Limit Points and the Setting for Function Limits

The notion of limit point is the quiet backbone underneath nearly every rigorous statement about limits, continuity, and convergence in mathematics.

• Real analysis proper: the $\varepsilon$–$\delta$ definition of $\lim_{x\to c} f(x)$ is stated with the explicit hypothesis that $c$ is a limit point of the domain — this is exactly the concept this article was built to supply.
• Topology: limit points generalize verbatim to metric spaces and topological spaces, where a set is defined to be closed precisely when it contains all its limit points, giving the modern definition of closure $\overline{D} = D \cup D'$.
• Differential equations and dynamical systems: fixed points and limit cycles of dynamical systems are analyzed via accumulation behavior of trajectories, a direct descendant of limit-point reasoning applied to orbits rather than static sets.
• Numerical analysis and optimization: convergence proofs for iterative algorithms (Newton's method, gradient descent) show that generated sequences accumulate at a fixed point, invoking Bolzano–Weierstrass-style compactness arguments.
• Probability and statistics: the theory of limit points of sequences of random variables underlies almost-sure convergence and the study of accumulation points of empirical distributions.
• Computer science and cryptography: discretized approximation algorithms (e.g., in computational geometry or numerical root-finding) rely on the existence of accumulation points of bounded search sequences to guarantee termination or correctness bounds.

Ask the tutor: How does the requirement that $c$ be a limit point of the domain get built explicitly into the $\varepsilon$–$\delta$ definition of $\lim_{x\to c} f(x)$, and what would go wrong in the uniqueness-of-limits proof if we dropped it?

Think of $D$ as a scattering of points, or a continuum, laid out on the real line. A limit point $c$ is a place where $D$ "piles up": no matter how tightly you zoom in around $c$, you keep finding other points of $D$ nearby — never running out. This is a strictly stronger requirement than merely containing $c$ itself: the definition explicitly discards $c$ from consideration by removing it via the "$\setminus \{c\}$." That single set-subtraction is the entire subtlety of the definition, and it is what makes limit points the right foundation for function limits — a limit describes behavior near $c$, deliberately excluding what happens at $c$. Contrast this with an isolated point: a point of $D$ that has some breathing room, some neighborhood entirely to itself (within $D$). If $D = \{1, 2, 3\}$, every point is isolated — you can draw a small disk around $1$ containing no other point of $D$. But if $D = \{1/n : n \in \mathbb{N}\}$, the point $0$ is a limit point of $D$ even though $0 \notin D$: every neighborhood of $0$, however small, contains infinitely many terms $1/n$. Meanwhile every point $1/n \in D$ is isolated in $D$, since consecutive terms $1/n$ and $1/(n+1)$ are bounded away from each other.

The upshot for calculus: when we later define $\lim_{x \to c} f(x)$, we are asking "as $x$ ranges over $D \setminus \{c\}$ and approaches $c$, does $f(x)$ approach some value $L$?" This question is vacuous or ill-posed unless $D$ actually has points other than $c$ arbitrarily close to $c$ — precisely the condition that $c$ is a limit point of $D$. This is why every honest statement of the $\varepsilon$–$\delta$ definition includes the hypothesis that $c$ is a limit point of the domain.

This corollary is not circular: it invokes Bolzano–Weierstrass, a prior and independent consequence of completeness (via nested intervals or monotone subsequences), not the theorem being proved here.

Example 1 — A mixed set with an interval, isolated points, and an "almost missed" limit point. Let

Determine the derived set $D'$ and identify the isolated points of $D$. Solution. We examine each piece. The interval $(0,1)$. Take any $c \in [0,1]$. We claim $c$ is a limit point of $D$. If $c \in (0,1)$, then for small enough $\varepsilon$ the whole neighborhood $V_\varepsilon(c)$ lies in $(0,1) \subseteq D$, and certainly contains points of $(0,1)$ other than $c$ (e.g. $c + \varepsilon/2$ when $\varepsilon$ small enough that $c+\varepsilon/2 <1$; more simply, $(0,1)$ has no isolated points because between any two reals there is another real — density of $\mathbb{R}$). If $c = 0$ or $c=1$ (the endpoints, not in $D$), every neighborhood of $c$ still meets $(0,1)$: e.g. for $c=0$, the sequence $x_n = 1/(n+2) \in (0,1)\subseteq D$ satisfies $x_n \to 0$, $x_n \ne 0$, so by the sequential characterization, $0 \in D'$. Similarly $1 \in D'$. For $c$ outside $[0,1]$, sufficiently small neighborhoods of $c$ miss $(0,1)$ entirely (since $(0,1)$ is bounded and $c$ is not in its closure), so such $c$ contributes no limit points from this piece. Hence this piece contributes exactly $[0,1]$ to $D'$. The isolated point $\{2\}$. Choose $\varepsilon = 1/2$. Then $V_{1/2}(2) = (1.5, 2.5)$. Does this meet $D \setminus \{2\}$? It does not meet $(0,1)$ (since $1.5 > 1$), and $3 + 1/n \geq 4$ for all $n\ge1$ so none of those points lie in $(1.5,2.5)$ either. Hence $V_{1/2}(2) \cap D = \{2\}$, confirming $2$ is an isolated point of $D$, not a limit point. The sequence $\{3 + 1/n\}$. As $n \to \infty$, $3 + 1/n \to 3$, and each term is in $D\setminus\{3\}$ (note $3 \notin D$, since the sequence never actually equals $3$). By the sequential characterization, $3$ is a limit point of $D$. Is $3$ the only limit point contributed here? Each individual term $3 + 1/n$ (for fixed $n$) is isolated in $D$: the gap to its neighbors $3+\frac{1}{n-1}$ and $3+\frac{1}{n+1}$ is strictly positive (a short computation: the gap to the next term is $\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} > 0$), so a small enough $\varepsilon$ isolates it. So this piece contributes only $\{3\}$ to $D'$, while every $3+1/n$ is an isolated point of $D$.

Conclusion:

Notice $D' \not\subseteq D$ (since $0,1,3 \notin D$) and $D \not\subseteq D'$ (since $2 \notin D'$, and none of $3+1/n \in D'$): limit points and domain membership are logically independent notions.

Example 2 — Why $\mathbb{Q}$ has every real number as a limit point. Let $D = \mathbb{Q}$, the rational numbers. Show that $D' = \mathbb{R}$, i.e., every real number is a limit point of $\mathbb{Q}$ — including irrational numbers, which are not even in $\mathbb{Q}$. Solution. Let $c \in \mathbb{R}$ be arbitrary (rational or irrational). We must produce, for every $\varepsilon>0$, a rational number $q$ with $0 < |q-c| < \varepsilon$. We use the density of $\mathbb{Q}$ in $\mathbb{R}$, itself a standard consequence of the Archimedean property: between any two distinct real numbers there lies a rational number. Concretely: fix $n \in \mathbb{N}$. Apply density of $\mathbb{Q}$ to the two distinct reals $c$ and $c + 1/n$ (distinct since $1/n \ne 0$) to obtain a rational $q_n$ with

Then $q_n \in \mathbb{Q}\setminus\{c\}$ (since $q_n > c$ strictly) and

By the squeeze theorem for sequences, $q_n \to c$. This is exactly a sequence in $D \setminus \{c\} = \mathbb{Q}\setminus\{c\}$ converging to $c$, so by the sequential characterization theorem, $c \in D' = \mathbb{Q}'$. Since $c \in \mathbb{R}$ was arbitrary, $\mathbb{Q}' = \mathbb{R}$. This is a striking illustration that a limit point of $D$ need not lie in $D$: no irrational number is in $\mathbb{Q}$, yet every irrational number is a limit point of $\mathbb{Q}$. It also shows $D'$ can be a strictly larger, "denser" set than $D$ itself.