Comparing the Open Sets
A topology is a collection of subsets declared to be open. Comparing two topologies on the same set therefore comes down to comparing their collections of open sets: one topology is contained in another precisely when every set open in the first is also open in the second. The cofinite, discrete, and indiscrete topologies give three simple but different collections to compare.
Let \(X\) be a set. The discrete topology contains every subset of \(X\). The indiscrete topology contains only \(\varnothing\) and \(X\). The cofinite topology contains \(\varnothing\) and every subset whose complement in \(X\) is finite. These are all topologies; the discrete and indiscrete cases were introduced earlier, and the cofinite topology was verified from the topology axioms in the earlier material on the cofinite topology.
The definition of the cofinite topology has two parts to keep in view. The empty set is open by definition, even if its complement is not finite. For a nonempty set \(U\), the test is whether the omitted points, \(X\setminus U\), form a finite set. It is not a test of whether \(U\) itself is finite or infinite.
The Inclusion Chain
Proof. First, \(\varnothing\in\mathcal{T}_{\mathrm{cof}}\) by definition, and \(X\in\mathcal{T}_{\mathrm{cof}}\) because \(X\setminus X=\varnothing\) is finite. These are the only open sets required for \(\mathcal{T}_{\mathrm{ind}}\), so \(\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}_{\mathrm{cof}}\). Every cofinite-open set is a subset of \(X\), and every subset of \(X\) is open in the discrete topology. Therefore \(\mathcal{T}_{\mathrm{cof}}\subseteq\mathcal{T}_{\mathrm{disc}}\).
Suppose now that \(X\) is infinite, and choose \(x\in X\). Since \(X\) is infinite, \(X\setminus\{x\}\) is nonempty. It is cofinite-open because its complement is the finite set \(\{x\}\). It is not in the indiscrete topology: it is neither empty nor all of \(X\). Hence \(\mathcal{T}_{\mathrm{ind}}\subsetneq\mathcal{T}_{\mathrm{cof}}\). Also \(\{x\}\) is open in the discrete topology. Its complement \(X\setminus\{x\}\) is infinite; otherwise \(X=(X\setminus\{x\})\cup\{x\}\) would be finite. Thus \(\{x\}\) is not cofinite-open, so \(\mathcal{T}_{\mathrm{cof}}\subsetneq\mathcal{T}_{\mathrm{disc}}\).
If \(X\) is finite, then \(X\setminus U\) is finite for every \(U\subseteq X\). Consequently every subset is cofinite-open, so \(\mathcal{T}_{\mathrm{cof}}=\mathcal{T}_{\mathrm{disc}}\). When \(X\) has at least two elements, a singleton is nonempty and proper, so it is discrete-open but not indiscrete-open; the first inclusion is strict. If \(X=\varnothing\), its only subset is \(\varnothing=X\), and all three topologies consist of that one set. If \(X\) has one element, its only subsets are \(\varnothing\) and \(X\), so again all three topologies are equal. \(\square\)
The proof separates the general inclusion argument from the strictness argument. The inclusions follow directly from the definitions and hold even for the empty set. To prove strictness, one must exhibit an open set that is missing from the smaller collection. On an infinite set, deleting one point gives such a set for the first inclusion, while keeping one point gives such a set for the second.
Worked Example: The Three Topologies on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,3,\dots\}\). Consider \(U=\mathbb{N}\setminus\{2,5\}\) and \(E=\{2,4,6,\dots\}\). Test each set against the three definitions.
For \(U\), its complement is \(\mathbb{N}\setminus U=\{2,5\}\), which is finite. Thus \(U\) is open in the cofinite topology, and it is also open in the discrete topology. Since \(U\) contains \(1\), it is nonempty; since it omits \(2\), it is not all of \(\mathbb{N}\). It is therefore not open in the indiscrete topology.
For \(E\), the complement is the set of odd natural numbers \(\{1,3,5,\dots\}\), which is infinite. Thus \(E\) is not cofinite-open. It is open in the discrete topology because it is a subset of \(\mathbb{N}\). It is neither empty nor all of \(\mathbb{N}\), so it is not indiscrete-open.
Check: \(U\) witnesses that a cofinite-open set need not be indiscrete-open, and \(E\) witnesses that a discrete-open set need not be cofinite-open. Both are proper inclusions, as the theorem predicts for an infinite set.
Worked Example: A Four-Point Set
Let \(X=\{a,b,c,d\}\). For any subset \(A\subseteq X\), the complement \(X\setminus A\) is a subset of a four-element set, so it is finite. Hence every subset is cofinite-open, and \(\mathcal{T}_{\mathrm{cof}}=\mathcal{T}_{\mathrm{disc}}=\mathcal{P}(X)\).
For instance, \(A=\{a,c\}\) has complement \(X\setminus A=\{b,d\}\), a finite set. It is cofinite-open and discrete-open. It is not indiscrete-open because it is nonempty and omits elements of \(X\). The singleton \(\{a\}\) gives another witness that the indiscrete topology is properly contained in the other two: it is nonempty and proper, so it is not one of the two indiscrete-open sets, but it is open in both the cofinite and discrete topologies.
Check: in this example the first inclusion is strict, but the second is an equality. The fact that \(X\) is finite is what makes the complement of every subset finite.
Worked Example: The Empty Set and a Singleton
First take \(X=\{p\}\). Its subsets are exactly \(\varnothing\) and \(\{p\}=X\). The discrete and indiscrete topologies both consist of these two sets. For the cofinite topology, \(\varnothing\) is included by definition, and \(X\) is included because \(X\setminus X=\varnothing\) is finite. There are no other subsets to include, so all three topologies are equal.
Now take \(X=\varnothing\). Its only subset is \(\varnothing\), which is also \(X\). The discrete and indiscrete definitions each give the collection \(\{\varnothing\}\). The cofinite definition gives that same collection: \(\varnothing\) is included, and there are no other subsets. Thus all three topologies are equal here as well.
Check: the distinction between “empty” and “all of \(X\)” causes no extra open sets in either case: when \(X=\varnothing\), those two sets coincide; when \(X=\{p\}\), they are the only two subsets.
A Common Mistake and a Practical Test
Common mistake. A finite set is not necessarily open in the cofinite topology. For example, in \(X=\mathbb{N}\), the set \(\{1,2\}\) is not cofinite-open: its complement \(\mathbb{N}\setminus\{1,2\}=\{3,4,5,\dots\}\) is infinite. The error is to test whether the set itself is finite. The correct test is whether its complement is finite. Conversely, \(\mathbb{N}\setminus\{1\}\) is cofinite-open even though it is infinite, because its complement is the one-element set \(\{1\}\).
For an infinite set, the three collections can be compared without listing every open set. The indiscrete topology has only the two required sets, the cofinite topology adds sets whose complements are finite, and the discrete topology includes every subset. The witness sets in the theorem show why neither inclusion can be reversed: \(X\setminus\{x\}\) belongs to the middle collection but not the first, while \(\{x\}\) belongs to the last but not the middle.
| Underlying set | Relationship | Reason |
|---|---|---|
| Infinite \(X\) | \(\mathcal{T}_{\mathrm{ind}}\subsetneq\mathcal{T}_{\mathrm{cof}}\subsetneq\mathcal{T}_{\mathrm{disc}}\) | Use \(X\setminus\{x\}\) and \(\{x\}\) as witnesses. |
| Finite \(X\) with at least two elements | \(\mathcal{T}_{\mathrm{ind}}\subsetneq\mathcal{T}_{\mathrm{cof}}=\mathcal{T}_{\mathrm{disc}}\) | Every subset has finite complement; a singleton is proper and nonempty. |
| \(X\) empty or a singleton | All three topologies are equal. | There are no subsets beyond \(\varnothing\) and \(X\). |
Check Your Understanding
Use the definitions to justify each answer, including whether the complement is finite.
- On an infinite set \(X\), explain why \(X\setminus\{x\}\) is cofinite-open but not indiscrete-open.
- On an infinite set \(X\), why is \(\{x\}\) not cofinite-open, even though it is discrete-open?
- If \(X\) has five elements, are the cofinite and discrete topologies equal? Are the indiscrete and discrete topologies equal?
- What are the three topologies on the empty set? Explain why the cofinite definition introduces no additional set.
- For \(X=\mathbb{N}\), decide whether \(\mathbb{N}\setminus\{3,7,9\}\) is open in each of the three topologies.