Closure by Sequences in First-Countable Spaces

Why Countable Local Bases Matter

Closure is defined through neighborhoods: a point belongs to the closure of a set when every open neighborhood of the point meets the set. In many familiar spaces, this neighborhood-by-neighborhood condition can be replaced by a sequence. The reason is that a countable local base lets us organize the relevant neighborhoods into a sequence that becomes progressively smaller.

Throughout, \((X,\mathcal{T})\) is a topological space. A local base at \(x\) is a family of open neighborhoods of \(x\) such that every open neighborhood of \(x\) contains at least one member of the family. The space is first-countable if every point has a countable local base. A sequence \((a_n)\) converges to \(x\) if every open neighborhood of \(x\) contains all terms \(a_n\) for sufficiently large \(n\).

Closure test: A point \(x\) belongs to \(\overline{A}\) if and only if every open neighborhood of \(x\) intersects \(A\). We use this neighborhood characterization of closure below.

The sequence test does not require \(X\) to be Hausdorff, nor does it require the terms of a sequence to be distinct. It requires first countability for the direction that constructs a sequence from the assumption \(x\in\overline{A}\).

The Sequence Characterization of Closure

Theorem: Let \(X\) be first-countable, let \(A\subseteq X\), and let \(x\in X\). Then \(x\in\overline{A}\) if and only if there is a sequence \((a_n)\) with every \(a_n\in A\) and \(a_n\to x\).

Proof. First suppose \(x\in\overline{A}\). Choose a countable local base \((U_n)_{n\in\mathbb{N}}\) at \(x\). If a local base is finite, its members can be repeated to obtain such an indexed sequence. The sets \(U_n\) need not be nested, so define, for each \(n\in\mathbb{N}\),

V_n=\bigcap_{i=1}^{n}U_i.

Each \(V_n\) is a finite intersection of open neighborhoods of \(x\), hence is itself an open neighborhood of \(x\). Also \(V_{n+1}\subseteq V_n\). The sequence \((V_n)\) is still a local base: if \(W\) is any open neighborhood of \(x\), then \(U_j\subseteq W\) for some \(j\), and therefore \(V_j\subseteq U_j\subseteq W\).

Since \(x\in\overline{A}\), the closure test says that \(V_n\cap A\ne\varnothing\) for every \(n\). Choose \(a_n\in V_n\cap A\) for each \(n\). This is the only selection needed in the construction; it uses countable choice, as is standard in undergraduate topology. To verify convergence, let \(W\) be any open neighborhood of \(x\). Because \((V_n)\) is a local base, \(V_j\subseteq W\) for some \(j\). For every \(n\ge j\), nesting gives \(V_n\subseteq V_j\), so \(a_n\in V_n\subseteq W\). Thus \(a_n\to x\).

Conversely, suppose \(a_n\in A\) for every \(n\) and \(a_n\to x\). Let \(W\) be any open neighborhood of \(x\). By convergence, there is an \(N\in\mathbb{N}\) such that \(a_n\in W\) whenever \(n\ge N\). In particular, \(a_N\in W\cap A\), so \(W\cap A\ne\varnothing\). The closure test now gives \(x\in\overline{A}\). \(\square\)

The nested neighborhoods in the first half of the proof are important: after a term has been chosen from \(V_n\), all later terms are chosen from subsets of \(V_n\). Consequently, once one of these neighborhoods lies inside a given neighborhood \(W\), every sufficiently late term lies in \(W\) as well.

Worked Examples

Worked Example: Closure in the Real Line

Take \(\mathbb{R}\) with its standard topology and let \(A=(0,1)\cup\{2\}\). We will show that \(\overline{A}=[0,1]\cup\{2\}\) using sequences for points in the closure and neighborhoods for points outside it.

If \(x\in(0,1)\), the constant sequence \(a_n=x\) has every term in \(A\) and converges to \(x\). For the endpoint \(0\), take \(a_n=1/(n+1)\). Since \(n\ge1\), we have \(n+1\ge2\), so \(0<1/(n+1)\le1/2<1\), and every \(a_n\in(0,1)\). Also \(1/(n+1)\to0\). For the endpoint \(1\), take \(a_n=1-1/(n+1)\). The inequality \(0<1/(n+1)\le1/2\) gives \(1/2\le a_n<1\), so \(a_n\in(0,1)\), and \(a_n\to1\). Finally, the constant sequence \(a_n=2\) lies in \(A\) and converges to \(2\). The sequence characterization shows that all points of \([0,1]\cup\{2\}\) lie in \(\overline{A}\).

Now let \(x\notin[0,1]\cup\{2\}\). If \(x<0\), set \(r=-x/2>0\). The interval \((x-r,x+r)\) has right endpoint \(x/2<0\), so it misses \(A\). If \(x>2\), set \(r=(x-2)/2>0\). The interval \((x-r,x+r)\) has left endpoint \((x+2)/2>2\), so it misses \(A\). If \(1<x<2\), take \(r=\min(x-1,2-x)/2>0\); then \((x-r,x+r)\subset(1,2)\), which misses \(A\). In each case there is a neighborhood of \(x\) disjoint from \(A\), so \(x\notin\overline{A}\).

Conclusion and check: \(\overline{A}=[0,1]\cup\{2\}\). Every point listed has a sequence from \(A\) converging to it, and every point outside the list has an open neighborhood missing \(A\), verifying both inclusions.

Worked Example: A Non-Hausdorff Space

Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and take \(A=\{a\}\). This space is first-countable: \(\{\{a\}\}\) is a local base at \(a\), and \(\{X\}\) is a local base at \(b\).

At \(a\), every open neighborhood contains \(a\), and therefore intersects \(A\). At \(b\), the only open neighborhood is \(X\), which also intersects \(A\). The closure test gives \(\overline{A}=X\). There is a sequence from \(A\) converging to each point: the constant sequence \(a_n=a\) converges to \(a\), and it also converges to \(b\), because the only open neighborhood of \(b\) is \(X\), which contains every term.

Check: The sequence characterization predicts that \(b\in\overline{A}\) because \(a_n=a\in A\) for every \(n\) and \(a_n\to b\). This example also shows why the theorem does not assert uniqueness of sequence limits: the same sequence converges to both \(a\) and \(b\).

Worked Example: The Discrete Topology

Let \(X=\mathbb{Z}\) with the discrete topology, in which every subset is open, and let \(A\subseteq\mathbb{Z}\). Each point \(x\) has the one-element local base \(\{\{x\}\}\), so this space is first-countable.

If \(x\in A\), the constant sequence \(a_n=x\) lies in \(A\) and converges to \(x\). If \(x\notin A\), the open neighborhood \(\{x\}\) of \(x\) misses \(A\), so the closure test gives \(x\notin\overline{A}\). Equivalently, no sequence whose terms all lie in \(A\) can converge to \(x\): convergence would require every sufficiently late term to lie in \(\{x\}\), hence to equal \(x\), contrary to \(x\notin A\).

Conclusion and check: \(\overline{A}=A\) for every \(A\subseteq\mathbb{Z}\). The sequence test gives the same result: exactly the points of \(A\) are limits of constant sequences from \(A\), and no point outside \(A\) can be a limit of a sequence from \(A\).

Closed Sets and Sequence Limits

Corollary: In a first-countable space \(X\), a subset \(A\) is closed if and only if every sequence in \(A\) that converges in \(X\) has its limit in \(A\).

Proof. Suppose \(A\) is closed and \((a_n)\) is a sequence in \(A\) converging to \(x\in X\). The sequence characterization gives \(x\in\overline{A}\). Since \(A\) is closed, \(\overline{A}=A\), so \(x\in A\).

Conversely, suppose every convergent sequence in \(A\) has its limit in \(A\). Let \(x\in\overline{A}\). By the sequence characterization there is a sequence \((a_n)\) in \(A\) converging to \(x\). The assumed property gives \(x\in A\). Thus \(\overline{A}\subseteq A\); also \(A\subseteq\overline{A}\) for every set. Hence \(A=\overline{A}\), which means \(A\) is closed. \(\square\)

This criterion is useful because it turns closedness into a test about limits. Its first-countability hypothesis matters: the theorem above is what guarantees that every point of the closure can be reached as the limit of a sequence from the set. Without that guarantee, testing sequences alone need not detect every point of the closure.

A Common Pitfall: Choosing from an Unrefined Local Base

It is not enough to take an arbitrary countable local base \((U_n)\), choose \(a_n\in A\cap U_n\), and assume the resulting sequence converges. The index \(n\) does not ensure that \(U_n\) becomes smaller as \(n\) increases.

For a concrete example in standard \(\mathbb{R}\), consider the point \(0\), the set \(A=\{1/2\}\cup\{1/m:m\ge2\}\), and the neighborhoods

U_{2k-1}=(-1,1),\qquad U_{2k}=\left(-\frac{1}{k+1},\frac{1}{k+1}\right) \quad (k\in\mathbb{N}).

This is a local base at \(0\): given any open neighborhood of \(0\), it contains \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and for sufficiently large \(k\), \(1/(k+1)<\varepsilon\). But choose \(a_{2k-1}=1/2\) and \(a_{2k}=1/(k+2)\). Every odd term lies in \(A\cap U_{2k-1}\). For each even term, \(1/(k+2)<1/(k+1)\), so it lies in \(A\cap U_{2k}\). Nevertheless, the sequence does not converge to \(0\): the neighborhood \((-1/4,1/4)\) contains none of its odd terms, which all equal \(1/2\).

The correction is to use the nested intersections \(V_n=\bigcap_{i=1}^{n}U_i\) before selecting terms, as in the theorem’s proof. Alternatively, in this example, \(b_n=1/(n+2)\) is a direct valid choice from \(A\) converging to \(0\). The test is eventual membership in every neighborhood of the limit, not merely membership of the \(n\)-th term in the \(n\)-th neighborhood.

Check Your Understanding

Use the sequence characterization and its hypotheses to answer each question.

In standard \(\mathbb{R}\), find a sequence from \(A=(0,1)\) converging to \(0\), and explain why \(0\in\overline{A}\).
In the Sierpinski space from the example, which open neighborhoods of \(b\) must a sequence eventually enter to converge to \(b\)?
Prove that in a first-countable space, if \(x\in\overline{A}\setminus A\), then a sequence from \(A\) converging to \(x\) cannot be constant.
Explain where the nestedness of \(V_n\) is used in the proof that the constructed sequence converges.
In a discrete space \(X\), determine \(\overline{A}\) for an arbitrary subset \(A\subseteq X\), and justify the result using neighborhoods.