When Do Sequences Detect Topology?
In “First-Countable Spaces,” we saw that sequences can characterize closure and continuity when the domain is first-countable. That leaves a natural question: is first-countability the exact condition that makes sequences useful, or can the same tests work in spaces without countable local bases? The answer leads to the broader notion of a sequential space.
The key test is whether a set can be recognized as closed by checking limits of sequences. This gives a useful extension of the first-countable results, but it is important not to overstate what sequences can do. Even in a sequential space, a point in the closure of a set need not be the limit of a sequence drawn directly from that set.
Sequentially Closed Sets and Sequential Spaces
Every closed set is sequentially closed, in any topological space. Indeed, if a sequence in a closed set \(A\) converges to \(x\notin A\), then the open complement \(X\setminus A\) is a neighborhood of \(x\). Convergence would force all sufficiently late terms into that neighborhood, contradicting that every term lies in \(A\). The definition of a sequential space asks whether the converse also holds for every subset.
For comparison, a space is called Fréchet–Urysohn if, whenever \(x\in\overline{A}\), there is a sequence of points of \(A\) converging to \(x\). This is a stronger property than being sequential. First-countability implies the Fréchet–Urysohn property, by the Sequential Characterization of Closure from “First-Countable Spaces.” A sequential space need only guarantee that sets passing the sequential closedness test are closed; it does not promise that every closure point is the limit of a sequence from the original set.
Worked Example: A Missing Endpoint in the Real Line
Consider \(A=(0,1)\) as a subset of the usual real line. The sequence \(a_n=1/(n+1)\) lies in \(A\) for every positive integer \(n\), and \(a_n\to 0\). Since \(0\notin A\), the set \(A\) is not sequentially closed and is not closed.
The same test detects the other endpoint: \(b_n=1-1/(n+1)\) lies in \(A\) and converges to \(1\notin A\). In fact, \(\overline{A}=[0,1]\). This is an instance of the first-countable closure characterization: in the real line, closure points are exactly limits of sequences from the set. The example illustrates the sequential test in a familiar space; the results below explain what the test means in more general spaces.
First-Countable Spaces Are Sequential
Proof. Let \(X\) be first-countable, and let \(A\subseteq X\) be sequentially closed. We show that \(A\) is closed. Take any \(x\in\overline{A}\). By the Sequential Characterization of Closure from “First-Countable Spaces,” there is a sequence \((a_n)\) in \(A\) that converges to \(x\). Since \(A\) is sequentially closed, this implies \(x\in A\). Thus \(\overline{A}\subseteq A\). Also \(A\subseteq\overline{A}\), so \(A=\overline{A}\), which means \(A\) is closed. Therefore every sequentially closed subset of \(X\) is closed, as required. \(\square\)
The proof uses a feature stronger than sequentiality: in a first-countable space, one sequence from \(A\) can witness each individual closure point of \(A\). Sequentiality is a broader condition, defined by the resulting test for closedness rather than by this point-by-point sequence construction.
Worked Example: Testing Closedness in a First-Countable Space
Let \(X\) be a first-countable space, and define \(A=\{x\in X:\text{there is a sequence in }X\setminus A\text{ converging to }x\}\) would be circular as a definition of a set, because \(A\) appears on both sides. Instead, take a specified subset \(E\subseteq X\), and suppose every sequence in \(E\) that converges in \(X\) has its limit in \(E\). This assumption says exactly that \(E\) is sequentially closed.
The theorem now gives a direct conclusion: \(E\) is closed. For example, in any first-countable space, if \(E\) is not closed, choose \(x\in\overline{E}\setminus E\). The closure characterization supplies a sequence \((e_n)\) in \(E\) converging to \(x\), so \(E\) fails the sequential closedness test. This is often an efficient way to prove a subset closed: assume a sequence in it converges, and verify that no limit outside the set is possible.
Continuity on Sequential Domains
A function \(f:X\to Y\) is sequentially continuous if, whenever \(x_n\to x\) in \(X\), the image sequence \(f(x_n)\) converges to \(f(x)\) in \(Y\). Every continuous function is sequentially continuous, with no countability assumption: the Continuity Through Open Sets theorem implies that the inverse image of each open neighborhood of \(f(x)\) is an open neighborhood of \(x\), and convergence then puts all sufficiently late \(x_n\) in that inverse image.
Proof. We have just recalled that continuity implies sequential continuity in arbitrary spaces. For the converse, suppose that \(f\) is sequentially continuous. Let \(F\) be closed in \(Y\). We show that \(f^{-1}(F)\) is closed in \(X\).
Take a sequence \((x_n)\) in \(f^{-1}(F)\) that converges in \(X\) to \(x\). Sequential continuity gives \(f(x_n)\to f(x)\). Every \(f(x_n)\) belongs to \(F\), and \(F\) is closed, so the closed-set argument above shows that \(f(x)\in F\). Hence \(x\in f^{-1}(F)\). This proves that \(f^{-1}(F)\) is sequentially closed. Since \(X\) is sequential, \(f^{-1}(F)\) is closed. The preimage of every closed subset of \(Y\) is therefore closed in \(X\), so \(f\) is continuous by the Continuity Through Closed Sets theorem. \(\square\)
The domain is where sequentiality is needed in this result; no separation or countability assumption is imposed on the target. In a first-countable domain, the theorem from “First-Countable Spaces” already gives the same continuity test. The sequential-space theorem applies beyond first-countable spaces because it uses the defining closed-set test instead.
Worked Example: A Sequentially Continuous Map That Is Not Continuous
This example shows why a general domain cannot be omitted from the theorem’s hypothesis. Let \(\omega_1\) denote the first uncountable ordinal, and give \(X=[0,\omega_1]\) the order topology. Set \(A=[0,\omega_1)\). This set is dense in \(X\): every neighborhood of \(\omega_1\) contains a terminal interval \((\alpha,\omega_1]\) for some \(\alpha<\omega_1\), and that interval contains \(\alpha+1<\omega_1\).
Nevertheless, no sequence in \(A\) converges to \(\omega_1\). The terms of any such sequence form a countable collection of countable ordinals, so their supremum \(\gamma\) is still a countable ordinal and satisfies \(\gamma<\omega_1\). The neighborhood \((\gamma,\omega_1]\) of \(\omega_1\) contains none of the sequence’s terms. Thus the sequence cannot converge to \(\omega_1\). It follows that \(A\) is sequentially closed: a sequence in \(A\) cannot converge to the only point of \(X\setminus A\). Since \(A\) is dense and proper, it is not closed. Therefore \(X\) is not sequential.
Now give \(\{0,1\}\) the discrete topology and define \(f:X\to\{0,1\}\) by \(f(\omega_1)=1\) and \(f(\alpha)=0\) for every \(\alpha<\omega_1\). This function is sequentially continuous. If a sequence converges to an ordinal \(\xi<\omega_1\), it is eventually in a neighborhood of \(\xi\) that excludes \(\omega_1\); its image is then eventually \(0=f(\xi)\). If a sequence converges to \(\omega_1\), it must be eventually equal to \(\omega_1\): if it had infinitely many terms below \(\omega_1\), those terms would have a countable supremum \(\gamma<\omega_1\), and the neighborhood \((\gamma,\omega_1]\) would fail to contain all sufficiently late terms. Its image is therefore eventually \(1=f(\omega_1)\).
But \(f\) is not continuous. The set \(\{1\}\) is open in the discrete target, while its preimage is \(\{\omega_1\}\), which is not open in the order topology: every neighborhood of \(\omega_1\) contains ordinals below it. Thus sequential continuity need not imply continuity when the domain is not sequential.
How to Use the Sequential Tests
The results give a practical hierarchy. In a first-countable space, sequences characterize closure point by point, and sequentially closed sets are closed. In any sequential space, sequentially closed sets are closed, and sequential continuity of a map out of that space implies continuity. In an arbitrary space, closedness still implies sequential closedness and continuity still implies sequential continuity, but the reverse implications can fail, as the ordinal example demonstrates.
A common pitfall is to treat “sequential” as if it meant “first-countable,” or to assume every closure point in a sequential space is a limit of a sequence from the original set. Neither inference is justified. First-countability is sufficient for sequentiality, but it is not the definition; and the stronger point-by-point property is Fréchet–Urysohn. The distinction matters when applying a sequence argument: first identify which hypothesis guarantees the particular sequential characterization being used.
Show that every convergent sequence in the set has its limit in the set. This establishes sequential closedness, which implies closedness.
Show that it preserves limits of convergent sequences. The continuity theorem then converts sequential continuity into continuity.
Check for first-countability, or another hypothesis that specifically guarantees such a sequence. Sequentiality alone does not provide that point-by-point conclusion.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why is every closed subset of a topological space sequentially closed?
- Which earlier theorem lets the proof that first-countable spaces are sequential find a sequence approaching a closure point?
- In the continuity theorem, where is the sequential-space hypothesis used?
- Why is the set \([0,\omega_1)\) dense in \([0,\omega_1]\), yet sequentially closed?
- What is the difference between a sequential space and a Fréchet–Urysohn space?
- Which direction of the sequential continuity test holds for every domain, even when the domain is not sequential?