When Can Sequences Detect the Topology?
In a metric space, neighborhoods of a point can be tested using balls of radius \(1/n\). A sequence can visit smaller and smaller such balls, so sequences often translate neighborhood statements into concrete choices of points. In an arbitrary topological space, there may be no countable collection of neighborhoods that captures all neighborhoods of a point. First-countability is the condition that supplies such a collection.
This condition is local: it concerns the neighborhoods available at each individual point, not a single countable collection that must describe the whole topology. We will see that it is enough to characterize closure by sequences and to test continuity by what happens to convergent sequences. Neither conclusion is valid in arbitrary topological spaces.
Countable Local Bases
A local base does not have to contain every neighborhood; it only has to provide a smaller one inside each neighborhood. A finite local base is also countable. When a local base is countable, we can list its members as \(U_1,U_2,\ldots\), allowing repetitions if it is finite. It is often useful to replace this list by a nested one.
Proof. List a countable local base as \(U_1,U_2,\ldots\). Define \(B_n=\bigcap_{j=1}^{n}U_j\). Each \(B_n\) is open, because it is a finite intersection of open sets, and contains \(x\), because every \(U_j\) contains \(x\). Also \(B_{n+1}\subseteq B_n\).
To check that the new collection is still a local base, let \(G\) be any open neighborhood of \(x\). Since the \(U_n\) form a local base, there is an index \(k\) such that \(U_k\subseteq G\). By construction, \(B_k\subseteq U_k\), so \(B_k\subseteq G\). Thus the \(B_n\) form a countable local base at \(x\), as required. \(\square\)
Worked Example: Metric Spaces Are First-Countable
Let \((X,d)\) be a metric space and fix \(x\in X\). Consider the open balls \(B_{1/n}(x)=\{y\in X:d(x,y)<1/n\}\), for \(n\geq 1\). They are open neighborhoods of \(x\). If \(G\) is any open neighborhood of \(x\), then some \(\varepsilon>0\) satisfies \(B_\varepsilon(x)\subseteq G\). Choose \(n\) so large that \(1/n<\varepsilon\). Then \(B_{1/n}(x)\subseteq B_\varepsilon(x)\subseteq G\).
Therefore the balls \(B_{1/n}(x)\) form a countable local base at \(x\). Since \(x\) was arbitrary, every metric space is first-countable. This recovers a useful feature of metric spaces without requiring that the space itself be countable.
Closure Characterized by Sequences
A point \(x\) belongs to the closure \(\overline{A}\) precisely when every open neighborhood of \(x\) meets \(A\). In a first-countable space, a nested countable local base lets us choose points of \(A\) that approach \(x\) in the topological sense of sequence convergence.
Proof. Suppose first that \(x\in\overline{A}\). Choose a nested countable local base \((B_n)\) at \(x\), using the proposition above. Since \(x\in\overline{A}\), every open neighborhood of \(x\) meets \(A\). In particular, \(B_n\cap A\ne\varnothing\) for every \(n\). Choose \(a_n\in B_n\cap A\).
We verify that \(a_n\to x\). Let \(G\) be any open neighborhood of \(x\). Since \((B_n)\) is a local base, there is \(k\) with \(B_k\subseteq G\). For every \(n\geq k\), nesting gives \(B_n\subseteq B_k\subseteq G\). Since \(a_n\in B_n\), it follows that \(a_n\in G\) for all \(n\geq k\). This is exactly convergence to \(x\).
Conversely, suppose a sequence \((a_n)\) in \(A\) converges to \(x\). Every open neighborhood \(G\) of \(x\) contains all sufficiently late terms of the sequence. At least one such term belongs to \(A\), so \(G\cap A\ne\varnothing\). Thus every open neighborhood of \(x\) meets \(A\), which means \(x\in\overline{A}\). \(\square\)
Worked Example: A Sequence Witnesses a Closure Point
In \(\mathbb{R}\) with its usual topology, let \(A=\{1/n:n\geq 1\}\) and take \(x=0\). For every \(\varepsilon>0\), choose \(n\) with \(1/n<\varepsilon\). Then \(1/n\in A\cap(-\varepsilon,\varepsilon)\), so every neighborhood of \(0\) meets \(A\), and \(0\in\overline{A}\).
The sequence \(a_n=1/n\) lies in \(A\) and converges to \(0\): given \(\varepsilon>0\), choose \(N\) with \(1/N<\varepsilon\). For \(n\geq N\), \(0<1/n\leq 1/N<\varepsilon\), so \(a_n\in(-\varepsilon,\varepsilon)\). This sequence gives the point-by-point witness promised by the theorem.
Sequential Continuity
Continuity at a point is defined through open neighborhoods: every neighborhood of the image must have a neighborhood of the original point mapped into it. First-countability lets us turn failure of that condition into a sequence. This gives a test for continuity that is expressed entirely in terms of convergent sequences.
Proof. First suppose \(f\) is continuous, and let \(x_n\to x\) in \(X\). Take any open neighborhood \(V\) of \(f(x)\). By continuity, \(f^{-1}(V)\) is open in \(X\) and contains \(x\). Since \(x_n\to x\), all sufficiently late \(x_n\) belong to \(f^{-1}(V)\). Thus all sufficiently late \(f(x_n)\) belong to \(V\), proving \(f(x_n)\to f(x)\).
Conversely, suppose \(f\) is sequentially continuous. We show that \(f\) is continuous at every \(x\in X\). If it were not continuous at some \(x\), the open-set characterization of continuity would give an open neighborhood \(V\) of \(f(x)\) such that no open neighborhood \(U\) of \(x\) satisfies \(f[U]\subseteq V\). Equivalently, every open neighborhood \(U\) of \(x\) contains some \(z\) with \(f(z)\notin V\).
Choose a nested countable local base \((B_n)\) at \(x\). For each \(n\), use the preceding property with \(U=B_n\) to choose \(x_n\in B_n\) such that \(f(x_n)\notin V\). The sequence \(x_n\) converges to \(x\): if \(U\) is any open neighborhood of \(x\), choose \(k\) with \(B_k\subseteq U\); then for \(n\geq k\), \(x_n\in B_n\subseteq B_k\subseteq U\).
Sequential continuity now implies \(f(x_n)\to f(x)\). Since \(V\) is an open neighborhood of \(f(x)\), all sufficiently late \(f(x_n)\) must lie in \(V\). This contradicts the choice \(f(x_n)\notin V\) for every \(n\). Therefore \(f\) is continuous at \(x\). Since \(x\) was arbitrary, \(f\) is continuous. \(\square\)
Worked Example: A Discontinuity Detected by a Sequence
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(0)=1\) and \(f(x)=0\) for \(x\ne 0\). The sequence \(x_n=1/n\) converges to \(0\), but \(f(x_n)=0\) for every \(n\). The image sequence is constantly \(0\), so it does not converge to \(f(0)=1\): for example, it never enters the open interval \((1/2,3/2)\) around \(1\).
Thus \(f\) is not sequentially continuous at \(0\), and consequently is not continuous there. The sequence supplies a direct witness to the failure that the neighborhood definition of continuity also detects.
Subspaces and the Limits of the Method
First-countability passes to subspaces. This matters because closure and continuity questions often concern a subset with its inherited topology rather than the whole space.
Proof. Let \(A\subseteq X\), where \(X\) is first-countable, and fix \(a\in A\). Take a countable local base \((B_n)\) at \(a\) in \(X\). The sets \(A\cap B_n\) are open neighborhoods of \(a\) in the subspace \(A\). If \(G\) is an open neighborhood of \(a\) in \(A\), the subspace topology gives an open set \(U\) in \(X\) with \(G=A\cap U\). Since \(a\in U\), some \(B_n\subseteq U\). Therefore \(A\cap B_n\subseteq A\cap U=G\). The sets \(A\cap B_n\) form a countable local base at \(a\) in \(A\). \(\square\)
The first-countability hypothesis in the two sequence theorems cannot simply be omitted. Here is a space in which closure points need not be limits of sequences from the set.
Worked Example: Closure Not Detected by Sequences
Let \(X\) be an uncountable set with the cocountable topology: its open sets are \(\varnothing\) and the subsets whose complements are countable. This space is not first-countable. To see why, fix \(x\in X\) and suppose \((U_n)\) were a countable local base at \(x\). Each complement \(C_n=X\setminus U_n\) is countable, so \(C=\bigcup_{n\geq 1}C_n\) is countable. Choose \(y\in X\setminus(C\cup\{x\})\), which is possible because \(X\) is uncountable. The open neighborhood \(X\setminus\{y\}\) of \(x\) must contain some \(U_n\), so \(U_n\subseteq X\setminus\{y\}\). This implies \(y\notin U_n\), or \(y\in C_n\), contradicting \(y\notin C\).
Now let \(A\subseteq X\) be uncountable and choose \(x\in X\setminus A\). Every open neighborhood of \(x\) has countable complement, so it must meet \(A\); otherwise \(A\) would be contained in that countable complement. Hence \(x\in\overline{A}\).
But no sequence in \(A\) converges to \(x\). The range \(D\) of any such sequence is countable and does not contain \(x\). Therefore \(X\setminus D\) is an open neighborhood of \(x\) that contains none of the sequence's terms. The sequence cannot converge to \(x\). Thus closure can contain a point that no sequence from \(A\) approaches in this topology.
What First-Countability Provides
The central tool is the countable local base: its members can be arranged to shrink, and one can make a choice at stage \(n\) that works inside the \(n\)-th neighborhood. This is precisely what turns a neighborhood condition into a sequence. It yields both the sequential characterization of closure and the sequential criterion for continuity, and it is inherited by subspaces.
A common pitfall is to assume that sequences detect all topological behavior merely because they do so in metric spaces. The cocountable example shows the issue: without a countable local base, a point can lie in a closure even though no sequence from the set converges to it. First-countability is a sufficient condition for the sequential tests proved here; it should not be silently assumed in an arbitrary topological space.
Check Your Understanding
Use the definitions and proved results to answer the following questions.
- What does a local base at \(x\) need to do for every open neighborhood of \(x\)?
- How does taking finite intersections turn a countable local base into a nested one?
- Where is first-countability used in the proof that closure can be characterized by sequences?
- Why does sequential continuity imply continuity when the domain is first-countable?
- Why does the uncountable set in the cocountable example have no countable local base at \(x\)?
- Does every sequence in an arbitrary topological space have at most one limit? Which separation condition guarantees uniqueness?