From Neighborhoods to a Sequence
A neighborhood argument can involve infinitely many conditions: a point belongs to the closure of a set when every open neighborhood meets that set. If there is a countable local base at the point, those conditions can be organized into a sequence of increasingly restrictive tests. Choosing one point to pass each test produces a sequence with a controlled limit.
Throughout, \((X,\mathcal{T})\) is a topological space. A sequence \((x_n)\) converges to \(x\) if every open neighborhood of \(x\) contains \(x_n\) for all sufficiently large \(n\). A point \(x\) lies in \(\overline{A}\) if every open neighborhood of \(x\) intersects \(A\).
The final sentence matters. The local-base condition says that at least one listed neighborhood refines any given neighborhood; it does not say that every later listed neighborhood does. For a sequence of choices to converge, we want the stronger tail condition that all sufficiently late neighborhoods lie inside any prescribed neighborhood. Finite intersections supply that condition.
Proof. Each \(V_n\) is open because it is a finite intersection of open sets, and it contains \(x\) because every \(B_i\) contains \(x\). Since the intersection defining \(V_{n+1}\) includes every set in the intersection defining \(V_n\), \(V_{n+1}\subseteq V_n\). Now let \(U\) be any open neighborhood of \(x\). By the local-base property, choose \(j\) with \(B_j\subseteq U\). For every \(n\ge j\), \(V_n\subseteq B_j\subseteq U\). Thus the \(V_n\) form a local base, and their tails eventually lie inside every neighborhood of \(x\). \(\square\)
Choosing Points to Witness Closure
Proof. Suppose first that \(x\in\overline{A}\), and refine a countable local base to the decreasing base \((V_n)\) above. Because \(V_n\) is an open neighborhood of \(x\), the closure condition gives \(V_n\cap A\ne\varnothing\) for every \(n\). Choose \(a_n\in V_n\cap A\). To verify convergence, let \(U\) be any open neighborhood of \(x\). There is \(j\) with \(V_j\subseteq U\), and for every \(n\ge j\), \(a_n\in V_n\subseteq V_j\subseteq U\). Hence \(a_n\to x\).
Conversely, suppose \(a_n\in A\) for every \(n\) and \(a_n\to x\). Let \(U\) be any open neighborhood of \(x\). By convergence, \(a_n\in U\) for all sufficiently large \(n\), so \(U\cap A\ne\varnothing\). Every open neighborhood of \(x\) therefore meets \(A\), which is exactly \(x\in\overline{A}\). \(\square\)
Worked Example: Approaching an Endpoint in the Real Line
Give a sequence in \(A=(0,1)\subseteq\mathbb{R}\) that witnesses \(0\in\overline{A}\), where \(\mathbb{R}\) has its standard topology.
For each \(n\in\mathbb{N}\), let \(V_n=(-1/n,1/n)\). These are open neighborhoods of \(0\), and \(V_{n+1}\subseteq V_n\). They form a local base: given any open neighborhood \(U\) of \(0\), it contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\); choose \(j\) with \(1/j<\varepsilon\), so \(V_j\subseteq U\).
Choose \(a_n=1/(n+1)\). For every \(n\ge1\), \(0<1/(n+1)<1\), so \(a_n\in A\). Also \(1/(n+1)<1/n\), so \(a_n\in V_n\). To check convergence directly, given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/(N+1)<\varepsilon\). For \(n\ge N\), \(0<a_n=1/(n+1)\le1/(N+1)<\varepsilon\), so \(|a_n-0|<\varepsilon\).
Check: The sequence lies in \(A\) and converges to \(0\). By the converse direction of the theorem, every open neighborhood of \(0\) meets \(A\), confirming \(0\in\overline{A}\).
Worked Example: A Local Base in the Sorgenfrey Line
Let \(\mathbb{R}_\ell\) denote the real line with the lower limit topology, whose basis consists of intervals \([a,b)\) with \(a<b\). Show that \(0\) lies in the closure of \(A=(0,1)\) and construct a sequence in \(A\) converging to \(0\) in this topology.
The basic neighborhoods of \(0\) are \([0,b)\) with \(b>0\). Set \(V_n=[0,1/n)\). These sets are open neighborhoods of \(0\), and \(V_{n+1}\subseteq V_n\). They form a local base: if \(U\) is an open neighborhood of \(0\), it contains a basic neighborhood \([0,b)\) for some \(b>0\); choose \(j\) with \(1/j\le b\), giving \(V_j\subseteq[0,b)\subseteq U\).
Choose \(a_n=1/(n+1)\). The inequalities \(0<1/(n+1)<1\) show \(a_n\in A\), and \(1/(n+1)<1/n\) shows \(a_n\in V_n\). If \(U\) is any open neighborhood of \(0\), the local-base property gives \(V_j\subseteq U\). For every \(n\ge j\), \(a_n\in V_n\subseteq V_j\subseteq U\), so \(a_n\to0\) in \(\mathbb{R}_\ell\). The sequence theorem then gives \(0\in\overline{A}\). In particular, the convergence check here uses the lower limit neighborhoods, not just the usual distance on \(\mathbb{R}\).
Using Sequences to Witness Discontinuity
The same selection idea can turn a failed continuity condition into a sequence. Continuity at \(x\) means that for every open neighborhood \(W\) of \(f(x)\), there is an open neighborhood \(U\) of \(x\) such that \(f(U)\subseteq W\). If continuity fails, one fixed neighborhood \(W\) fails this test for every neighborhood of \(x\). A countable local base lets us choose a counterexample point from each of its neighborhoods.
Proof. First suppose \(f\) is not continuous at \(x\). By the definition of continuity at a point, there is an open neighborhood \(W\) of \(f(x)\) such that no open neighborhood \(U\) of \(x\) satisfies \(f(U)\subseteq W\). Refine a countable local base at \(x\) to a decreasing one \((V_n)\). For each \(n\), \(f(V_n)\not\subseteq W\), so choose \(x_n\in V_n\) with \(f(x_n)\notin W\). The nested-base property proves \(x_n\to x\): every neighborhood of \(x\) contains all sufficiently late \(V_n\), and hence all sufficiently late \(x_n\). But \(f(x_n)\) never enters the neighborhood \(W\) of \(f(x)\), so it does not converge to \(f(x)\).
Now suppose \(f\) is continuous at \(x\) and \(x_n\to x\). Let \(W\) be any open neighborhood of \(f(x)\). Continuity provides an open neighborhood \(U\) of \(x\) with \(f(U)\subseteq W\). Since \(x_n\to x\), all sufficiently late \(x_n\) lie in \(U\), and their images then lie in \(W\). This holds for every such \(W\), proving \(f(x_n)\to f(x)\). \(\square\)
Worked Example: Detecting a Discontinuity at Zero
Define \(f:\mathbb{R}\to\mathbb{R}\), with both copies of \(\mathbb{R}\) carrying the standard topology, by \(f(0)=0\) and \(f(x)=1\) for \(x\ne0\). Use a sequence to verify that \(f\) is not continuous at \(0\).
Take \(V_n=(-1/n,1/n)\) at \(0\), and choose \(x_n=1/(n+1)\). The calculation \(0<1/(n+1)<1/n\) gives \(x_n\in V_n\) for every \(n\), and the neighborhoods form a decreasing local base. Thus \(x_n\to0\). Since \(x_n>0\), the definition of \(f\) gives \(f(x_n)=1\) for every \(n\), whereas \(f(0)=0\).
For a direct convergence check, use the open neighborhood \((-1/2,1/2)\) of \(0\). No term \(f(x_n)=1\) belongs to it, since \(1\notin(-1/2,1/2)\). Therefore \(f(x_n)\) does not converge to \(f(0)\), and the proposition confirms that \(f\) is not continuous at \(0\).
Check: The same neighborhood also verifies discontinuity from the definition: for every open neighborhood \(U\) of \(0\), some nonzero \(x\in U\) exists, and then \(f(x)=1\notin(-1/2,1/2)\). Thus no such \(U\) maps entirely into that target neighborhood.
A Common Pitfall: The Base Must Control the Tail
A frequent error is to choose one point from each member of a countable local base and assume the resulting sequence converges. The local-base condition alone does not justify that conclusion: a small neighborhood may occur early in the list, while a large one reappears arbitrarily late. The finite-intersection refinement prevents this by ensuring that every sufficiently late set lies inside each previously chosen base neighborhood.
Worked Example: A Non-nested Enumeration That Misleads
In standard \(\mathbb{R}\), define \(B_{2k}=(-1/k,1/k)\) and \(B_{2k-1}=(-1,1)\) for \(k\in\mathbb{N}\). This is a countable local base at \(0\): given any open neighborhood \(U\) of \(0\), choose \(\varepsilon>0\) with \((-\varepsilon,\varepsilon)\subseteq U\), then choose \(k\) with \(1/k<\varepsilon\). It follows that \(B_{2k}\subseteq U\).
Now choose \(a_{2k}=0\) and \(a_{2k-1}=1/2\). For every even index, \(a_{2k}=0\in(-1/k,1/k)=B_{2k}\). For every odd index, \(a_{2k-1}=1/2\in(-1,1)=B_{2k-1}\). Thus each chosen point belongs to its corresponding base member.
Nevertheless, \((a_n)\) does not converge to \(0\). The open neighborhood \((-1/4,1/4)\) excludes every odd term \(a_{2k-1}=1/2\), and there are arbitrarily large odd indices. So the sequence is not eventually in this neighborhood. The mistaken argument would be “\(a_n\in B_n\), and the \(B_n\) form a local base, therefore \(a_n\to0\).” The correction is to replace \(B_n\) by \(V_n=\bigcap_{i=1}^{n}B_i\), then select \(a_n\in V_n\) when such witnesses are available.
This construction is useful whenever a topological condition has a neighborhood quantifier and the point has a countable local base. For closure, the witness at stage \(n\) is a point of \(A\cap V_n\); for discontinuity, it is a point in \(V_n\) whose image avoids one fixed target neighborhood. The nested refinement is the step that turns separate local choices into one sequence with a verified limit.
Check Your Understanding
For each question, identify the neighborhood condition and explain how the sequence construction applies.
- Given a countable local base \((B_n)\) at \(x\), define a decreasing local base using finite intersections and prove its local-base property.
- If \(x\in\overline{A}\) and \(x\) has a countable local base, why must each set in a nested local base meet \(A\)?
- Why does a sequence in \(A\) converging to \(x\) imply \(x\in\overline{A}\) even when \(x\) has no countable local base?
- In the nonnested example, identify one open neighborhood of \(0\) that contains infinitely many terms but not all sufficiently late terms.
- In the discontinuity proposition, explain why avoiding one fixed open neighborhood of \(f(x)\) proves that the image sequence does not converge to \(f(x)\).