A Neighborhood-Based Diagnostic
Questions about closure, accumulation, isolation, and convergence can look similar because each concerns neighborhoods. Their tests are not interchangeable. In particular, closure concerns whether every neighborhood meets a set; accumulation excludes the point itself; and a sequence cluster point concerns every tail of a particular sequence. Keeping these tests separate prevents common errors about repeated terms, nonunique limits, and what sequences can detect.
Let \((X,\mathcal{T})\) be a topological space. An open neighborhood of \(x\in X\) is an open set containing \(x\). Recall that \(x\in\overline{A}\) exactly when every open neighborhood of \(x\) meets \(A\). A point \(x\) is an accumulation point of \(A\) when every open neighborhood of \(x\) meets \(A\setminus\{x\}\). A point \(x\in A\) is isolated in \(A\) when some open neighborhood of \(x\) meets \(A\) only in \(\{x\}\). Thus an isolated point belongs to \(A\) but is not an accumulation point of \(A\).
A sequence \((x_n)\) converges to \(x\) when, for every open neighborhood \(U\) of \(x\), there is \(N\in\mathbb{N}\) such that \(x_n\in U\) for all \(n\ge N\). A point \(x\) is a cluster point of \((x_n)\) when every open neighborhood of \(x\) meets every tail \(\{x_n:n\ge N\}\). A net is a family \((x_d)_{d\in D}\) indexed by a nonempty directed set \(D\): a preorder in which every two indices have a common upper bound. A net converges to \(x\) when it is eventually in every open neighborhood of \(x\).
Closure: Nets Always Work, Sequences Need a Hypothesis
Proof. If a net \((a_d)_{d\in D}\) in \(A\) converges to \(x\), every open neighborhood \(U\) of \(x\) eventually contains terms of the net. Since those terms belong to \(A\), \(U\cap A\ne\varnothing\). The neighborhood test for closure gives \(x\in\overline{A}\).
Conversely, suppose \(x\in\overline{A}\). Consider all pairs \((U,a)\) where \(U\) is an open neighborhood of \(x\) and \(a\in U\cap A\). This set of pairs is nonempty because every neighborhood meets \(A\). Order the pairs by \((U,a)\le(V,b)\) when \(V\subseteq U\). This is a preorder. Given \((U,a)\) and \((V,b)\), the open neighborhood \(U\cap V\) meets \(A\), so choose \(c\in (U\cap V)\cap A\). Then \((U\cap V,c)\) is an upper bound for both pairs. Thus the index set is directed. Define the net at \((U,a)\) to have value \(a\). All its terms lie in \(A\). For any open neighborhood \(W\) of \(x\), the index \((W,a)\) exists for some \(a\in W\cap A\), and every later pair \((V,b)\) has \(V\subseteq W\), hence \(b\in W\). The net is eventually in \(W\), so it converges to \(x\). \(\square\)
Sequences suffice under a countability condition. A space is first countable at \(x\) if \(x\) has a countable neighborhood base: a sequence \((B_n)\) of open neighborhoods such that every open neighborhood of \(x\) contains some \(B_n\).
Proof. If a sequence in \(A\) converges to \(x\), each neighborhood of \(x\) contains terms from \(A\), so the neighborhood test gives \(x\in\overline{A}\). For the converse, suppose \(x\in\overline{A}\), and let \((B_n)\) be a countable neighborhood base at \(x\). Set \(V_n=B_1\cap\cdots\cap B_n\). Each \(V_n\) is an open neighborhood of \(x\), so \(V_n\cap A\ne\varnothing\). Choose \(a_n\in V_n\cap A\) for each \(n\). If \(U\) is any open neighborhood of \(x\), there is \(k\) with \(B_k\subseteq U\). For every \(n\ge k\), \(a_n\in V_n\subseteq B_k\subseteq U\). Hence \(a_n\to x\). \(\square\)
The first theorem gives a net test in every space; the second explains why a sequence test is available in first-countable spaces. Without such a hypothesis, sequences can miss closure points.
Worked Example: Closure and Isolation in the Real Line
Use \(\mathbb{R}\) with its standard topology and let \(A=\{1/n:n\in\mathbb{N}\}\). Determine \(\overline{A}\), the accumulation points of \(A\), and the cluster points of the sequence \(x_n=1/n\).
First, \(0\) is an accumulation point. Every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\varepsilon\). Then \(0<1/n<\varepsilon\), so \(1/n\in A\setminus\{0\}\) lies in the neighborhood.
Each \(1/k\in A\) is isolated in \(A\). For \(k=1\), the interval \((3/4,5/4)\) meets \(A\) only at \(1\). For \(k\ge2\), the numbers immediately above and below \(1/k\) in the decreasing sequence are \(1/(k-1)\) and \(1/(k+1)\). The interval with endpoints \[ \frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+1}\right) \quad\text{and}\quad \frac{1}{2}\left(\frac{1}{k-1}+\frac{1}{k}\right) \] contains \(1/k\) and no other point of \(A\). Thus no point of \(A\) is an accumulation point of \(A\).
Now let \(y\in\mathbb{R}\setminus(A\cup\{0\})\). If \(y<0\), choose a sufficiently small interval about \(y\) lying below \(0\); it misses \(A\). If \(y>0\), choose \(N\) so large that \(1/N<y/2\). All terms \(1/n\) with \(n\ge N\) are below \(y/2\). The finitely many terms with \(n<N\) are all different from \(y\), so an interval about \(y\) can be chosen to miss those terms too. Shrinking it if necessary to lie above \(y/2\) gives a neighborhood missing all of \(A\). Therefore no such \(y\) lies in \(\overline{A}\).
Conclusion and check: \(\overline{A}=A\cup\{0\}\), and \(A\) has exactly one accumulation point, \(0\). The sequence \(x_n=1/n\) converges to \(0\), since for every \(\varepsilon>0\), \(n>1/\varepsilon\) implies \(|x_n-0|=1/n<\varepsilon\). It has no other cluster point: a neighborhood of any \(y\notin A\cup\{0\}\) misses all terms by the argument above, while each \(1/k\) has a neighborhood meeting the sequence only at the term \(x_k\), which is absent from every tail after \(k\). Thus its only cluster point is \(0\).
When Are Limits Unique?
Hausdorffness is the condition that any two distinct points have disjoint open neighborhoods. This separation condition guarantees uniqueness of limits; it is not a consequence of convergence alone.
Proof. Suppose a sequence or net converges to distinct points \(x\) and \(y\). Since the space is Hausdorff, there are disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Convergence to \(x\) means the sequence or net is eventually in \(U\); convergence to \(y\) means it is eventually in \(V\). For a sequence, take an index at least as large as both eventual thresholds. For a net, directedness gives an index above both eventual thresholds. The corresponding term must lie in \(U\cap V=\varnothing\), a contradiction. Thus the two limits cannot be distinct. \(\square\)
Worked Example: One Sequence with Two Limits
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). The constant sequence \(x_n=a\) converges both to \(a\) and to \(b\).
For \(a\), its open neighborhoods are \(\{a\}\) and \(X\), and every term lies in both. For \(b\), its only open neighborhood is \(X\), which also contains every term. Hence the sequence converges to each point. This does not contradict the proposition: the points cannot be separated by disjoint open neighborhoods, because every open neighborhood of \(b\) is \(X\), which intersects every neighborhood of \(a\).
There is a separate distinction between sequence cluster points and accumulation points of a set. The sequence is a net-like repeated list of terms, and both \(a\) and \(b\) are cluster points: every neighborhood of either point meets every tail, since each tail contains \(a\). But for the set \(A=\{a\}\), \(a\) is not an accumulation point: the neighborhood \(\{a\}\) contains no point of \(A\setminus\{a\}\). The point \(b\) is an accumulation point of \(A\), since its only neighborhood, \(X\), contains \(a\in A\setminus\{b\}\). In particular, repetition of a term does not make that term an accumulation point of the range.
A Space Where Sequences Miss a Closure Point
Worked Example: A Net Converges Where No Sequence Can
Let \(X=\mathbb{R}\) have the finite-complement topology: its open sets are \(\varnothing\) and all sets whose complement in \(\mathbb{R}\) is finite. Set \(p=0\) and \(A=\mathbb{R}\setminus\{0\}\).
Every open neighborhood \(U\) of \(0\) has finite complement. Since \(A\) is infinite, \(U\cap A\ne\varnothing\), so \(0\in\overline{A}\). In fact \(U\cap A\) is infinite: removing the finite set \(\mathbb{R}\setminus U\) from the infinite set \(A\) leaves an infinite set.
No sequence \((a_n)\) in \(A\) converges to \(0\). Its set of terms \(S=\{a_n:n\in\mathbb{N}\}\) is at most countable, so the set \(U=\mathbb{R}\setminus S\) is open and contains \(0\), because every \(a_n\ne0\). But \(U\) contains none of the sequence terms. Thus the sequence is not eventually in \(U\), contradicting convergence to \(0\).
For an explicit net, let \(D\) be the set of finite subsets \(F\subseteq A\), ordered by inclusion. It is directed because \(F\cup G\) is a finite subset of \(A\) containing any two indices \(F,G\). Define \(a_F\) to be the least positive integer not in \(F\). Such an integer exists because \(F\) is finite, and it lies in \(A\). Let \(U\) be any neighborhood of \(0\), with finite complement \(C=\mathbb{R}\setminus U\). Once \(F\) contains \(C\cap\mathbb{N}\), the least positive integer not in \(F\) cannot lie in \(C\); hence \(a_F\in U\). Therefore the net \((a_F)_{F\in D}\) converges to \(0\), as the net criterion predicts.
Check: the failure is not that \(0\) is outside the closure; every neighborhood of \(0\) meets \(A\). Rather, each proposed sequence has a single open neighborhood, the complement of its terms, that prevents convergence. The net can respond to each finite obstruction by moving beyond it.
Common Diagnostic Errors
Wrong calculation: treating “cluster point” as “limit.” In \(\mathbb{R}\), let \(x_n=(-1)^n\). It does not converge to \(1\): the neighborhood \((1/2,3/2)\) of \(1\) misses every even-indexed term, so no tail lies in that neighborhood. Yet \(1\) is a cluster point, because every neighborhood of \(1\) contains \(1\), and \(x_n=1\) for infinitely many indices. The correction is to test all sufficiently late terms for convergence, but only one term in every tail for a cluster point.
Wrong calculation: assuming every closure point is a sequential limit. In the finite-complement example, \(0\in\overline{A}\) but no sequence in \(A\) converges to \(0\). The correction is to use the sequence criterion only when first countability (or another sufficient hypothesis) is available; in an arbitrary space, the net criterion applies.
Wrong claim: limits are always unique. The constant sequence in the two-point example has two limits. The correction is to check a separation hypothesis: the uniqueness proposition uses Hausdorffness. Without it, neighborhoods of distinct points may fail to be disjoint.
For a new problem, first identify whether it asks about a set or a particular sequence or net. Then apply the relevant neighborhood test, recording whether the condition is “every neighborhood meets,” “meets away from the point,” “eventually lies in,” or “every tail meets.” Only after that should you invoke a sequence criterion or uniqueness result, with its hypotheses checked.
Check Your Understanding
For each question, state which neighborhood test or hypothesis you use.
- In the standard topology on \(\mathbb{R}\), find the closure and accumulation points of \(A=\{2+1/n:n\in\mathbb{N}\}\). Which points of \(A\) are isolated?
- Can a sequence converge to \(x\) while \(x\) is not an accumulation point of its range? Give a specific example and verify both claims.
- Why does the sequence criterion for closure require a condition such as first countability? Identify the closure point and the failed sequence test in the finite-complement example.
- Prove that if a sequence converges to \(x\), then \(x\) is a cluster point of that sequence.
- In the two-point topology above, determine whether the constant sequence \(x_n=b\) converges to \(a\), to \(b\), or to both.