Finite Complements, a Separation Contrast
The cofinite topology is defined by a simple rule: a set is open if it is empty or omits only finitely many points. On an infinite set, this rule gives a useful contrast between two separation properties. Every pair of distinct points can be distinguished by an open set, so the space is T1. Yet no two nonempty open sets are disjoint, so distinct points cannot have disjoint open neighborhoods. The space is therefore not Hausdorff.
Let \(X\) be an infinite set. A set \(U\subseteq X\) is cofinite when its complement \(X\setminus U\) is finite. The empty set is included as open whether or not it is described by the cofinite condition; since \(X\) is infinite, its complement \(X\) is not finite.
In this tutorial, T1 means that for any distinct \(x,y\in X\), there is an open neighborhood of \(x\) that does not contain \(y\). Hausdorff means that for any distinct \(x,y\in X\), there are disjoint open neighborhoods, one containing \(x\) and the other containing \(y\). The first condition asks to exclude the other point; the second asks two neighborhoods to avoid one another entirely.
Why These Sets Form a Topology
Before checking separation properties, we verify that the stated collection is a topology. A topology must contain \(\varnothing\) and \(X\), be closed under arbitrary unions, and be closed under finite intersections. The set \(X\) is in \(\mathcal{T}\), because \(X\setminus X=\varnothing\) is finite. The empty set is in \(\mathcal{T}\) by definition.
Proof. Let \(\{U_\alpha\}_{\alpha\in J}\) be any family of sets in \(\mathcal{T}\). If every \(U_\alpha\) is empty, their union is empty and belongs to \(\mathcal{T}\). Otherwise choose an index \(\beta\) for which \(U_\beta\ne\varnothing\). Its complement \(X\setminus U_\beta\) is finite. Since \(X\setminus\bigcup_{\alpha\in J}U_\alpha\subseteq X\setminus U_\beta\), the complement of the union is finite, so the union belongs to \(\mathcal{T}\).
Now take a finite family \(U_1,\ldots,U_n\) with \(n\ge1\). If one of its members is empty, the intersection is empty and belongs to \(\mathcal{T}\). Otherwise every \(X\setminus U_i\) is finite, and \(X\setminus\bigcap_{i=1}^{n}U_i=\bigcup_{i=1}^{n}(X\setminus U_i)\). A finite union of finite sets is finite, so the intersection belongs to \(\mathcal{T}\). The intersection of a family of zero sets is \(X\), which is already in \(\mathcal{T}\). All topology axioms hold. \(\square\)
The arbitrary-union condition is important: the union need not be formed from a finite number of open sets. The proof works because, as soon as one member is nonempty and cofinite, the complement of the entire union is contained in the finite complement of that member.
The Separation Properties
Proof. First fix \(x\in X\). The singleton \(\{x\}\) is closed precisely when its complement is open. That complement \(X\setminus\{x\}\) has finite complement \(\{x\}\), so it is open. Thus every singleton is closed. In particular, for distinct \(x,y\in X\), the open set \(X\setminus\{y\}\) contains \(x\) and excludes \(y\), proving the T1 condition.
For the Hausdorff condition, take any two nonempty open sets \(U,V\in\mathcal{T}\). Their complements \(F=X\setminus U\) and \(G=X\setminus V\) are finite. The intersection is \(U\cap V=X\setminus(F\cup G)\). The set \(F\cup G\) is finite. Since \(X\) is infinite, \(X\setminus(F\cup G)\) is nonempty: if it were empty, then \(X=F\cup G\) would be finite. Hence every two nonempty open sets intersect. In particular, no two distinct points can have disjoint open neighborhoods, and the space is not Hausdorff. \(\square\)
This proof identifies exactly what the finite-complement rule provides. Removing one point leaves an open set, which is enough for T1. But removing finitely many points cannot make two cofinite sets disjoint: together, two finite complements still omit only finitely many points from the infinite space.
Worked Example: The Cofinite Topology on \(\mathbb{N}\)
Give \(\mathbb{N}=\{1,2,3,\ldots\}\) the cofinite topology. The set \(U=\mathbb{N}\setminus\{2,7\}\) is open because its complement \(\{2,7\}\) is finite. It contains \(1\), excludes \(2\), and is a neighborhood of \(1\) that witnesses the T1 separation of \(1\) from \(2\). For any distinct \(m,n\in\mathbb{N}\), the same calculation with \(U=\mathbb{N}\setminus\{n\}\) gives an open neighborhood of \(m\) that excludes \(n\).
Now consider \(U=\mathbb{N}\setminus\{2,7\}\) and \(V=\mathbb{N}\setminus\{1,7,10\}\). Both are open, since their complements are finite. Their intersection is \(\mathbb{N}\setminus\{1,2,7,10\}\). For example, \(3\) belongs to this intersection: \(3\notin\{2,7\}\) and \(3\notin\{1,7,10\}\). Thus these particular nonempty open sets are not disjoint. The theorem proves the same for every pair of nonempty open sets, not just this choice.
Check: \(U\) contains \(1\) because \(1\notin\{2,7\}\), while \(V\) contains \(3\) because \(3\notin\{1,7,10\}\). The computed intersection contains \(3\), so the example also directly verifies that these two neighborhoods overlap.
Worked Example: Neighborhoods of Two Integers
Give \(X=\mathbb{Z}\) the cofinite topology and take the distinct points \(4\) and \(-3\). Consider \(U=\mathbb{Z}\setminus\{-3,6\}\) and \(V=\mathbb{Z}\setminus\{4,6,8\}\). The complements have respectively two and three elements, so \(U\) and \(V\) are open. Also \(4\in U\), since \(4\) is neither \(-3\) nor \(6\), and \(-3\in V\), since \(-3\) is none of \(4,6,8\). Thus these are neighborhoods of the chosen points.
The intersection removes the union of the two complements: \[ U\cap V=\mathbb{Z}\setminus\{-3,4,6,8\}. \] The integer \(0\) is not any of \(-3,4,6,8\), so \(0\in U\cap V\). These two neighborhoods therefore fail to separate the points by disjoint sets.
The general argument is stronger than this one calculation. For any open neighborhoods \(U_0\) of \(4\) and \(V_0\) of \(-3\), both complements are finite. Their union is finite, whereas \(\mathbb{Z}\) is infinite, so there is an integer outside that union. Such an integer lies in \(U_0\cap V_0\). No choice of open neighborhoods of these points can be disjoint.
Check: \(0\in U\) because \(0\notin\{-3,6\}\), and \(0\in V\) because \(0\notin\{4,6,8\}\). The displayed intersection excludes exactly the four listed integers, confirming the claimed witness.
Closed Sets and Closures
The closed sets also have a simple description. By definition, \(C\subseteq X\) is closed when \(X\setminus C\) is open. If \(C\) is finite, its complement has finite complement \(C\), so \(X\setminus C\) is open and \(C\) is closed. The set \(X\) is closed because its complement is empty and hence open. Conversely, if \(C\) is closed and \(C\ne X\), then \(X\setminus C\) is a nonempty open set, so it has finite complement \(C\). Therefore the closed sets are exactly the finite subsets and \(X\) itself.
Proof. The closure \(\overline{A}\) is the smallest closed set containing \(A\). If \(A\) is finite, then \(A\) is itself closed, so \(\overline{A}=A\), including when \(A=\varnothing\). If \(A\) is infinite, no finite closed set can contain it. The only remaining closed set is \(X\), so the smallest closed set containing \(A\) is \(X\), and \(\overline{A}=X\). \(\square\)
Worked Example: A Finite Set and an Infinite Set
In \(\mathbb{N}\) with the cofinite topology, let \(A=\{2,5,8\}\). Its complement \(\mathbb{N}\setminus A\) is open because its complement \(A\) is finite. Hence \(A\) is closed, and \(\overline{A}=A=\{2,5,8\}\).
Now let \(B=\{2,4,6,8,\ldots\}\), the set of positive even integers. It is infinite: for every \(k\in\mathbb{N}\), the distinct integers \(2,4,\ldots,2k\) belong to \(B\), so \(B\) cannot be finite. No finite closed set contains \(B\). The closed-set description therefore gives \(\overline{B}=\mathbb{N}\).
This closure calculation also follows from neighborhoods. Any nonempty open neighborhood \(U\) in this space has finite complement. It cannot miss all of \(B\), because that would put the infinite set \(B\) inside the finite set \(\mathbb{N}\setminus U\). Thus every open neighborhood of every point meets \(B\), so every point belongs to \(\overline{B}\). For \(A\), points outside \(A\) have the open neighborhood \(\mathbb{N}\setminus A\), which misses \(A\); points in \(A\) belong to its closure because \(A\subseteq\overline{A}\).
Check: \(3\notin A\), and the open set \(\mathbb{N}\setminus\{2,5,8\}\) contains \(3\) while missing \(A\), so \(3\notin\overline{A}\). On the other hand, any open neighborhood of \(3\) has finite complement and therefore cannot exclude every positive even integer. This agrees with \(\overline{A}=A\) and \(\overline{B}=\mathbb{N}\).
A Common Pitfall
Common pitfall. It is incorrect to infer that a T1 space must be Hausdorff. In \(\mathbb{N}\) with the cofinite topology, the open set \(\mathbb{N}\setminus\{2\}\) contains \(1\) but not \(2\); this verifies the required one-point exclusion. But the two open neighborhoods \(\mathbb{N}\setminus\{2\}\) and \(\mathbb{N}\setminus\{1\}\) have intersection \(\mathbb{N}\setminus\{1,2\}\), which is nonempty—for instance, \(3\) belongs to it. The wrong calculation would treat excluding \(1\) and \(2\) as if the two sets had no points left in common. In fact, they exclude only those two points, and infinitely many remain.
The distinction is also visible in the definitions: T1 lets a neighborhood of one point omit a specified other point, while Hausdorffness requires a pair of neighborhoods to have empty intersection. The cofinite topology meets the first requirement and fails the second.
Check Your Understanding
Use finite complements and the definitions of T1 and Hausdorffness to justify each answer.
- In the cofinite topology on an infinite set \(X\), why is \(X\setminus\{x\}\) open for each \(x\in X\)?
- For nonempty open sets \(U,V\), express \(U\cap V\) using \(X\setminus U\) and \(X\setminus V\). Why must this intersection be nonempty?
- What are the closed sets in the cofinite topology on an infinite set?
- Find the closure of a finite subset \(A\subseteq X\), and explain why it is unchanged.
- Find the closure of an infinite subset \(B\subseteq X\), and explain why no proper closed set can contain it.