Limit Points and Sequences

Limit points and sequential convergence are the load-bearing walls of analysis: every proof that a subset of $\mathbb{R}^n$ is closed, that a function is continuous, or that an optimization algorithm's iterates settle down, ultimately checks a limit-point or sequence condition. In real and complex analysis, the fact that closure equals "set plus accumulation points" underlies theorems about boundary values of holomorphic functions and the identity theorem (a holomorphic function's zero set having a limit point inside its domain forces the function to be identically zero). In dynamical systems, $\omega$-limit sets of trajectories are exactly the derived-set idea applied to orbits, describing where a system's long-run behavior accumulates. In numerical analysis and optimization, convergence proofs for gradient descent, Newton's method, and fixed-point iterations are direct applications of the sequential definition of convergence, often in Banach or Hilbert spaces rather than $\mathbb{R}$. In statistics and probability, almost-sure convergence of estimators and the Glivenko–Cantelli theorem rest on sequences of random variables converging pointwise, a probabilistic layer built atop this exact topological scaffolding. In cryptography and coding theory, metric-space limit arguments appear in the analysis of lattice-based systems and error-correcting code distances, where "closeness" in a metric governs decoding guarantees. And in pure topology, this article's careful separation of "always true" (Part 1) from "true because of first countability" (Part 2) is exactly the fork in the road that leads toward Sequential Compactness (where every sequence has a convergent subsequence) and toward Completeness and Cauchy Sequences (where convergence is characterized intrinsically, without needing to know the limit in advance).

Ask the tutor: Can you show me why the derived set of the rationals $\mathbb{Q}$ inside $\mathbb{R}$ is all of $\mathbb{R}$, and what that implies about $\overline{\mathbb{Q}}$?

Think of $A$ as a cloud of dust in space. A limit point of $A$ is a location — possibly itself dust, possibly empty space — that dust particles crowd toward, no matter how tightly you zoom in. If you can find some magnification at which a location has no other dust nearby (other than possibly itself), it is not a limit point; it is either isolated dust or ordinary empty space. Convergence of a sequence is the same crowding phenomenon, but staged in time rather than in space: a moving particle $x_n$ eventually stays inside every zoom-level neighborhood of its destination $L$. In a metric space these two pictures are forced to agree, because you can always build a sequence out of a limit point by zooming in with balls of radius $1/n$ and picking a witness point in each ball. This is precisely why metric spaces are called first-countable: at every point there is a countable "zoom sequence" of neighborhoods, $B(x,1), B(x,1/2), B(x,1/3), \dots$, that eventually sits inside any given neighborhood of $x$. It is this countable zoom sequence that manufactures a genuine sequence of points from purely "every-neighborhood" data — and it is exactly this feature that fails in more exotic (non-first-countable) spaces.

:::svg[Limit points vs. isolated points of $A = \{1/n : n \in \mathbb{N}\} \cup \{0\} \cup [3,4]$]

Example 1 — a bounded, non-symmetric set of rationals in $\mathbb{R}$. Let

with the standard metric $d(x,y) = |x-y|$. We find $A'$ and $\overline{A}$. The sequence $5 - 1/n$. As $n \to \infty$, $5 - 1/n \to 5$. Given $\varepsilon>0$, choosing $N > 1/\varepsilon$ gives $|(5-1/n) - 5| = 1/n < \varepsilon$ for $n \ge N$, confirming convergence to $5$ directly from the definition. Every ball $B(5,\varepsilon)$ contains infinitely many terms $5-1/n$ (all $n > 1/\varepsilon$), all distinct from $5$ itself and from each other; hence $5 \in A'$. Note $5 \notin A$ (the sequence never equals $5$), so $5$ is a limit point of $A$ that is not in $A$. Each individual point $5 - 1/n_0$ is isolated in $A$: choose $\varepsilon_0 < \min\left(\frac{1}{n_0} - \frac{1}{n_0+1},\ \frac{1}{n_0-1}-\frac{1}{n_0}\right)$ (for $n_0 \ge 2$; for $n_0=1$ use $\varepsilon_0 < 1 - \frac12 = \frac12$ and also small enough to avoid $7$ and $(-2,1)$, e.g. $\varepsilon_0 = 0.1$). Then $B(5-1/n_0,\varepsilon_0)$ contains no other point of the sequence, no point of $(-2,1)$ (since $5-1/n_0 > 4$ while the interval tops out below $1$), and not $7$ (distance $\ge 1.9$). So $B(5-1/n_0,\varepsilon_0) \cap A = \{5-1/n_0\}$, confirming isolation. The point $7$. Choosing $\varepsilon = 1$, the ball $B(7,1) = (6,8)$ contains no other point of $A$ (the sequence $5-1/n$ stays below $5$, and $(-2,1)$ is far away). So $7$ is isolated in $A$; $7 \in A$ but $7 \notin A'$. The interval $(-2,1)\cap\mathbb{Q}$. Let $x \in [-2,1]$ be arbitrary (including irrational $x$, and including the endpoints). Given any $\varepsilon > 0$, the ball $B(x,\varepsilon) = (x-\varepsilon,x+\varepsilon)$ overlaps $(-2,1)$ in a nonempty open subinterval (since $x \in [-2,1]$ is at distance $0$ from the closed interval $[-2,1]$, standard density arguments show this overlap is nonempty for any $\varepsilon>0$), and every nonempty open interval contains a rational number distinct from $x$ (density of $\mathbb{Q}$ in $\mathbb{R}$). Hence $(B(x,\varepsilon)\setminus\{x\}) \cap A \ne \varnothing$, so $x \in A'$ for every $x \in [-2,1]$ — including the two endpoints $-2$ and $1$, which are not themselves in $A$ (since $A$ only takes rationals strictly between $-2$ and $1$). Conversely, if $x \notin [-2,1]$, then $x$ has positive distance $\delta$ from the closed interval $[-2,1]$, and $B(x,\delta/2)$ misses $(-2,1)$ entirely, misses $7$ if $x \ne 7$ appropriately small $\delta/2$ chosen, and misses all but finitely many (possibly zero) terms $5-1/n$; shrinking $\delta$ further if needed to avoid these finitely many terms and the point $7$ (each handled by taking the minimum of finitely many positive distances) shows $x \notin A'$ unless $x = 5$ (already covered) or $x=7$ (shown isolated) or $x$ is one of the isolated $5-1/n$ (also shown isolated, hence not a limit point of $A$). Putting it together:

This matches Theorem, Part (1) exactly, and illustrates that limit points can lie outside $A$ (the endpoints $-2,1$ and the point $5$), while points of $A$ itself can be isolated (each $5-1/n$, and $7$).

Example 2 — sequential characterization in $\mathbb{R}^2$ with a non-convex set. Let $X = \mathbb{R}^2$ with the Euclidean metric, and let

the classic "topologist's sine curve" piece. We show the origin $(0,0)$ is in $\overline{A}$ by explicitly building a convergent sequence, per Theorem Part (2). Take $x_n = \dfrac{1}{n\pi + \pi/2}$ for $n = 1,2,3,\dots$ so that $\sin(1/x_n) = \sin(n\pi + \pi/2) = \cos(n\pi) = (-1)^n$. Define

(Check $0 < x_n \le 1/\pi$ for all $n\ge1$: indeed $x_1 = 1/(3\pi/2) = 2/(3\pi) < 1/\pi$, and $x_n$ decreases as $n$ grows, so the bound holds for all $n \ge 1$.) As $n \to \infty$, $x_n \to 0$ but the $y$-coordinate $(-1)^n$ oscillates between $-1$ and $1$ and does not converge. So this particular sequence $a_n$ does not converge to $(0,0)$ — it doesn't converge at all, since the $y$-coordinates form the non-convergent sequence $(-1)^n$. This is instructive: not every sequence built from $A$ works, and we must choose more carefully to witness $(0,0) \in \overline{A}$. Instead take $b_n = \left(\dfrac{1}{2n\pi},\ \sin(2n\pi)\right) = \left(\dfrac{1}{2n\pi},\ 0\right) \in A$ for $n \ge 1$ (valid since $\sin(2n\pi)=0$ and $0 < \frac{1}{2n\pi} \le \frac{1}{2\pi} < \frac1\pi$). Now

as $n\to\infty$: given $\varepsilon>0$, pick $N > \frac{1}{2\pi\varepsilon}$, so $n \ge N$ gives $d(b_n,(0,0)) = \frac{1}{2n\pi} < \varepsilon$. Hence $b_n \to (0,0)$, with every $b_n \in A$. By Theorem Part (2) ($\Leftarrow$ direction), this proves $(0,0) \in \overline{A}$ — even though $(0,0) \notin A$ (since $A$ requires $x>0$), so $(0,0)$ is a genuine limit point not belonging to $A$, matching Part (1) as well: $(0,0) \in A' \subseteq \overline{A}$. This example also shows the value of Part (2) as a tool: rather than checking "every open set meets $A$" directly (a universally-quantified, harder-to-visualize condition), we exhibit one explicit sequence, which is often easier to construct and verify by direct $\varepsilon$–$N$ estimates.