Why Sequences Can Detect Closure
Closure is defined using all neighborhoods of a point, while a sequence gives only a countable list of points. A countable local base bridges these descriptions: it provides a countable list of neighborhoods sufficient to test every neighborhood. The key step is to arrange that list so each neighborhood in it is contained in the previous ones.
Throughout, \((X,\mathcal{T})\) is a topological space, \(A\subseteq X\), and an open neighborhood of \(x\) means an open set containing \(x\). A sequence \((a_n)\) converges to \(x\) if every open neighborhood of \(x\) contains all terms from some point onward. A sequence in \(A\) means that \(a_n\in A\) for every \(n\in\mathbb{N}\).
The condition is local: to characterize closure at a particular point, it is enough for that point to have a countable local base. First-countability of the whole space is a convenient sufficient condition, but the proof below uses no local base at any other point.
Making the Local Base Decreasing
Proof. List a countable local base as \((U_n)_{n\in\mathbb{N}}\). If the base is finite, repeat its members to make an infinite list. Define \(V_n=\bigcap_{k=1}^{n}U_k\). This is a finite intersection of open sets, so it is open, and it contains \(x\) because each \(U_k\) contains \(x\). Also, \(V_{n+1}=V_n\cap U_{n+1}\subseteq V_n\). To verify the local-base property, let \(U\) be any open neighborhood of \(x\). Some \(U_j\) is contained in \(U\), by the local-base property of \((U_n)\). Then \(V_j\subseteq U_j\subseteq U\). Thus \((V_n)\) is a decreasing countable local base at \(x\). \(\square\)
The finite intersections matter: the original local base need not be nested. The refinement guarantees that once a selected sequence term lies in \(V_n\), every later term selected from \(V_{n+1},V_{n+2},\ldots\) lies there as well.
The Sequence Characterization of Closure
Recall that \(\overline{A}\) is the intersection of all closed subsets of \(X\) that contain \(A\). The neighborhood test says that \(x\in\overline{A}\) exactly when every open neighborhood of \(x\) intersects \(A\). Indeed, if a neighborhood \(U\) misses \(A\), then the closed set \(X\setminus U\) contains \(A\) but not \(x\), so \(x\notin\overline{A}\). Conversely, if \(x\notin\overline{A}\), the open set \(X\setminus\overline{A}\) contains \(x\) and misses \(A\).
Proof. Suppose first that \(x\in\overline{A}\). Let \((V_n)\) be a decreasing countable local base at \(x\), obtained from the lemma. By the neighborhood test, each \(V_n\) intersects \(A\). For every \(n\in\mathbb{N}\), choose \(a_n\in V_n\cap A\). Let \(U\) be any open neighborhood of \(x\). Because \((V_n)\) is a local base, there is an index \(j\) such that \(V_j\subseteq U\). For every \(n\ge j\), decreasingness gives \(V_n\subseteq V_j\subseteq U\), and therefore \(a_n\in U\). This is exactly \(a_n\to x\).
Conversely, suppose a sequence \((a_n)\) in \(A\) converges to \(x\). Let \(U\) be any open neighborhood of \(x\). By convergence, there is an index \(N\) such that \(a_n\in U\) for every \(n\ge N\); in particular, \(a_N\in U\cap A\). Thus every open neighborhood of \(x\) intersects \(A\), and the neighborhood test gives \(x\in\overline{A}\). If \(A=\varnothing\), the neighborhood test shows that no point is in \(\overline{A}\), and there cannot be a sequence in \(A\); so the equivalence also covers this case. \(\square\)
The forward direction is where countability enters: it lets us choose one point of \(A\) for each of a countable list of neighborhoods. The reverse direction needs no countability assumption. Whenever a sequence in \(A\) converges to \(x\), it forces every neighborhood of \(x\) to meet \(A\).
Worked Example: Approaching an Endpoint in the Standard Real Line
Let \(\mathbb{R}\) have its standard topology, let \(A=(0,1)\), and consider \(x=0\). The intervals \(V_n=(-1/n,1/n)\), for \(n\in\mathbb{N}\), form a decreasing local base at \(0\). Each is an open neighborhood of \(0\), and every open neighborhood of \(0\) contains one of these intervals.
For each \(n\in\mathbb{N}\), set \(a_n=1/(n+1)\). Since \(n+1\ge2\), we have \(0<1/(n+1)\le1/2<1\), so \(a_n\in A\). Also, \(1/(n+1)<1/n\), so \(a_n\in V_n\).
To check convergence directly, let \(\varepsilon>0\). Choose \(N\in\mathbb{N}\) such that \(1/(N+1)<\varepsilon\). For every \(n\ge N\), \(0<|a_n-0|=1/(n+1)\le1/(N+1)<\varepsilon\). Thus \(a_n\to0\), and the theorem confirms \(0\in\overline{A}\).
Check: Every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\). The convergence calculation shows that \(a_n\) eventually belongs to that interval, so it intersects \(A\), as required by the neighborhood test.
Worked Example: The Sorgenfrey Line
Let \(\mathbb{R}_\ell\) denote the real line with the topology having basis all intervals \([a,b)\), where \(a<b\). Consider \(A=\{1/k:k\in\mathbb{N}\}\) and \(x=0\). The intervals \(V_n=[0,1/n)\) form a decreasing local base at \(0\). In fact, every basic neighborhood \([0,b)\) of \(0\), where \(b>0\), contains \(V_n\) once \(n\) is chosen sufficiently large that \(1/n\le b\).
Set \(a_n=1/(n+1)\). This is a point of \(A\), since its denominator \(n+1\) is in \(\mathbb{N}\). The inequality \(n+1>n>0\) gives \(0<1/(n+1)<1/n\), so \(a_n\in[0,1/n)=V_n\). If \(U\) is any open neighborhood of \(0\), it contains a basic interval \([0,b)\) for some \(b>0\). Choose \(N\) so that \(1/(N+1)<b\). For every \(n\ge N\), \(0\le a_n=1/(n+1)\le1/(N+1)<b\), so \(a_n\in[0,b)\subseteq U\). Hence \(a_n\to0\) in \(\mathbb{R}_\ell\), and \(0\in\overline{A}\).
This example uses the topology’s actual neighborhoods, not just distances in the usual metric. The decreasing local base makes the sequence test work in this topology as well.
Worked Example: Closure and Convergence in a Discrete Space
Let \(X=\{p,q\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), and let \(A=\{p\}\). For \(x=p\), the constant sequence \(a_n=p\) lies in \(A\) and converges to \(p\): every open neighborhood of \(p\) contains \(p\), so it contains every term. Thus \(p\in\overline{A}\).
For \(x=q\), the open neighborhood \(\{q\}\) misses \(A\), so \(q\notin\overline{A}\). No sequence in \(A\) can converge to \(q\): every term of such a sequence equals \(p\), and the neighborhood \(\{q\}\) contains none of its terms. Therefore \(\overline{A}=\{p\}=A\), in agreement with the theorem.
Check: In a discrete space, every singleton is open. A sequence converges to \(x\) only if it is eventually equal to \(x\), because it must eventually lie in the neighborhood \(\{x\}\). Conversely, if it is eventually equal to \(x\), it converges to \(x\), since every neighborhood of \(x\) contains \(x\).
Why First-Countability Cannot Be Omitted
The reverse implication in the theorem is valid in every topological space, but the forward implication can fail if the point has no countable local base. The following example verifies both the failure of the sequence test and the missing hypothesis.
Worked Example: An Uncountable Cocountable Space
Let \(X\) be an uncountable set with the cocountable topology: the 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, so \(U\) is uncountable. Removing the single point \(p\) still leaves points in \(U\cap A\). Thus every open neighborhood of \(p\) meets \(A\), and \(p\in\overline{A}\).
Now take any sequence \((a_n)\) in \(A\), and let \(R=\{a_n:n\in\mathbb{N}\}\) be its range. The set \(R\) is countable and does not contain \(p\). Therefore \(U=X\setminus R\) is an open neighborhood of \(p\). It contains no term \(a_n\), so the sequence does not converge to \(p\). Since this applies to every sequence in \(A\), closure at \(p\) is not detected by sequences here.
For completeness, \(p\) has no countable local base in this topology. Suppose \((V_n)_{n\in\mathbb{N}}\) were one. Each \(V_n\) is an open neighborhood of \(p\), hence uncountable, so choose \(y_n\in V_n\setminus\{p\}\). The set \(U=X\setminus\{y_n:n\in\mathbb{N}\}\) is an open neighborhood of \(p\), because its complement is countable. But \(y_n\in V_n\) and \(y_n\notin U\) for every \(n\), so no \(V_n\) is contained in \(U\). This contradicts the local-base property.
Common pitfall. The invalid inference is: “\(x\in\overline{A}\), so choose a sequence in \(A\) converging to \(x\).” Closure guarantees that each individual neighborhood intersects \(A\), but without a countable local base there may be no countable list of neighborhoods from which to make the choices sufficient for convergence. In the cocountable example, any proposed sequence has countable range \(R\); the explicitly computed neighborhood \(X\setminus R\) contains \(p\) and misses every term. The correction is to check the hypothesis: use the sequence characterization at a point with a countable local base, not at an arbitrary point.
The criterion is useful because sequence arguments are often concrete: convergence can be checked by testing eventual membership in neighborhoods, and closure can then be tested by constructing a sequence. The first-countability condition is precisely what licenses that replacement here; it is not a dispensable technical detail.
Check Your Understanding
Use the neighborhood definition of convergence and the closure test to answer the following.
- Given a countable local base \((U_n)\) at \(x\), define a decreasing local base using finite intersections and prove that it is still a local base.
- In the standard topology on \(\mathbb{R}\), let \(A=\{1/n:n\in\mathbb{N}\}\). Give a sequence in \(A\) converging to \(0\), and verify its convergence.
- In a discrete space, characterize the sequences that converge to a given point \(x\).
- In the cocountable example, explain why the complement of the range of any sequence in \(A\) is an open neighborhood of \(p\).
- Which direction of the sequence characterization holds in every topological space, even when no countable local base is available?