,c\},\{c\},\varnothing\). The complements of the \(\mathcal{S}\)-open sets give \(\mathcal{S}\)-closed sets \(X,\{b,c\},\{a,b\},\{b\},\varnothing\). The set \(\{c\}\) is \(\mathcal{T}\)-closed but not \(\mathcal{S}\)-closed, and \(\{a,b\}\) is \(\mathcal{S}\)-closed but not \(\mathcal{T}\)-closed. The closed-set test therefore confirms failure of inclusion in both directions.
Common Pitfall: Reversing the Closed-Set Test
Here is a concrete way to catch the reversed condition. On \(X=\{a,b,c\}\), let \(\mathcal{T}=\{\varnothing,X\}\), the indiscrete topology, and let \(\mathcal{S}=\{\varnothing,\{a\},\{a,b\},X\}\). The listed members of \(\mathcal{S}\) form a nested chain, so \(\mathcal{S}\) is a topology; \(\mathcal{T}\) is a topology because it contains only \(\varnothing\) and \(X\). Since both \(\mathcal{T}\)-open sets occur in \(\mathcal{S}\), we have \(\mathcal{T}\subseteq\mathcal{S}\).
A mistaken closed-set test might say that every \(\mathcal{S}\)-closed set must be \(\mathcal{T}\)-closed. But \(\{b,c\}=X\setminus\{a\}\) is \(\mathcal{S}\)-closed and is not \(\mathcal{T}\)-closed: the only \(\mathcal{T}\)-closed sets are \(\varnothing\) and \(X\). The correct test goes the other way. The \(\mathcal{T}\)-closed sets \(\varnothing\) and \(X\) are both \(\mathcal{S}\)-closed, as the theorem requires.
For a final consistency check, equality of topologies means inclusion in both directions. Indeed, if \(\mathcal{T}\subseteq\mathcal{S}\) and \(\mathcal{S}\subseteq\mathcal{T}\), every set in either collection belongs to the other, so \(\mathcal{T}=\mathcal{S}\). Conversely, equal collections are contained in each other. For finite examples, a reliable procedure is therefore: verify each listed collection is a topology; test each inclusion separately; and, for strictness or incomparability, write down the set or sets that witness the result.
Check Your Understanding
For each question, justify the comparison by naming the relevant open-set witnesses or by checking all the listed sets.
- Let \(X=\{p,q\}\), \(\mathcal{T}=\{\varnothing,X\}\), and \(\mathcal{S}=\{\varnothing,\{p\},X\}\). Which topology is finer, and is the inclusion strict?
- On \(X=\{a,b,c\}\), compare \(\{\varnothing,\{a\},X\}\) with \(\{\varnothing,\{a\},\{a,b\},X\}\). State a witness to strictness.
- Give one set witnessing each direction of incomparability for the two topologies in the second worked example.
- If \(\mathcal{T}\subseteq\mathcal{S}\), which topology’s closed sets must also be closed in the other topology?
- Can the topology with fewer open sets be finer? Explain what the inclusion definition requires, regardless of the number of sets.