The Timpson Lyceum · General Topology (Point Set Topology)

Limit Points & Boundaries – Lecture 11 (Topology)

Overview

This lecture develops the topological notion of a limit point and uses it to give a clean characterization of the closure. We then port the calculus idea of a limit into general topology through convergent sequences, prove the prototype result $\lim_{n\to\infty} \tfrac{1}{n} = 0$ two ways (topologically and with the $\varepsilon$-style definition on $\mathbb{R}$), and connect limit points to sequences in standard $\mathbb{R}^n$. The lecture closes with Hausdorff spaces, where limits are unique, and a first look at the boundary of a set. The recurring theme is that limits, continuity, derivatives, and integrals are all special cases of ideas that topology states in full generality.

Key Concepts

Recap (interior, closure, Theorem 2.5).

• The interior $A^\circ$ is the largest open set contained in $A$ (the union of all open subsets of $A$); the closure $\overline{A}$ is the smallest closed set containing $A$ (the intersection of all closed sets containing $A$). Always $A^\circ \subseteq A \subseteq \overline{A}$.
• $A$ is open $\iff A = A^\circ$; $A$ is closed $\iff A = \overline{A}$.
• Theorem 2.5: $x \in \overline{A} \iff$ every neighborhood of $x$ meets $A$, i.e. $\forall\, U_x,\ U_x \cap A \neq \varnothing$. Contrapositive: $x \notin \overline{A} \iff$ some neighborhood of $x$ is disjoint from $A$.
• Complement rules: $(A^\circ)^c = \overline{A^c}$ and $(\overline{A})^c = (A^c)^\circ$.

Limit point. Let $A \subseteq X$. A point $x \in X$ is a limit point of $A$ if every neighborhood of $x$ meets $A$ at a point other than $x$: $x \in A' \iff \forall\, U_x,\ \big(U_x \setminus \{x\}\big) \cap A \neq \varnothing.$ The set of all limit points is denoted $A'$. The only difference from being a closure point is that the shared point may not be $x$ itself — so a limit point need not belong to $A$.

Examples (standard topology on $\mathbb{R}$).

• $A = \{\,1/n : n \in \mathbb{Z}^+\,\}$: the only limit point is $0$. Each point $1/n$ is not a limit point (a small enough interval isolates it), but every interval around $0$ catches infinitely many terms. Note $0 \notin A$, yet $0 \in A'$.
• $A = [0,1)$: every point of $[0,1]$ is a limit point, so $A' = [0,1]$ (including the missing endpoint $1$).
• $A = \mathbb{Q}$: every real number is a limit point, $A' = \mathbb{R}$, because the rationals are dense — every interval contains infinitely many rationals.

Theorem 2.8 (closure via limit points). $\overline{A} = A \cup A'$.

Corollary 2.9. $A$ is closed $\iff A' \subseteq A$ (a set is closed exactly when it contains all of its limit points).

Sequence. A sequence in $X$ is a function $f : \mathbb{Z}^+ \to X$, written $(x_n)$ where $x_n = f(n)$. Order matters (parentheses, not braces).

Convergence (general topology). $(x_n)$ converges to $x$ if $\lim_{n\to\infty} x_n = x \iff \forall\, U_x\ \exists\, N \in \mathbb{Z}^+\ \text{such that}\ x_n \in U_x\ \forall\, n \ge N.$ Intuition: for any neighborhood of $x$, the sequence is eventually trapped inside it. On standard $\mathbb{R}$ this specializes (open balls) to: $\forall\, \varepsilon > 0\ \exists\, N$ with $|x_n - x| < \varepsilon$ for all $n \ge N$ — exactly the calculus definition.

Limit points via sequences (standard $\mathbb{R}^n$). If $x$ is a limit point of $A$, then there exists a sequence in $A$ converging to $x$ (take $x_k \in B(x, 1/k) \cap A$).

Hausdorff spaces. $X$ is Hausdorff if any two distinct points have disjoint neighborhoods. In a Hausdorff space a convergent sequence has a unique limit. (The trivial/chaotic topology is not Hausdorff: there every sequence converges to every point.) All standard topologies on $\mathbb{R}^n$ are Hausdorff.

Boundary. The boundary of $A$ is $\partial A = \overline{A} \setminus A^\circ,$ the closure minus the interior — the points wedged between "in $A$" and "not in $A$." It is always a closed set (and in nice $\mathbb{R}^2$ sets, a closed loop), regardless of which boundary points belong to $A$.

Worked Example

Claim. In standard $\mathbb{R}$, $\displaystyle \lim_{n\to\infty} \frac{1}{n} = 0$.

Goal: for every $\varepsilon > 0$, find $N \in \mathbb{Z}^+$ so that $\left|\tfrac{1}{n} - 0\right| < \varepsilon$ for all $n \ge N$. Fix $\varepsilon > 0$. By the Archimedean property there is a positive integer $N$ with $N > \tfrac{1}{\varepsilon}$, equivalently $\tfrac{1}{N} < \varepsilon$. For any $n \ge N$, a larger denominator gives a smaller value, so $\tfrac{1}{n} \le \tfrac{1}{N} < \varepsilon$:

Thus $\left|\tfrac{1}{n} - 0\right| < \varepsilon$ for all $n \ge N$, so the sequence converges to $0$. $\blacksquare$

The same engine handles other limits, e.g. $\displaystyle \lim_{n\to\infty}\frac{n^2}{2n^2 + n} = \frac{1}{2}$: the $+n$ is negligible against $n^2$, so the ratio is driven arbitrarily close to (but never equal to) $\tfrac12$.

Summary

• A limit point of $A$ is a point every neighborhood of which meets $A$ at some other point; the set of these is $A'$, and limit points need not lie in $A$.
• Closure decomposition: $\overline{A} = A \cup A'$, and $A$ is closed $\iff A' \subseteq A$ (Theorem 2.8 / Corollary 2.9).
• A sequence $(x_n)$ converges to $x$ when it is eventually inside every neighborhood of $x$; on $\mathbb{R}$ this is the familiar $\varepsilon$-condition $|x_n - x| < \varepsilon$, proved via the Archimedean property for $1/n \to 0$.
• In standard $\mathbb{R}^n$, $x$ is a limit point of $A$ exactly when some sequence in $A$ converges to $x$.
• In a Hausdorff space, limits are unique; the trivial topology fails this, with every sequence converging everywhere.
• The boundary is $\partial A = \overline{A} \setminus A^\circ$ — always closed, sitting precisely between the interior and the exterior of $A$.