When Sequences Do Not Detect Closure
In a metric space, closure can be tested using sequences: a point is in the closure of a set precisely when some sequence from that set converges to the point. This test does not hold in every topological space. We will construct a space in which every open neighborhood of a particular point meets a set, but no sequence in that set converges to the point. The construction also gives a concrete reason that first countability matters when using sequences to study closure.
Let \(X\) be an uncountable set. A set is countable here if it is finite or can be listed as a sequence. Define a collection \(\mathcal{T}\) of subsets of \(X\) by declaring that \(\varnothing\) is open and that a nonempty subset \(U\subseteq X\) is open when \(X\setminus U\) is countable. This is called the cocountable topology.
We check that this really is a topology. The empty set is included by definition, and \(X\) is included because its complement is empty. For a union of open sets, if every set in the union is empty, the union is empty. Otherwise choose a nonempty member \(U\); the complement of the union is contained in \(X\setminus U\), so it is countable. For a finite intersection, if any member is empty the intersection is empty. Otherwise the complement of the intersection is a finite union of countable complements, hence countable. These are exactly the arbitrary-union and finite-intersection conditions for a topology.
We will use the neighborhood test for closure: \(x\in\overline{A}\) if and only if every open neighborhood of \(x\) intersects \(A\). A local base at \(x\) is a collection of open neighborhoods of \(x\) such that every open neighborhood of \(x\) contains a member of the collection. A space is first countable if every point has a countable local base.
A Point in the Closure Missed by Every Sequence
Worked Example: Delete One Point
Take \(X=\mathbb{R}\) with the cocountable topology, fix \(p=0\), and let \(A=\mathbb{R}\setminus\{0\}\). We determine \(\overline{A}\) and check whether any sequence in \(A\) converges to \(0\).
First consider an arbitrary open neighborhood \(U\) of \(0\). It is nonempty, so \(C=\mathbb{R}\setminus U\) is countable. Since \(0\in U\), we have \(0\notin C\). The set \(C\cup\{0\}\) is countable, while \(\mathbb{R}\) is uncountable. Therefore there is a point in \(\mathbb{R}\setminus(C\cup\{0\})\). This point belongs to \(U\) and to \(A\), so \(U\cap A\ne\varnothing\). Every open neighborhood of \(0\) meets \(A\), and the neighborhood test gives \(0\in\overline{A}\). Every point of \(A\) is already in its closure, so \(\overline{A}=\mathbb{R}\).
Now let \((a_n)\) be any sequence in \(A\). Its range \(S=\{a_n:n\in\mathbb{N}\}\) is countable, and \(0\notin S\). Thus \(V=\mathbb{R}\setminus S\) is an open neighborhood of \(0\). It contains no term \(a_n\). In particular, the sequence is not eventually in \(V\), so it does not converge to \(0\).
Check: The closure conclusion used every neighborhood of \(0\), not just a selected collection: each such neighborhood omits only a countable set, and cannot omit all of the uncountable set \(A\). The sequence conclusion uses a neighborhood chosen from that sequence’s own countable range.
The two tests in this example concern different quantifiers. Closure asks whether every neighborhood meets the set. Convergence of a particular sequence asks whether, for each neighborhood, all sufficiently late terms lie in it. A countable range can be removed by one cocountable open set, even though no single countable complement can remove all of \(A\).
Proof. Let \(U\) be any open neighborhood of \(p\). Its complement \(C=X\setminus U\) is countable, and \(p\notin C\). Since \(C\cup\{p\}\) is countable and \(X\) is uncountable, there is a point \(x\in X\setminus(C\cup\{p\})\). Then \(x\in U\cap A\). Thus every open neighborhood of \(p\) meets \(A\), so \(p\in\overline{A}\) by the neighborhood test. For any sequence \((a_n)\) in \(A\), its range \(S=\{a_n:n\in\mathbb{N}\}\) is countable and does not contain \(p\). Hence \(X\setminus S\) is an open neighborhood of \(p\) containing none of the terms. The sequence cannot converge to \(p\). \(\square\)
Which Sequences Converge Here?
The preceding argument works for every sequence in \(A\). More generally, convergence in the cocountable topology has a particularly restrictive form: a sequence converges to \(x\) exactly when it is eventually equal to \(x\). “Eventually equal” means that there is an \(N\in\mathbb{N}\) such that \(x_n=x\) for every \(n\ge N\).
Proof. Suppose first that \(x_n\to x\). The set \(C=\{x_n:n\in\mathbb{N},\ x_n\ne x\}\) is countable and does not contain \(x\). Consequently \(U=X\setminus C\) is an open neighborhood of \(x\). Convergence means that there is an \(N\in\mathbb{N}\) such that \(x_n\in U\) for every \(n\ge N\). Every term different from \(x\) belongs to \(C\), and therefore lies outside \(U\). Thus \(x_n=x\) for every \(n\ge N\). Conversely, suppose \(x_n=x\) for all \(n\ge N\). Every open neighborhood of \(x\) contains \(x\), so all terms with index at least \(N\) lie in that neighborhood. This is convergence to \(x\). \(\square\)
Worked Example: A Sequence of Distinct Real Numbers
In \(\mathbb{R}\) with the cocountable topology, consider the sequence \(x_n=1/n\). Each term is nonzero, so this sequence lies in \(A=\mathbb{R}\setminus\{0\}\) from the preceding example. The terms are distinct: if \(1/m=1/n\) for positive integers \(m,n\), multiplication by \(mn\) gives \(n=m\). Thus the sequence is not eventually equal to any fixed real number.
The proposition shows that the sequence has no limit in this topology. To check directly that it does not converge to \(0\), take \(U=\mathbb{R}\setminus\{1/n:n\in\mathbb{N}\}\). Its complement is countable, and \(0\notin\{1/n:n\in\mathbb{N}\}\), so \(U\) is an open neighborhood of \(0\). Every term of the sequence is outside \(U\), which rules out convergence to \(0\).
If a proposed limit is \(x=1/k\) for some \(k\in\mathbb{N}\), use \(V=\mathbb{R}\setminus(\{1/n:n\in\mathbb{N}\}\setminus\{1/k\})\). The removed set is countable and does not contain \(1/k\), so \(V\) is an open neighborhood of \(1/k\). For every \(n\ne k\), the term \(1/n\) lies outside \(V\); there are terms with indices larger than any chosen index, so the sequence is not eventually in \(V\). If \(x\) is not one of the terms, the neighborhood \(U\) already given contains \(x\) and excludes every term. These cases cover every candidate limit.
Check: The constant sequence \(x_n=0\) does converge to \(0\), because every neighborhood of \(0\) contains every term. The distinction is not that sequences never converge in this space; rather, convergence forces eventual equality with the limit.
The Space Is Not First Countable
The example also explains why the sequence test for closure can fail: the cocountable topology on an uncountable set is not first countable. We prove this at an arbitrary point, rather than only at the point used in the example.
Proof. Fix \(p\in X\), and suppose for contradiction that \((U_n)_{n\in\mathbb{N}}\) is a countable local base at \(p\). Each \(U_n\) is an open neighborhood of \(p\), so \(C_n=X\setminus U_n\) is countable and does not contain \(p\). The union \(C=\bigcup_{n\in\mathbb{N}}C_n\) is countable. Since \(X\) is uncountable, choose \(y\in X\setminus(C\cup\{p\})\). Then \(V=X\setminus\{y\}\) is an open neighborhood of \(p\). The local-base property gives some \(n\) with \(U_n\subseteq V\). But \(y\notin C_n\), so \(y\in U_n\), contradicting \(U_n\subseteq X\setminus\{y\}\). Therefore no countable local base at \(p\) exists. Since \(p\) was arbitrary, the space is not first countable. \(\square\)
In this argument, the countable family of neighborhoods omits altogether only the countable set \(C\). We choose \(y\) outside that set, then use the neighborhood that omits just \(y\) to contradict the proposed base property. The proof depends on \(X\) being uncountable; it is that assumption which guarantees such a \(y\) exists.
Countable and Uncountable Sets Have Different Closures
Worked Example: Rational and Irrational Numbers
Again take \(X=\mathbb{R}\), now with the cocountable topology. We compute the closures of \(\mathbb{Q}\) and \(\mathbb{R}\setminus\mathbb{Q}\). Recall that the closure of a set is the intersection of all closed sets containing it, and a set is closed when its complement is open.
The set \(\mathbb{Q}\) is countable. Its complement \(\mathbb{R}\setminus\mathbb{Q}\) is therefore open in the cocountable topology, so \(\mathbb{Q}\) is closed. Since it is itself a closed set containing \(\mathbb{Q}\), its closure is \(\overline{\mathbb{Q}}=\mathbb{Q}\).
The set \(\mathbb{R}\setminus\mathbb{Q}\) is uncountable: if it were countable, then its union with the countable set \(\mathbb{Q}\) would make \(\mathbb{R}\) countable. A proper closed subset \(F\) of \(X\) must be countable, because \(X\setminus F\) is a nonempty open set and hence has countable complement \(F\). No countable closed set can contain the uncountable set \(\mathbb{R}\setminus\mathbb{Q}\). Thus the only closed set containing it is \(X\), and \(\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\).
Check: These closures differ from the familiar closures in the standard topology on \(\mathbb{R}\). The topology in this example is cocountable, not standard, so the standard-topology density of \(\mathbb{Q}\) does not determine its closure here.
Common pitfall. A tempting but incorrect conclusion is that \(p\notin\overline{A}\) in the first example because no sequence in \(A\) converges to \(p\). That calculation substitutes a sequence test for the neighborhood test without checking its hypotheses. The direct closure calculation shows the error: for every open neighborhood \(U\) of \(p\), the complement \(X\setminus U\) is countable, while \(A=X\setminus\{p\}\) is uncountable; hence \(U\cap A\ne\varnothing\) and \(p\in\overline{A}\). The sequence argument proves only that no individual sequence from \(A\) converges to \(p\). In this space, closure and sequential limits genuinely give different answers.
The example matters because it marks the boundary of a useful method. In first-countable spaces, countable local bases allow sequences to test closure. Without first countability, even an uncountable set can have a point in its closure that no sequence from the set reaches as a limit. Before applying a sequence-based closure test, check that the space has the required countability property.
Check Your Understanding
Use the definitions of the cocountable topology, closure, convergence, and local base to answer the following.
- For uncountable \(X\) with the cocountable topology, show directly that every open neighborhood of \(p\) meets \(X\setminus\{p\}\).
- Let \(A\subseteq X\) be countable. Show that \(A\) is closed and determine \(\overline{A}\).
- In the cocountable topology on \(\mathbb{R}\), determine whether the sequence \(x_n=2\) for every \(n\) converges, and identify its limit.
- Explain why the proof that the space is not first countable needs \(X\) to be uncountable.
- Give the neighborhood that rules out convergence to \(p\) for an arbitrary sequence in \(X\setminus\{p\}\).