The Finite-Complement Test
In the discrete topology, every subset is open. In the cofinite topology on an infinite set, openness has a different test: a nonempty set is open precisely when only finitely many points of the space are left out. The complement must always be taken relative to the space being considered. This makes it important to identify the ambient set before testing a candidate.
The definition has two parts. The empty set is included explicitly. Every other open set must have finite complement. Since \(X\) is infinite, \(X\setminus\varnothing=X\) is infinite, so the empty set would not qualify through the finite-complement condition. The whole space \(X\), by contrast, qualifies because \(X\setminus X=\varnothing\), and the empty set is finite.
Thus, for a proposed subset \(U\subseteq X\), the practical test is: first decide whether \(U=\varnothing\); if not, compute \(X\setminus U\) and check whether that complement is finite. An infinite candidate set need not be open: its complement could also be infinite. Likewise, a finite nonempty candidate set is not open when \(X\) is infinite, because removing finitely many points from \(X\) still leaves infinitely many.
Two Basic Consequences
Proof. By definition, \(C\) is closed if and only if \(X\setminus C\) is open. If \(C\) is finite, then \(X\setminus C\) has finite complement \(C\), so \(X\setminus C\) is open. If \(C=X\), its complement is \(\varnothing\), which is open. This proves that every finite \(C\), and also \(X\), is closed. Conversely, suppose \(C\) is closed. Then \(X\setminus C\) is open. If \(X\setminus C=\varnothing\), then \(C=X\). Otherwise, the definition of cofinite openness says that \(X\setminus(X\setminus C)=C\) is finite. Therefore \(C\) is finite or \(C=X\). \(\square\)
Proof. Let \(U\) be nonempty and open, and let \(A\subseteq X\) be infinite. Since \(U\) is nonempty and open, \(X\setminus U\) is finite. If \(U\cap A=\varnothing\), then every element of \(A\) lies outside \(U\), so \(A\subseteq X\setminus U\). A subset of a finite set is finite, which would make \(A\) finite. This contradicts the assumption that \(A\) is infinite. Hence \(U\cap A\ne\varnothing\). \(\square\)
These statements provide useful checks on a proposed answer. A finite nonempty subset of an infinite \(X\) is closed but not open: its complement is infinite, and the set itself is not empty. An infinite subset can be open or not open, depending on whether the points it leaves out form a finite set.
Worked Tests on Infinite Sets
Worked Example: Subsets of the Natural Numbers
Give the cofinite topology to \(X=\mathbb{N}=\{1,2,3,\dots\}\). Test the following sets: \(U=\mathbb{N}\setminus\{2,5,9\}\), \(E=\{2,4,6,\dots\}\), \(F=\{2,5\}\), and \(\varnothing\).
The set \(U\). Its complement relative to \(\mathbb{N}\) is \[ \mathbb{N}\setminus U=\{2,5,9\}. \] This complement has exactly three elements, so it is finite. Also \(U\ne\varnothing\), since \(1\in U\). Therefore \(U\) is open by the finite-complement condition.
The set \(E\). Its complement in \(\mathbb{N}\) is the set of odd positive integers, \(\{1,3,5,\dots\}\). This complement is infinite: for every \(k\in\mathbb{N}\), \(2k-1\) is a positive odd integer, and these values are distinct as \(k\) varies. Thus \(E\) is nonempty but its complement is not finite, so \(E\) is not open.
The set \(F\). It is nonempty. Its complement contains every natural number \(n\ge 6\), so the complement is infinite. Therefore \(F\) is not open.
The empty set. It is open by the separately stated \(\{\varnothing\}\) part of the definition. It does not pass the finite-complement test: \(\mathbb{N}\setminus\varnothing=\mathbb{N}\), which is infinite.
Check: the whole set \(\mathbb{N}\) is also open, since its complement is \(\varnothing\), a finite set. These answers distinguish the two exceptional sets: \(\varnothing\) is included directly, while \(\mathbb{N}\) satisfies the finite-complement condition.
Worked Example: Subsets of the Integers
Give \(X=\mathbb{Z}\) the cofinite topology. Decide whether \(A=\{n\in\mathbb{Z}:-3\le n\le 3\}\), \(B=\mathbb{Z}\setminus\{-2,0,7\}\), and \(P=\{1,2,3,\dots\}\) are open.
The set \(A\). Listing its elements gives \[ A=\{-3,-2,-1,0,1,2,3\}. \] It is nonempty, but \(\mathbb{Z}\setminus A\) contains every integer \(n\ge4\). Those integers form an infinite set, since the values \(4,5,6,\dots\) are distinct. Therefore \(A\) is not open.
The set \(B\). Its complement in the ambient space \(\mathbb{Z}\) is exactly \[ \mathbb{Z}\setminus B=\{-2,0,7\}. \] This is a set of three integers, hence finite. The set \(B\) is nonempty, for example \(1\in B\), so \(B\) is open.
The set \(P\). Its complement in \(\mathbb{Z}\) is \[ \mathbb{Z}\setminus P=\{\dots,-3,-2,-1,0\}. \] This is infinite because it contains the distinct integers \(-1,-2,-3,\dots\). Since \(P\) is nonempty and its complement is infinite, \(P\) is not open in this topology.
Check: the test uses complements in \(\mathbb{Z}\), not in \(\mathbb{N}\). For instance, the complement of \(P\) includes all nonpositive integers; it is not a finite set of omitted positive integers.
Worked Example: Subsets of the Real Line
Give \(X=\mathbb{R}\) the cofinite topology, rather than assuming the standard topology. Test \(V=\mathbb{R}\setminus\{0,1,2\}\), the interval \(J=(0,1)\), and the singleton \(S=\{4\}\).
The set \(V\). Its complement in \(\mathbb{R}\) is \(\{0,1,2\}\), which has exactly three elements. Also \(V\ne\varnothing\), since \(3\in V\). Hence \(V\) is open in the cofinite topology.
The interval \(J\). Its complement is \((-\infty,0]\cup[1,\infty)\). This complement is infinite; for example, it contains every integer \(n\ge1\). The interval is nonempty, so it is not open in the cofinite topology.
The singleton \(S\). It is nonempty, and its complement contains every integer other than \(4\), an infinite set. Thus \(S\) is not open. Notice that this conclusion is about the cofinite topology: it follows directly from the finite-complement test, without relying on whether the set is open in some other topology on \(\mathbb{R}\).
Check: \(\mathbb{R}\) itself is open because its complement is empty, while \(\varnothing\) is open by definition even though its complement \(\mathbb{R}\) is infinite.
A Common Mistake: Looking at the Set Instead of Its Complement
A frequent error is to assume that an infinite set must be open. In the cofinite topology, the number of points in the set is not the test; the number of points omitted from the ambient space is the test.
Wrong calculation: “The even natural numbers \(E=\{2,4,6,\dots\}\) are infinite, so \(E\) is open in \(\mathbb{N}\).” This uses the size of \(E\), but the definition asks about \(\mathbb{N}\setminus E\). That complement is \(\{1,3,5,\dots\}\), which is infinite, so the conclusion does not follow and \(E\) is not open.
Correction: compute the complement first. For \(U=\mathbb{N}\setminus\{2,5,9\}\), the complement is exactly \(\{2,5,9\}\), a finite set, and \(U\) is nonempty. Therefore \(U\) is open. For \(E\), the complement is infinite, so \(E\) is not open. The same check also prevents a related error: a nonempty finite subset of an infinite space is not open just because it has only finitely many elements.
The cofinite topology is therefore best approached as a complement-counting rule. In every example, name the ambient set, form its complement of the candidate, and distinguish the empty set from nonempty candidates before applying the finite-complement test.
Check Your Understanding
For each question, state the ambient set and justify the answer by computing the complement when the candidate is nonempty.
- In the cofinite topology on \(\mathbb{N}\), are \(\mathbb{N}\setminus\{1,4\}\), \(\{1,4\}\), and \(\varnothing\) open?
- On an infinite set \(X\), explain why \(X\) is open even though its complement is not nonempty.
- In the cofinite topology on \(\mathbb{Z}\), is the set of negative integers open? Is \(\mathbb{Z}\setminus\{-1,0,1\}\) open?
- Prove that no nonempty finite subset of an infinite set \(X\) is open in the cofinite topology.
- Let \(A\) be an infinite subset of an infinite set \(X\), and let \(U\) be nonempty and open in the cofinite topology. Must \(A\cap U\) be nonempty? Give the reason.