When Sequences Detect Topological Behavior
In a metric space, sequences often give a practical way to test whether a point is near a set or whether a function behaves continuously. In an arbitrary topological space, however, neighborhoods need not be captured by sequences. First countability is a useful condition that supplies enough local neighborhoods for sequence tests to work. We will use it to characterize closure and continuity, and to construct a specific sequence when continuity fails.
Throughout, \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) denote topological spaces. A sequence \((x_n)\) in \(X\) converges to \(x\in X\) if every open neighborhood of \(x\) contains all but finitely many terms: for each open neighborhood \(U\) of \(x\), there is \(N\in\mathbb{N}\) such that \(x_n\in U\) whenever \(n\ge N\).
First countability is a local condition: to test a claim at \(x\), it is enough that \(x\) have a countable neighborhood base. The whole domain need not be first-countable for a sequential test at that particular point.
Closure Can Be Tested by Sequences
Proof. Suppose first that \(x\in\overline{A}\). Choose a countable neighborhood base at \(x\), and list its members as \(B_1,B_2,\ldots\). If the base is finite, repeat its members so that it is listed as a sequence. Define \(U_n=B_1\cap\cdots\cap B_n\). Each \(U_n\) is an open neighborhood of \(x\), and the \(U_n\) form a neighborhood base: if \(V\) is any open neighborhood of \(x\), some \(B_k\subseteq V\), and then \(U_k\subseteq B_k\subseteq V\). Since \(x\in\overline{A}\), every open neighborhood of \(x\) meets \(A\). Choose \(a_n\in U_n\cap A\) for each \(n\). Given any open neighborhood \(V\) of \(x\), choose \(k\) with \(U_k\subseteq V\). For every \(n\ge k\), \(U_n\subseteq U_k\subseteq V\), so \(a_n\in V\). Thus \(a_n\to x\).
Conversely, suppose a sequence \((a_n)\) in \(A\) converges to \(x\). Every open neighborhood \(V\) of \(x\) contains all sufficiently late terms, and therefore contains at least one term \(a_n\in A\). So every open neighborhood of \(x\) meets \(A\), which is the neighborhood criterion for \(x\in\overline{A}\). This direction does not require first countability. If \(A=\varnothing\), the forward direction has no possible \(x\in\overline{A}\), so the equivalence still holds. \(\square\)
The theorem says that a point is in the closure when a sequence from the set can approach it, provided the point has a countable neighborhood base. It does not say that every sequence in \(A\) must converge, or that every point in \(A\) is a limit point of \(A\).
Worked Example: The Closure of a Sequence’s Range
Take \(X=\mathbb{R}\) with its standard topology and \(A=\{1/n:n\in\mathbb{N}\}\). We find \(\overline{A}\) using the sequential criterion. The real line is first-countable: at each \(x\), the intervals \((x-1/n,x+1/n)\), for \(n\in\mathbb{N}\), form a countable neighborhood base.
The sequence \(a_n=1/n\) is in \(A\) and converges to \(0\). Indeed, for any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For \(n\ge N\), \(0<1/n\le 1/N<\varepsilon\). Thus \(0\in\overline{A}\). Each \(1/k\in A\) is also in \(\overline{A}\), since the constant sequence with every term \(1/k\) converges to \(1/k\).
Now let \(x<0\). The interval \((x-|x|/2,x+|x|/2)\) lies strictly below \(0\), so it misses \(A\). Hence \(x\notin\overline{A}\). Let \(x>0\) and suppose \(x\notin A\). Choose \(N\in\mathbb{N}\) with \(1/N<x/2\). The tail terms satisfy \(1/n\le1/N<x/2\) for \(n\ge N\). The finitely many numbers \(1/n\), \(1\le n<N\), are all different from \(x\). If \(N>1\), set \[ \delta=\min\bigl(\{x/4\}\cup\{|x-1/n|/2:1\le n<N\}\bigr); \] if \(N=1\), set \(\delta=x/4\). In either case \(\delta>0\). The interval \((x-\delta,x+\delta)\) misses every head term \(1/n\) with \(n<N\), because its radius is at most half the distance to each such term. It misses every tail term because \(x-\delta\ge 3x/4>x/2>1/n\) for \(n\ge N\). Thus \(x\notin\overline{A}\).
Conclusion and check: \(\overline{A}=A\cup\{0\}\). The constructed sequence proves \(0\in\overline{A}\), and each point of \(A\) is in its closure by a constant sequence. For every other real number, the intervals exhibited above are open neighborhoods disjoint from \(A\), so no additional point belongs to the closure.
Continuity Is Equivalent to Preserving Convergent Sequences
Proof. Suppose \(f\) is continuous at \(x\), and let \(x_n\to x\). If \(V\) is an open neighborhood of \(f(x)\), continuity at \(x\) gives an open neighborhood \(U\) of \(x\) with \(f(U)\subseteq V\). Since \(x_n\to x\), all sufficiently late terms lie in \(U\), and their images lie in \(V\). Thus \(f(x_n)\to f(x)\).
For the converse, suppose \(f\) is not continuous at \(x\). By the neighborhood definition of continuity, there is an open neighborhood \(V\) of \(f(x)\) such that no open neighborhood \(U\) of \(x\) satisfies \(f(U)\subseteq V\). As in the proof of Theorem 1, choose a decreasing countable neighborhood base \(U_1,U_2,\ldots\) at \(x\). For each \(n\), the failure of \(f(U_n)\subseteq V\) supplies a point \(x_n\in U_n\) with \(f(x_n)\notin V\). The base is decreasing, so for each neighborhood \(W\) of \(x\), some \(U_k\subseteq W\), and \(x_n\in U_n\subseteq U_k\subseteq W\) whenever \(n\ge k\). Hence \(x_n\to x\). But no \(f(x_n)\) lies in \(V\), so \(f(x_n)\) does not converge to \(f(x)\). This contradicts sequential continuity at \(x\). Therefore sequential continuity implies continuity at \(x\). Applying the pointwise result at every point proves the global statement when \(X\) is first-countable. \(\square\)
The construction in the converse is useful in practice: to witness discontinuity, fix one neighborhood \(V\) of the proposed image that cannot be reached by mapping any neighborhood of \(x\) entirely inside \(V\). Then choose a point outside \(f^{-1}(V)\) from each successively smaller basic neighborhood of \(x\). The resulting sequence approaches \(x\), while its images all stay outside \(V\).
Worked Example: Building a Sequence That Witnesses Discontinuity
Let \(X=Y=\mathbb{R}\), both with their standard topologies, and define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(0)=0\) and \(f(x)=1\) for \(x\ne0\). We test continuity at \(0\).
Take the open neighborhood \(V=(-1/2,1/2)\) of \(f(0)=0\). Every open neighborhood \(U\) of \(0\) contains an interval \((-\delta,\delta)\) for some \(\delta>0\). The point \(x=\delta/2\) lies in this interval and is nonzero, so \(f(x)=1\notin V\). Therefore no neighborhood \(U\) of \(0\) has \(f(U)\subseteq V\), and \(f\) is not continuous at \(0\).
The first-countable construction gives the simpler explicit sequence \(x_n=1/n\). For every \(\varepsilon>0\), choose \(N>1/\varepsilon\); then \(n\ge N\) gives \(|x_n-0|=1/n\le1/N<\varepsilon\), so \(x_n\to0\). For every \(n\), \(x_n\ne0\), and hence \(f(x_n)=1\). The image sequence does not converge to \(f(0)=0\), since none of its terms belongs to \(V=(-1/2,1/2)\). This directly checks the failure of sequential continuity.
At any \(x\ne0\), the interval \((x-|x|/2,x+|x|/2)\) excludes \(0\). On that neighborhood \(f\) is constantly \(1\), so \(f\) is continuous at \(x\). Thus \(0\) is the only point of discontinuity.
Sequentially Closed Sets and the Need for First Countability
A set \(A\subseteq X\) is called sequentially closed if every convergent sequence whose terms lie in \(A\) has its limit in \(A\). In a first-countable space, this test agrees with ordinary closedness.
Proof. Suppose \(A\) is closed and \(x_n\in A\) with \(x_n\to x\). If \(x\notin A\), the open set \(X\setminus A\) is a neighborhood of \(x\). Convergence would force all sufficiently late \(x_n\) into \(X\setminus A\), contradicting \(x_n\in A\). Thus \(x\in A\), and \(A\) is sequentially closed.
Conversely, suppose \(A\) is sequentially closed. If \(A\) were not closed, some \(x\in\overline{A}\setminus A\) would exist. First countability and Theorem 1 give a sequence in \(A\) converging to \(x\). Its limit would not lie in \(A\), contradicting sequential closedness. Hence \(A\) is closed. \(\square\)
Worked Example: Why the Hypothesis Cannot Be Dropped
Let \(X\) be an uncountable set with the countable-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are countable. Fix \(p\in X\) and set \(A=X\setminus\{p\}\). Every open neighborhood \(U\) of \(p\) has countable complement. Since \(A\) is uncountable, \(U\cap A\ne\varnothing\). Thus \(p\in\overline{A}\), even though \(p\notin A\), so \(A\) is not closed: \(X\setminus A=\{p\}\) is not open because its complement \(A\) is uncountable.
No sequence in \(A\) converges to \(p\). Indeed, let \((a_n)\) be any sequence in \(A\), and let \(C=\{a_n:n\in\mathbb{N}\}\) be its range. The set \(C\) is countable, so \(U=X\setminus C\) is open. It contains \(p\), since every \(a_n\ne p\), but it contains none of the terms \(a_n\). This violates the definition of convergence to \(p\). The set \(A\) is nevertheless sequentially closed: if a sequence in \(A\) converges to some point of \(X\), that point cannot be \(p\), by the argument just given, and so it belongs to \(A\).
To check the missing hypothesis, suppose there were a countable neighborhood base \(U_1,U_2,\ldots\) at \(p\). The union \(C=\bigcup_{n\in\mathbb{N}}(X\setminus U_n)\) is countable. Since \(X\) is uncountable, choose \(y\in X\setminus(C\cup\{p\})\). The set \(X\setminus\{y\}\) is an open neighborhood of \(p\), but every \(U_n\) contains \(y\), so no \(U_n\) is contained in \(X\setminus\{y\}\). This contradicts the definition of a neighborhood base. Thus \(X\) is not first-countable, and here a sequentially closed set need not be closed.
A Reliable Way to Use the Criteria
The hypotheses in these results matter. Continuity always implies preservation of convergent sequences, with no first-countability assumption. The reverse implication uses a countable neighborhood base at the point in the domain. Likewise, a sequence converging to \(x\) from \(A\) always proves \(x\in\overline{A}\), but first countability is needed for the reverse implication.
Common pitfall. It is incorrect to claim that every point of the closure of a set is the limit of a sequence from that set in every topological space. In the countable-complement example, \(p\in\overline{A}\), but no sequence in \(A\) converges to \(p\). The correction is to verify first countability at the point under consideration before using a sequence to detect closure or to turn sequential continuity into continuity.
For a continuity proof, start with an arbitrary sequence converging to the point and use continuity to show its images converge. For a discontinuity proof in a first-countable domain, start with a neighborhood \(V\) of the image that defeats continuity and choose, in each basic neighborhood \(U_n\), a point whose image lies outside \(V\). The directions are different: preservation of sequences proves continuity, while carefully chosen sequences expose its failure.
Check Your Understanding
For each question, identify where first countability is used and distinguish a necessary test from an equivalent criterion.
- State the sequential criterion for \(x\in\overline{A}\), including the local hypothesis at \(x\).
- If \(f:X\to Y\) is continuous and \(x_n\to x\), why must \(f(x_n)\to f(x)\) even when \(X\) is not first-countable?
- Describe how to construct a sequence witnessing that \(f\) is not continuous at a first-countable point \(x\).
- In the countable-complement example, why is \(A=X\setminus\{p\}\) sequentially closed but not closed?
- Give the distinction between a set being closed and being sequentially closed in an arbitrary topological space.