What a Sequence Argument Needs
In a metric space, sequences often provide an effective way to test closure and continuity. In a general topological space, however, the neighborhoods of a point may be too numerous to be controlled by a sequence. The absence of a sequence with a desired property therefore does not, by itself, prove that the property fails. Two hypotheses are particularly useful to check: first countability, which gives a local countable family of neighborhoods, and sequentiality, which says that sequences detect closed sets.
Throughout, \((X,\mathcal{T})\) denotes a topological space. A sequence \((x_n)\) converges to \(x\) if, for every open neighborhood \(U\) of \(x\), there is \(N\in\mathbb{N}\) such that \(x_n\in U\) whenever \(n\ge N\). The closure \(\overline{A}\) of \(A\subseteq X\) consists of the points whose every open neighborhood intersects \(A\).
These conditions are related but not interchangeable. First countability supplies a sequence test for closure at each point. Sequentiality guarantees that sequences detect whether a whole set is closed; it does not, in general, supply a sequence from \(A\) converging to every particular point of \(\overline{A}\).
When Sequences Detect Closure
Proof. Suppose first that \(x\in\overline{A}\). If \(x\in A\), the constant sequence \(x_n=x\) lies in \(A\) and converges to \(x\). If \(x\notin A\), choose a countable neighborhood base \(U_1,U_2,\ldots\) at \(x\). Define \(V_n=U_1\cap\cdots\cap U_n\). Each \(V_n\) is an open neighborhood of \(x\), and the sets \(V_n\) are decreasing. Because \(x\in\overline{A}\), each \(V_n\) intersects \(A\); choose \(a_n\in V_n\cap A\). For any open neighborhood \(W\) of \(x\), the base property gives some \(k\) with \(U_k\subseteq W\). For \(n\ge k\), \(a_n\in V_n\subseteq U_k\subseteq W\), so \(a_n\to x\).
Conversely, suppose \(a_n\in A\) for every \(n\) and \(a_n\to x\). Every open neighborhood \(W\) of \(x\) contains all terms from some index onward, so in particular it contains a term \(a_n\in A\). Thus every such \(W\) intersects \(A\), which is exactly \(x\in\overline{A}\). \(\square\)
The countable base is used in the first direction to make one sequence meet every neighborhood requirement. The reverse direction needs no countability assumption: any convergent sequence in \(A\) shows that its limit belongs to \(\overline{A}\).
Worked Example: A Valid Sequence Test in the Real Line
Use sequences to check that \(0\in\overline{A}\) for \(A=(0,1)\subseteq\mathbb{R}\) with the standard topology.
For each \(n\in\mathbb{N}\), let \(a_n=1/(n+1)\). Since \(n+1\ge2\), \(0<a_n\le1/2<1\), so \(a_n\in(0,1)=A\). Also, for any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N+1>1/\varepsilon\). Then for every \(n\ge N\), \(0<a_n=1/(n+1)\le1/(N+1)<\varepsilon\), hence \(a_n\to0\). Theorem 1 applies because the standard topology on \(\mathbb{R}\) is first countable: intervals centered at a point with radii \(1/n\) form a countable neighborhood base. Therefore \(0\in\overline{A}\).
Check: Directly, every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and the convergence calculation shows that some \(a_n\) lies in this interval and in \(A\).
A Space Where Sequences Miss a Closure Point
Consider an uncountable set \(X\) with the cocountable topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are countable. A countable set is allowed to be finite or empty. For this topology, closure is still tested by neighborhoods; the question is whether those neighborhoods can be tested by a single sequence.
Worked Example: Closure in the Cocountable Topology
Let \(X=\mathbb{R}\) with the cocountable topology, let \(p=0\), and set \(A=\mathbb{R}\setminus\{0\}\). We show that \(0\in\overline{A}\), but no sequence in \(A\) converges to \(0\).
Every open neighborhood \(U\) of \(0\) has countable complement \(\mathbb{R}\setminus U\). The set \(A=\mathbb{R}\setminus\{0\}\) is uncountable, so it cannot be contained in the countable set \(\mathbb{R}\setminus U\). Therefore \(U\cap A\ne\varnothing\). This holds for every open neighborhood \(U\) of \(0\), proving \(0\in\overline{A}\).
Now take any sequence \((a_n)\) in \(A\), and let \(D=\{a_n:n\in\mathbb{N}\}\) be its range. The set \(D\) is countable and does not contain \(0\). Thus \(U=\mathbb{R}\setminus D\) is an open neighborhood of \(0\). It contains none of the terms \(a_n\). In particular, the sequence is not eventually in \(U\), so it does not converge to \(0\). Since the sequence was arbitrary, no sequence in \(A\) converges to \(0\).
Check: The conclusion is consistent with the neighborhood definition of closure: \(0\) belongs to every neighborhood’s closure test because each neighborhood intersects \(A\), even though each individual sequence has a countable range that one neighborhood can avoid.
This example also shows that the space is not first countable at \(0\). Indeed, if it were, Theorem 1 would produce a sequence in \(A\) converging to \(0\), contrary to the calculation above. The example makes a useful diagnostic concrete: before using a sequence to characterize closure, check that first countability (or another theorem providing the same conclusion) is available.
Nets Recover the Neighborhood Information
A net is a family \((x_d)_{d\in D}\) indexed by a directed set \(D\). Here “directed” means that \(D\) has an order \(\le\) such that for any \(d,e\in D\), there is \(f\in D\) with \(d\le f\) and \(e\le f\). A net converges to \(x\) if, for every open neighborhood \(U\) of \(x\), there is \(d_0\in D\) such that \(x_d\in U\) whenever \(d\ge d_0\). A sequence is a net indexed by \(\mathbb{N}\), but a general directed set can record more neighborhood information.
Proof. If a net in \(A\) converges to \(x\), every open neighborhood of \(x\) contains all terms from some index onward, and hence intersects \(A\). Thus \(x\in\overline{A}\).
For the reverse implication, suppose \(x\in\overline{A}\). Let \(D\) be the set of all open neighborhoods of \(x\), ordered by reverse inclusion: \(U\le V\) means \(V\subseteq U\). This is directed, since \(U\cap V\) is an open neighborhood of \(x\) and is an upper bound for \(U\) and \(V\) in this order. For each \(U\in D\), choose \(a_U\in U\cap A\), which is possible because \(x\in\overline{A}\). This defines a net in \(A\). Given any open neighborhood \(W\) of \(x\), whenever \(U\ge W\) in the reverse-inclusion order, \(U\subseteq W\), so \(a_U\in U\subseteq W\). The net therefore converges to \(x\). \(\square\)
The construction selects one point from each neighborhood intersection; as usual, such a simultaneous selection is made using the axiom of choice. In the cocountable example, this theorem applies at \(0\): its net follows the neighborhoods themselves, rather than attempting to encode them with countably many terms.
Sequentiality and Continuity Tests
First countability gives a pointwise sequence test for closure. Sequentiality gives a different, set-level test: a space is sequential precisely when its sequentially closed subsets are closed. This distinction matters when sequences are used to test continuity.
Proof. If \(f\) is continuous and \(x_n\to x\), let \(V\) be any open neighborhood of \(f(x)\). The preimage \(f^{-1}(V)\) is open and contains \(x\), so \(x_n\) is eventually in \(f^{-1}(V)\). Hence \(f(x_n)\) is eventually in \(V\), proving sequential continuity.
Conversely, suppose \(f\) is sequentially continuous. Let \(F\subseteq Y\) be closed. We show \(f^{-1}(F)\) is closed. If \(x_n\in f^{-1}(F)\) and \(x_n\to x\), then \(f(x_n)\in F\) for every \(n\), and sequential continuity gives \(f(x_n)\to f(x)\). A closed set contains the limit of each convergent sequence of its points: if \(f(x)\notin F\), the open neighborhood \(Y\setminus F\) of \(f(x)\) would have to contain all sufficiently late terms, contradicting that every term belongs to \(F\). Thus \(f(x)\in F\), so \(x\in f^{-1}(F)\). This proves that \(f^{-1}(F)\) is sequentially closed. Since \(X\) is sequential, it is closed. The preimage of every closed set is therefore closed, which is equivalent to continuity: taking complements gives the open-set preimage condition. \(\square\)
Worked Example: A Sequentially Continuous Map That Is Not Continuous
Let \(X=\mathbb{R}\) with the cocountable topology, let \(Y=\mathbb{R}\) with the standard topology, and take \(f:X\to Y\) to be the identity map. We verify that \(f\) is sequentially continuous but not continuous.
First, every convergent sequence \((x_n)\) in \(X\) is eventually constant at its limit \(x\). To see this, let \(D=\{x_n:x_n\ne x\}\), a countable set not containing \(x\). The set \(U=X\setminus D\) is an open neighborhood of \(x\). Convergence implies that \(x_n\in U\) for all sufficiently large \(n\). Since every term is either \(x\) or belongs to \(D\), those sufficiently late terms must equal \(x\). Applying \(f\) to an eventually constant sequence leaves it eventually constant at \(f(x)=x\), so \(f(x_n)\to f(x)\) in the standard topology. This proves sequential continuity.
However, the standard open interval \(V=(-1,1)\) has preimage \(f^{-1}(V)=(-1,1)\). Its complement in \(X\) is uncountable, so \((-1,1)\) is not open in the cocountable topology. Thus \(f\) is not continuous. In particular, \(X\) cannot be sequential: otherwise Theorem 3 would make this sequentially continuous map continuous.
Check: The failure is in the domain hypothesis of Theorem 3, not in the computation of convergent sequences. In this domain, sequences see only eventual constancy, which is too weak to detect all open sets.
A Common Mistake and a Practical Test
Common pitfall. The false inference “there is no sequence in \(A\) converging to \(x\), so \(x\notin\overline{A}\)” fails for \(X=\mathbb{R}\) with the cocountable topology, \(A=\mathbb{R}\setminus\{0\}\), and \(x=0\). The neighborhood calculation gives \(0\in\overline{A}\), while the countable-range argument shows that no sequence in \(A\) converges to \(0\). The correction is to identify the hypothesis that justifies the sequence test. First countability suffices for the pointwise closure criterion; without it, test neighborhoods directly or use nets.
Before relying on a sequence argument, ask what its conclusion requires. If the claim concerns a closure point, check first countability or use the neighborhood definition. If it concerns closed sets or continuity, check whether the domain is sequential. These checks prevent a valid-looking sequence calculation from being used beyond what it proves.
Check Your Understanding
For each question, identify the relevant hypothesis before applying a sequence test.
- Which direction of Theorem 1 uses first countability, and where does the countable neighborhood base enter?
- In the cocountable topology on an uncountable set \(X\), show that every point of a nonempty set \(A\) belongs to \(\overline{A}\). Is this true for \(A=\varnothing\)?
- Why does the set \(D=\{a_n:n\in\mathbb{N}\}\) in the cocountable example give an open neighborhood of \(0\) that prevents convergence?
- State the difference between a first-countable space and a sequential space in terms of what sequences are guaranteed to detect.
- For the identity map from the cocountable real line to the standard real line, identify an open set in the target whose preimage is not open in the domain.