A set can fail to contain a point and still press against it from every direction allowed by the surrounding topology. In $\mathbb{R}$, consider the set \begin{align*} \{1, 1/2, 1/3, \dots\}. \end{align*} It never reaches $0$, but any open interval around $0$ catches some of its terms. The point $0$ is therefore not a member of the set, not an endpoint chosen by algebra, and not a visible dot in the list; it is a point forced by the way neighbourhoods inspect the set. This is the problem limit points solve. Topology does not begin with distances, formulas, or parametrisations. It begins with the question of which subsets count as neighbourhoods of a point. Once that question is fixed, a second question becomes unavoidable: when does a point have other points of a set arbitrarily close to it? The answer is the notion of a limit point, also called an accumulation point or cluster point in many texts. [example: The Missing Point of $\{1/n\}$] Let $A \subsetneq \mathbb{R}$ be given by \begin{align*} A = \{1/n : n \in \mathbb{N}\} \end{align*} with the usual topology. Since $1/n > 0$ for every $n \in \mathbb{N}$, the point $0$ is not in $A$. We show that $0$ is nevertheless a limit point of $A$. Let $\varepsilon > 0$. Choose $n \in \mathbb{N}$ with \begin{align*} n > \frac{1}{\varepsilon}. \end{align*} Because $\varepsilon > 0$ and $n>0$, multiplying the inequality by $\varepsilon/n$ gives \begin{align*} \frac{1}{n} < \varepsilon. \end{align*} Also $1/n \in A$ and $1/n \ne 0$, so \begin{align*} \frac{1}{n} \in A \setminus \{0\}. \end{align*} Since $0 < 1/n < \varepsilon$, we have \begin{align*} \frac{1}{n} \in (-\varepsilon,\varepsilon). \end{align*} Therefore \begin{align*} (-\varepsilon,\varepsilon) \cap (A \setminus \{0\}) \ne \varnothing. \end{align*} As this holds for every $\varepsilon > 0$, every usual open interval around $0$ contains a point of $A$ different from $0$, so $0$ is a limit point of $A$. By contrast, $1 \in A$ because $1 = 1/1$, but $1$ is not a limit point of $A$. The interval $(3/4,5/4)$ is a neighbourhood of $1$. If $1/n \in A$ and $n=1$, then $1/n=1$; if $n \ge 2$, then \begin{align*} \frac{1}{n} \le \frac{1}{2} < \frac{3}{4}, \end{align*} so $1/n \notin (3/4,5/4)$. Hence \begin{align*} (3/4,5/4) \cap (A \setminus \{1\}) = \varnothing. \end{align*} Thus membership in $A$ and being a limit point of $A$ are different local properties. [/example] The example separates two ideas that are often conflated at first contact. Belonging to a set is a membership statement. Being a limit point is a neighbourhood statement. A point of the set may fail to be a limit point, and a point outside the set may be a limit point. The rest of the chapter is about making that distinction precise and then using it to recognize closure, convergence, density, compactness, and separation phenomena. ## Local Language ### Neighbourhoods Before a point can be called a limit point, we need to say what it means to look near a point without necessarily using a metric. The topology supplies open sets, and open sets supply the local tests that replace intervals and balls. [definition: Neighbourhood] Let $(X, \tau)$ be a [topological space](/page/Topological%20Space) and let $x \in X$. A subset $N \subset X$ is a neighbourhood of $x$ if there exists an [open set](/page/Open%20Set) $U \in \tau$ such that \begin{align*} x \in U \subset N. \end{align*} [/definition] ### Deleted Neighbourhoods If every point of $A$ counted as accumulated merely because it belongs to $A$, the concept would not detect crowding. We need a local test that looks near $x$ while ignoring $x$ itself; this is what separates an isolated member from a point approached by other members. [definition: Deleted Neighbourhood] Let $(X, \tau)$ be a topological space, let $x \in X$, and let $N \subset X$ be a neighbourhood of $x$. The corresponding deleted neighbourhood of $x$ is \begin{align*} N \setminus \{x\}. \end{align*} [/definition] ## Definition ### The Topological Test Deleted neighbourhoods give the exact language for the opening sequence approaching $0$. We want to know whether every local view around $x$ contains a point of $A$ that is not merely $x$ being counted again. [definition: Limit Point] Let $(X, \tau)$ be a topological space and let $A \subset X$. A point $x \in X$ is a limit point of $A$ if every neighbourhood $N$ of $x$ satisfies \begin{align*} (N \setminus \{x\}) \cap A \ne \varnothing. \end{align*} [/definition] The definition uses arbitrary neighbourhoods, but most calculations start with open sets. Since every neighbourhood of $x$ contains an open set containing $x$, the definition is equivalent to the following open-neighbourhood test: $x$ is a limit point of $A$ if and only if every open set $U \in \tau$ with $x \in U$ satisfies \begin{align*} (U \setminus \{x\}) \cap A \ne \varnothing. \end{align*} ### Metric Ball Test In metric spaces, checking every open neighbourhood directly is usually too broad for computation. Since the usual topology is generated by open balls, the local question becomes whether every sufficiently small ball around the candidate point contains a point of the set other than the candidate itself. This converts the topological definition into the inequality language used throughout analysis. [quotetheorem:5010] The metric ball test is not a different concept; it is the topological definition written with the topology generated by open balls. It is useful because inequalities can then replace general neighbourhood language. [example: An Isolated Member and an External Limit Point] Let \begin{align*} A = \{0\} \cup \{1 + 1/n : n \in \mathbb{N}\} \subsetneq \mathbb{R} \end{align*} with the usual topology. The point $0$ belongs to $A$, but it is not a limit point of $A$. Indeed, \begin{align*} A \setminus \{0\} = \{1 + 1/n : n \in \mathbb{N}\}. \end{align*} If $n \in \mathbb{N}$, then $n \ge 1$, so \begin{align*} \frac{1}{n} > 0 \end{align*} and therefore \begin{align*} 1 + \frac{1}{n} > 1 > \frac{1}{2}. \end{align*} Thus no point of $A \setminus \{0\}$ lies in $(-1/2,1/2)$, and hence \begin{align*} (-1/2,1/2) \cap (A \setminus \{0\}) = \varnothing. \end{align*} Since $(-1/2,1/2)$ is a neighbourhood of $0$, the point $0$ fails the limit point test. By contrast, $1 \notin A$: it is not $0$, and if $1 + 1/n = 1$, then subtracting $1$ gives \begin{align*} \frac{1}{n} = 0, \end{align*} which is impossible for $n \in \mathbb{N}$. We show that $1$ is nevertheless a limit point of $A$. Let $\varepsilon > 0$. Choose $n \in \mathbb{N}$ with \begin{align*} n > \frac{1}{\varepsilon}. \end{align*} Because $n>0$ and $\varepsilon>0$, this implies \begin{align*} \frac{1}{n} < \varepsilon. \end{align*} Also $1/n>0$, so \begin{align*} 1 < 1+\frac{1}{n} < 1+\varepsilon. \end{align*} Since $\varepsilon>0$, we also have \begin{align*} 1-\varepsilon < 1 < 1+\frac{1}{n}. \end{align*} Combining the two inequalities gives \begin{align*} 1-\varepsilon < 1+\frac{1}{n} < 1+\varepsilon, \end{align*} so \begin{align*} 1+\frac{1}{n} \in (1-\varepsilon,1+\varepsilon). \end{align*} The same point lies in $A$, and it is not equal to $1$ because $1/n>0$. Hence \begin{align*} (1-\varepsilon,1+\varepsilon) \cap (A \setminus \{1\}) \ne \varnothing. \end{align*} As this holds for every $\varepsilon>0$, every usual open interval around $1$ contains a point of $A$ different from $1$, so $1$ is a limit point of $A$. Thus this set contains an isolated member at $0$ while forcing an external limit point at $1$. [/example] This example gives the basic mental model: limit points measure local crowding, not membership. A set can contain lonely points, and it can force nearby points that it never contains. ## Closure and Isolated Points The closure of a set records all points that cannot be separated from the set by an open neighbourhood. Limit points refine this idea by separating the points already in the set from points approached by other points of the set. This refinement is what allows closure to be decomposed into membership plus accumulation. A point of a set that fails the limit point test deserves its own name, because such points behave like removable atoms of the set. We need terminology for members that are present by membership alone rather than forced by the surrounding members. [definition: Isolated Point] Let $(X, \tau)$ be a topological space and let $A \subset X$. A point $x \in A$ is an isolated point of $A$ if there exists a neighbourhood $N$ of $x$ such that \begin{align*} N \cap A = \{x\}. \end{align*} [/definition] An isolated point is a member of the set that has a private neighbourhood relative to that set. It is the opposite local behaviour from being a limit point while inside $A$. [example: Integers Inside the Real Line] In $\mathbb{R}$ with the usual topology, every integer is isolated as a point of $\mathbb{Z}$. Let $m \in \mathbb{Z}$ and consider the open interval \begin{align*} I_m = (m-1/2,m+1/2). \end{align*} Since $m-1/2 < m < m+1/2$, we have $m \in I_m \cap \mathbb{Z}$. Conversely, if $z \in I_m \cap \mathbb{Z}$, then \begin{align*} m-\frac{1}{2} < z < m+\frac{1}{2}. \end{align*} Subtracting $m$ from each part gives \begin{align*} -\frac{1}{2} < z-m < \frac{1}{2}. \end{align*} Because $z,m \in \mathbb{Z}$, the difference $z-m$ is an integer. The only integer strictly between $-1/2$ and $1/2$ is $0$, so \begin{align*} z-m = 0, \end{align*} and hence $z=m$. Therefore \begin{align*} (m-1/2,m+1/2) \cap \mathbb{Z} = \{m\}. \end{align*} Thus each integer has a neighbourhood whose intersection with $\mathbb{Z}$ is only itself. Now let $x \in \mathbb{R} \setminus \mathbb{Z}$. Choose $k \in \mathbb{Z}$ with \begin{align*} k < x < k+1. \end{align*} Set \begin{align*} \delta = \frac{1}{2}\min\{x-k,k+1-x\}. \end{align*} Both $x-k$ and $k+1-x$ are positive, so $\delta>0$. Since $\delta \le (x-k)/2$, \begin{align*} x-\delta \ge x-\frac{x-k}{2} = \frac{x+k}{2} > k. \end{align*} Since $\delta \le (k+1-x)/2$, \begin{align*} x+\delta \le x+\frac{k+1-x}{2} = \frac{x+k+1}{2} < k+1. \end{align*} Hence \begin{align*} (x-\delta,x+\delta) \subset (k,k+1), \end{align*} and no integer lies strictly between two consecutive integers $k$ and $k+1$. Therefore \begin{align*} (x-\delta,x+\delta) \cap \mathbb{Z} = \varnothing. \end{align*} Every real number has a neighbourhood whose deleted intersection with $\mathbb{Z}$ is empty, so $\mathbb{Z}$ has no limit points in $\mathbb{R}$, even though it is infinite. [/example] The integers show that infinitude alone does not create limit points in a non-compact ambient space. To study the accumulation behaviour of a set as a set in its own right, we need to collect all its limit points into a single object. [definition: Derived Set] Let $(X, \tau)$ be a topological space and let $A \subset X$. The derived set of $A$ is the set \begin{align*} A' = \{x \in X : x \text{ is a limit point of } A\}. \end{align*} [/definition] The derived set may intersect $A$, contain points outside $A$, or be empty. To compare this accumulation set with the more familiar operation of adding all unavoidable points, we need the closure operation in the same neighbourhood language. [definition: Closure] Let $(X, \tau)$ be a topological space and let $A \subset X$. The closure of $A$ is \begin{align*} \overline{A} = \{x \in X : U \cap A \ne \varnothing \text{ for every open set } U \in \tau \text{ with } x \in U\}. \end{align*} [/definition] The closure test allows the point $x$ itself to account for the intersection with $A$, while the limit point test deletes $x$. The obstruction is that a point of $A$ may pass the closure test simply because it is already present, even if no other point of $A$ comes near it. To compare closure with accumulation, we must separate these two ways a neighbourhood can meet $A$. [quotetheorem:1006] This theorem says that every point in the closure has one of two reasons for being there: it is already in the set, or it is approached by other points of the set. The theorem is a compact way to remember why isolated points matter. [example: Closure of a Convergent Sequence Set] Let $A \subsetneq \mathbb{R}$ be given by \begin{align*} A = \{1/n : n \in \mathbb{N}\}. \end{align*} We first compute its derived set. The point $0$ is a limit point of $A$: if $\varepsilon>0$, choose $n \in \mathbb{N}$ with \begin{align*} n > \frac{1}{\varepsilon}. \end{align*} Since $n>0$ and $\varepsilon>0$, this is equivalent to \begin{align*} \frac{1}{n}<\varepsilon. \end{align*} Also $1/n \in A$ and $1/n \ne 0$, so \begin{align*} \frac{1}{n} \in (-\varepsilon,\varepsilon)\cap (A\setminus\{0\}). \end{align*} Thus every interval around $0$ contains a point of $A$ different from $0$. Now fix $k \in \mathbb{N}$. The point $1/k$ is not a limit point of $A$. If $k=1$, then every point of $A\setminus\{1\}$ has the form $1/n$ with $n\ge 2$, and hence \begin{align*} \frac{1}{n}\le \frac{1}{2}<\frac{3}{4}, \end{align*} so \begin{align*} (3/4,5/4)\cap (A\setminus\{1\})=\varnothing. \end{align*} If $k\ge 2$, set \begin{align*} r=\frac{1}{2}\min\left\{\frac{1}{k-1}-\frac{1}{k},\frac{1}{k}-\frac{1}{k+1}\right\}. \end{align*} Both numbers inside the minimum are positive, so $r>0$. If $n<k$, then \begin{align*} \frac{1}{n}\ge \frac{1}{k-1} \end{align*} and therefore \begin{align*} \left|\frac{1}{n}-\frac{1}{k}\right|\ge \frac{1}{k-1}-\frac{1}{k}>r. \end{align*} If $n>k$, then \begin{align*} \frac{1}{n}\le \frac{1}{k+1} \end{align*} and therefore \begin{align*} \left|\frac{1}{k}-\frac{1}{n}\right|\ge \frac{1}{k}-\frac{1}{k+1}>r. \end{align*} Hence \begin{align*} \left(\frac{1}{k}-r,\frac{1}{k}+r\right)\cap (A\setminus\{1/k\})=\varnothing, \end{align*} so $1/k$ is isolated in $A$. It remains to rule out points outside $A\cup\{0\}$. Let $x\in \mathbb{R}\setminus (A\cup\{0\})$. If $x<0$, then with $r=-x/2>0$ we have \begin{align*} x+r=\frac{x}{2}<0, \end{align*} so \begin{align*} (x-r,x+r)\cap A=\varnothing. \end{align*} If $x>0$, choose $N\in\mathbb{N}$ with \begin{align*} \frac{1}{N}<\frac{x}{2}. \end{align*} For $n\ge N$, \begin{align*} \frac{1}{n}\le \frac{1}{N}<\frac{x}{2}, \end{align*} so \begin{align*} \left|x-\frac{1}{n}\right|>\frac{x}{2}. \end{align*} For the finitely many indices $1\le n<N$, none of the numbers $1/n$ equals $x$, so \begin{align*} d=\min_{1\le n<N}\left|x-\frac{1}{n}\right|>0. \end{align*} Taking \begin{align*} r=\frac{1}{2}\min\left\{\frac{x}{2},d\right\} \end{align*} gives $r>0$ and \begin{align*} (x-r,x+r)\cap A=\varnothing. \end{align*} Thus no point other than $0$ is a limit point of $A$, and therefore \begin{align*} A'=\{0\}. \end{align*} By *Closure Decomposition by Limit Points*, \begin{align*} \overline{A}=A\cup A'=A\cup\{0\}. \end{align*} So the closure adds exactly the missing accumulation point $0$, while every listed point $1/n$ is present only as an isolated member of $A$. [/example] The closure decomposition turns closedness into a statement about missing accumulation points. We need the next criterion because many arguments prove a set is closed by showing that it has already included every limit point it creates. [quotetheorem:5011] Closedness therefore means there are no missing accumulation points. This is often the most intuitive way to read closure in analysis: a closed set contains the limits it forces. ## Metric and Sequential Tests In metric spaces, limit points can be detected by sequences. This is powerful because a topological condition involving every neighbourhood can be replaced by the construction of a countable list of points approaching the candidate limit. The replacement is special to spaces with enough countable local information, and metric spaces have that structure built in through balls of radius $1/n$. The sequential version deserves its own name because it appears throughout analysis, especially when compactness and completeness are studied through subsequences. [definition: Sequential Limit Point] Let $(X,d)$ be a [metric space](/page/Metric%20Space) and let $A \subset X$. A point $x \in X$ is a sequential limit point of $A$ if there exists a sequence $(a_n)_{n=1}^{\infty}$ in $A \setminus \{x\}$ such that \begin{align*} a_n \to x \end{align*} with respect to the metric $d$. [/definition] This definition asks for one organized stream of points from $A$ approaching $x$. The topological definition asks for points in every neighbourhood. In metric spaces these are equivalent because the radii \begin{align*} 1, 1/2, 1/3, \dots \end{align*} form a countable scale of tests around $x$. [quotetheorem:5012] The theorem is the technical bridge from topology to sequential analysis. It is why many first courses define accumulation points using sequences in $\mathbb{R}^n$, then later replace that definition by neighbourhoods in general topology. [example: Building the Sequence from Balls] Let $A=\mathbb{Q}\subsetneq \mathbb{R}$ and let $x=\sqrt{2}$. For each $n\in\mathbb{N}$, the interval \begin{align*} \left(\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n}\right) \end{align*} is an open interval around $\sqrt{2}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, this interval contains some rational number $a_n$, so \begin{align*} a_n\in\mathbb{Q} \end{align*} and \begin{align*} \sqrt{2}-\frac{1}{n}<a_n<\sqrt{2}+\frac{1}{n}. \end{align*} Subtracting $\sqrt{2}$ from each part gives \begin{align*} -\frac{1}{n}<a_n-\sqrt{2}<\frac{1}{n}, \end{align*} and therefore \begin{align*} |a_n-\sqrt{2}|<\frac{1}{n}. \end{align*} Also $a_n\ne \sqrt{2}$, because $a_n\in\mathbb{Q}$ while $\sqrt{2}\notin\mathbb{Q}$. Hence \begin{align*} a_n\in A\setminus\{\sqrt{2}\} \end{align*} for every $n\in\mathbb{N}$. We now verify convergence. Let $\varepsilon>0$. Choose $N\in\mathbb{N}$ with \begin{align*} N>\frac{1}{\varepsilon}. \end{align*} If $n\ge N$, then \begin{align*} n\ge N>\frac{1}{\varepsilon}, \end{align*} so, since $n>0$ and $\varepsilon>0$, \begin{align*} \frac{1}{n}<\varepsilon. \end{align*} Using the construction of $a_n$, \begin{align*} |a_n-\sqrt{2}|<\frac{1}{n}<\varepsilon. \end{align*} Thus $a_n\to\sqrt{2}$. By the [Metric Sequence Characterization of Limit Points](/theorems/5012), $\sqrt{2}$ is a limit point of $\mathbb{Q}$ in $\mathbb{R}$. [/example] This example is not about the special number $\sqrt{2}$; it is about density. The rationals accumulate at every real point, including those outside the rationals. More importantly, the construction succeeded because the infinitely many neighbourhood tests around $\sqrt{2}$ could be organized into the single countable scale $1,1/2,1/3,\dots$. That countable scale is not automatic in a general topological space. If the neighbourhoods around a point cannot be controlled by a countable local list, then the instruction "choose one point from each shrinking neighbourhood" may have no sequential meaning. We therefore need a name for spaces where countably many neighbourhoods do control all local behaviour at each point. [definition: First Countable Space] A topological space $(X,\tau)$ is first countable if for every $x \in X$ there exists a sequence $(U_n)_{n=1}^{\infty}$ of neighbourhoods of $x$ such that for every neighbourhood $N$ of $x$, there exists $n \in \mathbb{N}$ with \begin{align*} U_n \subset N. \end{align*} [/definition] Equivalently, many texts require the local base sets $U_n$ to be open neighbourhoods of $x$. The formulation above is the neighbourhood-base version; replacing each $U_n$ by an open set contained in it gives the open-neighbourhood version. First countability is the hidden reason that sequences work in metric spaces. Without it, a point may be topologically pressed by a set in every neighbourhood while no single sequence from that set manages to converge to the point. [example: The First Uncountable Ordinal] Let $\omega_1$ be the first uncountable ordinal, equip $X=\omega_1+1$ with the order topology, and let \begin{align*} A=\omega_1=\{\beta:\beta<\omega_1\}\subset X. \end{align*} We first show that $\omega_1$ is a limit point of $A$. Let $N$ be a neighbourhood of $\omega_1$. By the definition of neighbourhood, there is an open set $U$ such that \begin{align*} \omega_1\in U\subset N. \end{align*} In the order topology on $\omega_1+1$, every open neighbourhood of the maximum point $\omega_1$ contains a basic final interval, so there is some $\alpha<\omega_1$ with \begin{align*} (\alpha,\omega_1]\subset U. \end{align*} Since $\alpha<\omega_1$, the ordinal $\alpha$ is countable, and therefore its successor $\alpha+1$ is also countable. Hence \begin{align*} \alpha<\alpha+1<\omega_1, \end{align*} so \begin{align*} \alpha+1\in A \end{align*} and \begin{align*} \alpha+1\ne \omega_1. \end{align*} Also $\alpha+1\in(\alpha,\omega_1]$, and therefore \begin{align*} \alpha+1\in N\cap (A\setminus\{\omega_1\}). \end{align*} Thus every neighbourhood of $\omega_1$ contains a point of $A$ different from $\omega_1$, so $\omega_1$ is a limit point of $A$. Now let $(\alpha_n)_{n=1}^{\infty}$ be any sequence in $A$. Each $\alpha_n$ is a countable ordinal. Set \begin{align*} \alpha=\sup_{n\in\mathbb{N}}\alpha_n. \end{align*} For ordinals, the supremum is the union: \begin{align*} \alpha=\bigcup_{n\in\mathbb{N}}\alpha_n. \end{align*} Each $\alpha_n$ is countable as a set of smaller ordinals, and a [countable union of countable sets](/theorems/755) is countable, so $\alpha$ is a countable ordinal. Hence \begin{align*} \alpha<\omega_1. \end{align*} The final interval \begin{align*} (\alpha,\omega_1]=\{\beta\in X:\alpha<\beta\le \omega_1\} \end{align*} is a neighbourhood of $\omega_1$. But by the definition of $\alpha$ as an upper bound, \begin{align*} \alpha_n\le \alpha \end{align*} for every $n\in\mathbb{N}$, so no term $\alpha_n$ lies in $(\alpha,\omega_1]$. Therefore the sequence cannot eventually lie in every neighbourhood of $\omega_1$, and hence no sequence in $A$ converges to $\omega_1$. This shows that $\omega_1$ is genuinely forced by the neighbourhoods of $A$, even though no countable sequence of points from $A$ can approach it. [/example] This example is the standard warning behind the general definition. The neighbourhood condition sees all local tests at once; sequences see only countably many stages of approach. [remark: Why Neighbourhoods Come First] In a general topological space, a point may be forced by every neighbourhood of a set without being the limit of any sequence from that set. [Nets and filters](/page/Nets%20and%20Filters) repair this mismatch, but the definition of limit point itself is already local and does not need those tools. [/remark] The metric theorem should therefore be read as an additional privilege of metric spaces, not as the meaning of limit point in every topological space. ## Dense and Perfect Sets Some sets have isolated pieces and a small derived set; others accumulate everywhere. Limit points give a precise way to distinguish sparse sets, dense sets, and sets with no isolated members. These distinctions are central in real analysis, descriptive set theory, and algebraic geometry, where the closure of a set often carries more information than the set first presented. A set is dense when every point of the ambient space lies in its closure. Rephrased through limit points, density means that every outside point is a limit point, while points inside the set are either isolated members or internal limit points. [definition: Dense Subset] Let $(X, \tau)$ be a topological space. A subset $A \subset X$ is dense in $X$ if \begin{align*} \overline{A} = X. \end{align*} [/definition] The closure definition of density is concise, but it can hide the operational test used in examples. To verify density in practice, we need to know that every nonempty open region must meet the set. [quotetheorem:1005] This theorem explains why dense sets can be small from the viewpoint of size but unavoidable from the viewpoint of topology. The rationals are countable, but no real interval avoids them. [example: Rationals and Irrationals Accumulate Everywhere] In $\mathbb{R}$ with the usual topology, we show that every real number is a limit point of both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$. Fix $x\in\mathbb{R}$ and $\varepsilon>0$. First find a rational point in the deleted interval. By the *Archimedean integer-bracketing property*, since \begin{align*} x < x+\varepsilon, \end{align*} there exists $q\in\mathbb{Q}$ such that \begin{align*} x<q<x+\varepsilon. \end{align*} Thus $q\ne x$, and the two inequalities give \begin{align*} -\varepsilon < 0 < q-x < \varepsilon. \end{align*} Hence \begin{align*} q\in (x-\varepsilon,x+\varepsilon)\cap(\mathbb{Q}\setminus\{x\}). \end{align*} Now find an irrational point in the same deleted interval. Choose $n\in\mathbb{N}$ with \begin{align*} n>\frac{\sqrt{2}}{\varepsilon}. \end{align*} Since $n>0$ and $\varepsilon>0$, this gives \begin{align*} 0<\frac{\sqrt{2}}{n}<\varepsilon. \end{align*} Again by the *Archimedean integer-bracketing property*, choose $r\in\mathbb{Q}$ with \begin{align*} x-\frac{\sqrt{2}}{n}<r<x. \end{align*} Set \begin{align*} y=r+\frac{\sqrt{2}}{n}. \end{align*} From $r>x-\sqrt{2}/n$, adding $\sqrt{2}/n$ gives \begin{align*} y>x. \end{align*} From $r<x$, adding $\sqrt{2}/n$ gives \begin{align*} y<x+\frac{\sqrt{2}}{n}<x+\varepsilon. \end{align*} Therefore \begin{align*} x<y<x+\varepsilon, \end{align*} so $y\ne x$ and \begin{align*} y\in (x-\varepsilon,x+\varepsilon). \end{align*} Also $y$ is irrational: if $y\in\mathbb{Q}$, then \begin{align*} \frac{\sqrt{2}}{n}=y-r\in\mathbb{Q}, \end{align*} and multiplying by $n$ would give $\sqrt{2}\in\mathbb{Q}$, a contradiction. Hence \begin{align*} y\in (x-\varepsilon,x+\varepsilon)\cap((\mathbb{R}\setminus\mathbb{Q})\setminus\{x\}). \end{align*} Since every $\varepsilon$-interval around $x$ contains a rational point different from $x$ and an irrational point different from $x$, every real number is a limit point of $\mathbb{Q}$ and also of $\mathbb{R}\setminus\mathbb{Q}$. [/example] This example is a warning against interpreting limit points visually as endpoints or boundary points. A dense set has limit points everywhere, including deep inside regions where the set and its complement are both interlaced. The next condition removes isolated points from a set. It is weaker than being closed, but it captures the idea that every member of the set is locally accompanied by other members. [definition: Dense-in-Itself Set] Let $(X, \tau)$ be a topological space. A subset $A \subset X$ is dense-in-itself if every point of $A$ is a limit point of $A$. [/definition] A dense-in-itself set has no isolated points relative to itself. The phrase does not mean dense in the whole ambient space; it means internally crowded at every one of its own points. [example: The Open Interval Has No Isolated Points] The interval $(0,1) \subsetneq \mathbb{R}$ is dense-in-itself. Fix $x \in (0,1)$ and let $\varepsilon>0$. Since $0<x<1$, both $x$ and $1-x$ are positive. Set \begin{align*} \delta=\frac{1}{2}\min\{\varepsilon,x,1-x\} \end{align*} and choose \begin{align*} y=x+\delta. \end{align*} Because $\varepsilon>0$, $x>0$, and $1-x>0$, their minimum is positive, so \begin{align*} \delta>0. \end{align*} Hence \begin{align*} y=x+\delta>x, \end{align*} so $y\ne x$, and in particular $y>0$. Also, \begin{align*} \delta \le \frac{1}{2}(1-x), \end{align*} so \begin{align*} y=x+\delta \le x+\frac{1}{2}(1-x)=\frac{x+1}{2}. \end{align*} Since $x<1$, \begin{align*} \frac{x+1}{2}<1, \end{align*} and therefore \begin{align*} 0<y<1. \end{align*} Thus $y\in(0,1)$. Finally, \begin{align*} |y-x|=|x+\delta-x|=|\delta|=\delta. \end{align*} Since $\delta>0$ and \begin{align*} \delta\le \frac{\varepsilon}{2}<\varepsilon, \end{align*} we have \begin{align*} |y-x|<\varepsilon. \end{align*} Therefore \begin{align*} y\in (x-\varepsilon,x+\varepsilon)\cap((0,1)\setminus\{x\}). \end{align*} Since this works for every $x\in(0,1)$ and every $\varepsilon>0$, every point of $(0,1)$ is a limit point of $(0,1)$, so $(0,1)$ has no isolated points. [/example] Dense-in-itself sets may fail to be closed: $(0,1)$ has every point internally accumulated, but it omits the limit points $0$ and $1$. We therefore need a name for sets that have no isolated points and also omit none of their accumulation points. [definition: Perfect Set] Let $(X, \tau)$ be a topological space. A subset $P \subset X$ is perfect if $P$ is closed and every point of $P$ is a limit point of $P$. [/definition] A perfect set has no isolated points and no missing limit points. It is topologically complete with respect to its own accumulation behaviour. [example: Closed Intervals Are Perfect] In $\mathbb{R}$ with the usual topology, we show that $[0,1]$ is perfect by verifying both parts of the definition. First, $[0,1]$ is closed. Indeed, if $x<0$, set \begin{align*} r=-\frac{x}{2}. \end{align*} Then $r>0$ and \begin{align*} x+r=x-\frac{x}{2}=\frac{x}{2}<0, \end{align*} so $(x-r,x+r)\subset \mathbb{R}\setminus[0,1]$. If $x>1$, set \begin{align*} r=\frac{x-1}{2}. \end{align*} Then $r>0$ and \begin{align*} x-r=x-\frac{x-1}{2}=\frac{x+1}{2}>1, \end{align*} so again $(x-r,x+r)\subset \mathbb{R}\setminus[0,1]$. Thus every point of $\mathbb{R}\setminus[0,1]$ has an open interval contained in $\mathbb{R}\setminus[0,1]$, so $\mathbb{R}\setminus[0,1]$ is open and $[0,1]$ is closed. Now we show that every point of $[0,1]$ is a limit point of $[0,1]$. Let $\varepsilon>0$. If $x\in(0,1)$, then $1-x>0$, so \begin{align*} \delta=\frac{1}{2}\min\{\varepsilon,1-x\} \end{align*} satisfies $\delta>0$. Set \begin{align*} y=x+\delta. \end{align*} Since $\delta>0$, \begin{align*} y=x+\delta>x, \end{align*} so $y\ne x$. Also $x>0$, hence $y>0$, and \begin{align*} \delta\le \frac{1}{2}(1-x), \end{align*} so \begin{align*} y=x+\delta\le x+\frac{1}{2}(1-x)=\frac{x+1}{2}<1. \end{align*} Thus $y\in[0,1]$. Finally, \begin{align*} |y-x|=|x+\delta-x|=|\delta|=\delta\le \frac{\varepsilon}{2}<\varepsilon. \end{align*} At the endpoint $0$, set \begin{align*} y=\frac{1}{2}\min\{\varepsilon,1\}. \end{align*} Then $y>0$, and since $\min\{\varepsilon,1\}\le 1$, \begin{align*} y\le \frac{1}{2}<1, \end{align*} so $y\in[0,1]$ and $y\ne 0$. Also, \begin{align*} |y-0|=y=\frac{1}{2}\min\{\varepsilon,1\}\le \frac{\varepsilon}{2}<\varepsilon. \end{align*} At the endpoint $1$, set \begin{align*} \eta=\frac{1}{2}\min\{\varepsilon,1\}, \qquad y=1-\eta. \end{align*} Then $\eta>0$, so \begin{align*} y=1-\eta<1, \end{align*} and since $\eta\le 1/2$, \begin{align*} y=1-\eta\ge \frac{1}{2}>0. \end{align*} Thus $y\in[0,1]$ and $y\ne 1$. Moreover, \begin{align*} |y-1|=|1-\eta-1|=|-\eta|=\eta\le \frac{\varepsilon}{2}<\varepsilon. \end{align*} Therefore every $\varepsilon$-interval around every point of $[0,1]$ contains a different point of $[0,1]$. Hence every point of $[0,1]$ is a limit point of $[0,1]$, and the closed interval $[0,1]$ is perfect. [/example] Perfect sets are important because they are closed sets whose structure is not explained by isolated atoms. They can be intervals, fractals, or much more complicated closed sets. [example: Cantor-Type Perfect Behaviour] The middle-thirds [Cantor set](/page/Cantor%20Set) $C \subsetneq [0,1]$ is perfect. Let $C_k$ denote the union of the retained closed intervals after $k$ deletion stages, so \begin{align*} C=\bigcap_{k=0}^{\infty} C_k. \end{align*} Each $C_k$ is a finite union of closed intervals, hence is closed in $\mathbb{R}$, and an arbitrary intersection of closed sets is closed. Therefore $C$ is closed. It remains to show that $C$ has no isolated points. Fix $x\in C$ and let $\varepsilon>0$. Choose $k\in\mathbb{N}$ with \begin{align*} 3^{-k}<\varepsilon. \end{align*} Since $x\in C\subset C_k$, the point $x$ lies in some retained interval \begin{align*} I=[a,b]\subset C_k. \end{align*} Every retained interval at stage $k$ has length $3^{-k}$, so \begin{align*} b-a=3^{-k}>0. \end{align*} Thus $a\ne b$. The endpoints of a retained interval are never removed at later stages, so \begin{align*} a,b\in C. \end{align*} Choose \begin{align*} y= \begin{cases} a, & \text{if } x\ne a,\\ b, & \text{if } x=a. \end{cases} \end{align*} Then $y\in C$ and $y\ne x$. Also $x,y\in[a,b]$, so \begin{align*} |y-x|\le b-a=3^{-k}<\varepsilon. \end{align*} Hence \begin{align*} y\in (x-\varepsilon,x+\varepsilon)\cap(C\setminus\{x\}). \end{align*} Since this works for every $x\in C$ and every $\varepsilon>0$, every point of $C$ is a limit point of $C$. Therefore $C$ is closed and has no isolated points, so [the Cantor set is perfect](/theorems/1199), despite containing no interval of positive length. [/example] The Cantor set shows that perfect does not mean smooth, interval-like, or large in the sense of length. It means closed and everywhere internally accumulated. [illustration:cantor-stages-limit-point] ## Compactness and Infinite Sets Limit points become especially powerful when the ambient space prevents points from escaping indefinitely. Compactness is the condition that turns infinitude into accumulation. Without compactness, an infinite set can be locally sparse everywhere, as $\mathbb{Z} \subsetneq \mathbb{R}$ demonstrates. With compactness, an infinite subset must cluster somewhere. The classical real-variable form is the [Bolzano-Weierstrass theorem](/theorems/628). It is the first major result many readers encounter where limit points are not merely definitions but consequences of boundedness and closedness. [quotetheorem:628] The theorem says that bounded infinite subsets of Euclidean space cannot remain separated forever. Some point of the ambient space must be approached by infinitely much of the set. [example: Boundedness Cannot Be Dropped] The set $\mathbb{N} \subsetneq \mathbb{R}$ is infinite, but it has no limit point in $\mathbb{R}$. We show that every real number $x$ has an open interval around it whose deleted intersection with $\mathbb{N}$ is empty. First suppose $x\in\mathbb{N}$. Take $r=1/2$. If $m\in\mathbb{N}$ and $m\ne x$, then $m-x$ is a nonzero integer, so \begin{align*} |m-x|\ge 1>\frac{1}{2}=r. \end{align*} Thus no natural number different from $x$ lies in $(x-r,x+r)$, and hence \begin{align*} (x-r,x+r)\cap(\mathbb{N}\setminus\{x\})=\varnothing. \end{align*} Now suppose $x\notin\mathbb{N}$. If $x<1$, set \begin{align*} r=\frac{1-x}{2}. \end{align*} Then $r>0$ and \begin{align*} x+r=x+\frac{1-x}{2}=\frac{x+1}{2}<1. \end{align*} Since every $m\in\mathbb{N}$ satisfies $m\ge 1$, we get \begin{align*} (x-r,x+r)\cap\mathbb{N}=\varnothing. \end{align*} It remains to consider $x>1$ with $x\notin\mathbb{N}$. By the integer-bracketing property, there is a natural number $k$ such that \begin{align*} k<x<k+1. \end{align*} Set \begin{align*} r=\frac{1}{2}\min\{x-k,k+1-x\}. \end{align*} Both $x-k$ and $k+1-x$ are positive, so $r>0$. Since \begin{align*} r\le \frac{x-k}{2}, \end{align*} we have \begin{align*} x-r\ge x-\frac{x-k}{2}=\frac{x+k}{2}>k. \end{align*} Since \begin{align*} r\le \frac{k+1-x}{2}, \end{align*} we have \begin{align*} x+r\le x+\frac{k+1-x}{2}=\frac{x+k+1}{2}<k+1. \end{align*} Therefore \begin{align*} (x-r,x+r)\subset (k,k+1). \end{align*} No natural number lies strictly between the consecutive natural numbers $k$ and $k+1$, so \begin{align*} (x-r,x+r)\cap\mathbb{N}=\varnothing. \end{align*} In every case, there is some $r>0$ such that \begin{align*} ((x-r,x+r)\setminus\{x\})\cap\mathbb{N}=\varnothing. \end{align*} Thus no real number is a limit point of $\mathbb{N}$. The set is infinite because it escapes to infinity, not because it accumulates locally. [/example] This is the standard failure mode: compactness blocks escape. Once escape is blocked, infinite sets must either repeat location in a limiting sense or contradict compactness. The topological form uses compact spaces rather than bounded subsets of Euclidean space. To avoid finite sets producing artificial counterexamples in spaces with poor separation, it is common to state the clean version for $T_1$ spaces, where singletons are closed and finite sets have no accidental topological crowding. [definition: $T_1$ Space] A topological space $(X,\tau)$ is a $T_1$ space if for every pair of distinct points $x,y \in X$, there exists an open set $U \in \tau$ such that \begin{align*} x \in U, \qquad y \notin U. \end{align*} [/definition] The $T_1$ condition ensures that points can be separated from each other by open neighbourhoods one at a time. It is weak compared with Hausdorffness, but strong enough to make finite sets behave as expected. The fully topological [compactness theorem](/theorems/2748) is stated using nets, which are sequence-like objects indexed by a directed set rather than by $\mathbb{N}$. This is the right level of generality because ordinary sequences do not detect compactness in arbitrary topological spaces. [definition: Net, Cluster Point, and Subnet] A directed set is a preorder $(D,\leq)$ such that for every $\alpha,\beta\in D$ there exists $\gamma\in D$ with $\alpha\leq\gamma$ and $\beta\leq\gamma$. A net in a topological space $X$ is a function $D\to X$, written $(x_\alpha)_{\alpha\in D}$, where $D$ is a directed set. The net $(x_\alpha)$ converges to $x\in X$ if for every neighbourhood $N$ of $x$ there exists $\alpha_0\in D$ such that $x_\alpha\in N$ whenever $\alpha\geq\alpha_0$. A point $x\in X$ is a cluster point of the net if for every neighbourhood $N$ of $x$ and every $\alpha_0\in D$, there exists $\alpha\geq\alpha_0$ with $x_\alpha\in N$. A subnet is a net obtained from $(x_\alpha)$ by passing to a monotone cofinal reindexing; it is convergent if it converges in the preceding sense. [/definition] With this language in place, compactness can be recognized by the behaviour of all nets. The theorem below is broader than the infinite-subset principle, but it is the general mechanism behind it: compactness prevents a directed family of points from escaping every neighbourhood system. [quotetheorem:1051] This theorem is the conceptual heart of the subject: compactness converts infinite membership into forced local accumulation. [example: Compactness Creates a Limit Point] Let \begin{align*} A=\{1/n:n\in\mathbb{N}\}. \end{align*} For every $n\in\mathbb{N}$ we have $n\ge 1$, so \begin{align*} 0<\frac{1}{n}\le 1. \end{align*} Hence $A\subset[0,1]$. The set $A$ is infinite: if $m,n\in\mathbb{N}$ and \begin{align*} \frac{1}{m}=\frac{1}{n}, \end{align*} then multiplying both sides by $mn>0$ gives \begin{align*} n=m. \end{align*} Thus different natural numbers give different points of $A$. Inside the compact interval $[0,1]$, the point $0$ is a limit point of $A$. Let $r>0$. Choose $n\in\mathbb{N}$ with \begin{align*} n>\frac{1}{r}. \end{align*} Since $n>0$ and $r>0$, this implies \begin{align*} \frac{1}{n}<r. \end{align*} Also $1/n\in A$ and $1/n\ne 0$, so \begin{align*} \frac{1}{n}\in [0,1]\cap(0,r)=B_{[0,1]}(0,r)\setminus\{0\}. \end{align*} Therefore every ball around $0$ in the subspace $[0,1]$ contains a point of $A$ different from $0$, so $0$ is a limit point of $A$ in $[0,1]$. If the same formula for $A$ is viewed as a subset of $(0,1]$, then $0$ cannot be a limit point there because $0\notin(0,1]$ and limit points must belong to the ambient topological space. In fact, the missing endpoint is exactly where the sequence accumulates: for each $r>0$ the point $1/n$ can be forced into $(0,r)$ by choosing $n>1/r$, but $(0,1]$ does not contain the point $0$ that these neighbourhood tests approach. Thus the compact ambient interval $[0,1]$ contains the forced accumulation point, while the non-compact space $(0,1]$ has removed it. [/example] The ambient space matters. A limit point must be a point of the topological space under discussion, even if the same formula suggests a missing point in a larger completion. ## Separation Pathologies In familiar metric spaces, finite sets have no limit points, and a point cannot be forced merely because open sets are too coarse to distinguish it from another point. General topology allows coarser behaviours. Limit points therefore also diagnose how much separation the topology provides. The simplest warning is the indiscrete topology, where the only nonempty open set is the whole space. In such a space, neighbourhoods cannot focus on a single point. [definition: Indiscrete Topology] Let $X$ be a set. The indiscrete topology on $X$ is \begin{align*} \tau = \{\varnothing, X\}. \end{align*} [/definition] The indiscrete topology is the coarsest topology on $X$. It sees no local distinction between points, so limit point behaviour becomes extremely broad. [example: A Finite Set with Limit Points] Let $X=\{a,b\}$ with the indiscrete topology \begin{align*} \tau=\{\varnothing,X\}, \end{align*} and let $A=\{a\}$. We first determine the neighbourhoods of $b$. If $N$ is a neighbourhood of $b$, then there is an open set $U\in\tau$ such that \begin{align*} b\in U\subset N. \end{align*} Since $\varnothing$ does not contain $b$, the only possible open set is $U=X$. Hence \begin{align*} X\subset N\subset X, \end{align*} so $N=X$. Thus the only neighbourhood of $b$ is $X$. Now \begin{align*} X\setminus\{b\} &=\{a,b\}\setminus\{b\}\\ &=\{a\}, \end{align*} and therefore \begin{align*} (X\setminus\{b\})\cap A &=\{a\}\cap\{a\}\\ &=\{a\}\ne\varnothing. \end{align*} Since the only neighbourhood of $b$ has deleted intersection with $A$ nonempty, $b$ is a limit point of $A$, even though $A=\{a\}$ is finite. The same neighbourhood calculation shows that the only neighbourhood of $a$ is also $X$. For this neighbourhood, \begin{align*} X\setminus\{a\} &=\{a,b\}\setminus\{a\}\\ &=\{b\}, \end{align*} so \begin{align*} (X\setminus\{a\})\cap A &=\{b\}\cap\{a\}\\ &=\varnothing. \end{align*} Thus $a$ is not a limit point of $A$: its only deleted neighbourhood contains no point of $A$. [/example] This example explains why many theorems about finite sets and limit points assume at least $T_1$. To see the opposite behaviour, we need the topology in which points can be isolated as much as possible. [definition: Discrete Topology] Let $X$ be a set. The discrete topology on $X$ is \begin{align*} \tau = \mathcal{P}(X), \end{align*} where $\mathcal{P}(X)$ denotes the power set of $X$. [/definition] The discrete topology is the finest topology on $X$. Every singleton is open, so every point has a neighbourhood that excludes all other points. Thus the limit point test faces the strongest possible obstruction: around any candidate point, there is an open set whose deleted part is empty. The derived set should therefore vanish for every subset. [quotetheorem:5013] This theorem shows that accumulation is not a property of a set alone. It depends on the topology placed on the ambient space. [example: Same Set, Different Topology] Let $X=\mathbb{R}$ and let \begin{align*} A=\{1/n:n\in\mathbb{N}\}. \end{align*} The underlying set $A$ will be kept fixed, but we will compare two different topologies on $X$. First give $X$ the usual topology. We show that $0$ is a limit point of $A$. Since $1/n>0$ for every $n\in\mathbb{N}$, no element of $A$ is equal to $0$, so \begin{align*} 0\notin A. \end{align*} Let $\varepsilon>0$. Choose $n\in\mathbb{N}$ such that \begin{align*} n>\frac{1}{\varepsilon}. \end{align*} Because $n>0$ and $\varepsilon>0$, multiplying by $\varepsilon$ gives \begin{align*} n\varepsilon>1. \end{align*} Dividing by $n>0$ gives \begin{align*} \varepsilon>\frac{1}{n}, \end{align*} so \begin{align*} 0<\frac{1}{n}<\varepsilon. \end{align*} Thus \begin{align*} \frac{1}{n}\in(-\varepsilon,\varepsilon). \end{align*} Also $1/n\in A$, and $1/n\ne0$, hence \begin{align*} \frac{1}{n}\in(-\varepsilon,\varepsilon)\cap(A\setminus\{0\}). \end{align*} Therefore \begin{align*} (-\varepsilon,\varepsilon)\cap(A\setminus\{0\})\ne\varnothing. \end{align*} Since this holds for every $\varepsilon>0$, every usual open interval around $0$ contains a point of $A$ different from $0$, so $0$ is a limit point of $A$ in the usual topology. Now give the same underlying set $X=\mathbb{R}$ the discrete topology. In the discrete topology every subset of $X$ is open, so the singleton $\{0\}$ is an open neighbourhood of $0$. Its deleted neighbourhood is \begin{align*} \{0\}\setminus\{0\}=\varnothing. \end{align*} Hence \begin{align*} (\{0\}\setminus\{0\})\cap A &=\varnothing\cap A\\ &=\varnothing. \end{align*} So there is a neighbourhood of $0$ whose deleted intersection with $A$ is empty. Therefore $0$ is not a limit point of $A$ in the discrete topology. The set $A$ has not changed; only the topology on the ambient set $\mathbb{R}$ changed. Limit point behaviour is therefore topological, not purely set-theoretic. [/example] The underlying elements have not changed, and the subset has not changed. Only the topology changed. Limit points are therefore topological data, not purely set-theoretic data. ## Accumulation for Sequences and Stronger Notions The phrase limit point is most often attached to a subset, but the same local idea appears when a sequence, family, or collection of sets visits neighbourhoods repeatedly. This broader language connects limit points to convergence theory and dynamical systems. For sequences, it is useful to distinguish the limit of the whole sequence from a point that is reached along some subsequence. Such points record recurring behaviour rather than final behaviour of every term. [definition: Subsequential Limit] Let $(X,d)$ be a metric space and let $(x_n)_{n=1}^{\infty}$ be a sequence in $X$. A point $x \in X$ is a subsequential limit of $(x_n)_{n=1}^{\infty}$ if there exists a strictly increasing sequence of natural numbers $(n_k)_{k=1}^{\infty}$ such that \begin{align*} x_{n_k} \to x. \end{align*} [/definition] A subsequential limit is a limit point of the sequence as a process, not necessarily a limit point of the set of values. Repetition can matter for sequences even when it disappears from the underlying set. [example: Repetition Creates a Subsequential Limit] Let $x_n=(-1)^n$ as a sequence in $\mathbb{R}$. Since \begin{align*} (-1)^{2k}=1, \qquad (-1)^{2k-1}=-1 \end{align*} for every $k\in\mathbb{N}$, the set of values of the sequence is \begin{align*} \{-1,1\}. \end{align*} This two-point set has no limit points in the usual topology. For the point $1$, the interval $(1/2,3/2)$ is a neighbourhood of $1$, and \begin{align*} (1/2,3/2)\cap(\{-1,1\}\setminus\{1\}) &=(1/2,3/2)\cap\{-1\}\\ &=\varnothing, \end{align*} because $-1<1/2$. Hence $1$ is not a limit point of $\{-1,1\}$. Similarly, for the point $-1$, the interval $(-3/2,-1/2)$ is a neighbourhood of $-1$, and \begin{align*} (-3/2,-1/2)\cap(\{-1,1\}\setminus\{-1\}) &=(-3/2,-1/2)\cap\{1\}\\ &=\varnothing, \end{align*} because $1>-1/2$. Thus $-1$ is not a limit point of $\{-1,1\}$. The sequence nevertheless has two subsequential limits. The even-indexed subsequence is constant: \begin{align*} x_{2k}=(-1)^{2k}=1 \end{align*} for every $k\in\mathbb{N}$. Therefore, for every $\varepsilon>0$ and every $k\in\mathbb{N}$, \begin{align*} |x_{2k}-1|=|1-1|=0<\varepsilon, \end{align*} so $x_{2k}\to 1$. The odd-indexed subsequence is also constant: \begin{align*} x_{2k-1}=(-1)^{2k-1}=-1 \end{align*} for every $k\in\mathbb{N}$. Hence, for every $\varepsilon>0$ and every $k\in\mathbb{N}$, \begin{align*} |x_{2k-1}-(-1)|=|-1+1|=0<\varepsilon, \end{align*} so $x_{2k-1}\to -1$. Thus $1$ and $-1$ are subsequential limits of the sequence, even though neither is a limit point of the set of values. The difference is that subsequential limits remember repeated visits, while limit points of a set ignore multiplicity and order. [/example] This distinction prevents a common mistake: a sequence is not the same object as its range. Limit points of sets ignore multiplicity and order; subsequential limits retain enough of both to detect repeated visits. A related set-level notion asks that every neighbourhood meet a set infinitely many times, not merely once. This stronger wording is useful in compactness arguments because it rules out local contact caused by a finite number of nearby points. [definition: Strong Accumulation Point] Let $(X,\tau)$ be a topological space and let $A \subset X$. A point $x \in X$ is a strong accumulation point of $A$ if every neighbourhood $N$ of $x$ satisfies \begin{align*} |N \cap A| = \infty. \end{align*} [/definition] A usual limit point only asks for at least one other point in every neighbourhood. In $T_1$ spaces, finite obstructions can be removed one point at a time, so the ordinary limit point condition for a subset already forces infinitely many nearby points. [quotetheorem:5014] The careful hypothesis matters. In non-$T_1$ spaces, a finite set can create a limit point outside itself because the topology cannot separate the outside point from the finite set. ## Beyond and Connected Topics Limit points are the entry point to [closure](/page/Closure), [closed sets](/page/Closed%20Set), and [continuity](/page/Continuity). For maps between topological spaces, continuity preserves the appropriate image convergence: if a sequence, net, or filter converges to $x$, then its image converges to the image of $x$. For set limit points one must be more careful, because a continuous map can collapse nearby distinct points to a single value; additional hypotheses such as local injectivity, or a formulation using non-eventually-constant nets, are needed to turn accumulation of a set into accumulation of its image. This is one reason topology phrases continuity through preimages of open sets rather than through formulas. In metric spaces, limit points lead directly to [compact spaces](/page/Compact%20Space), [sequential compactness](/page/Sequential%20Compactness), and completeness. The Bolzano-Weierstrass theorem is the prototype: bounded infinite sets in Euclidean space must accumulate. Later courses replace sequences by open covers, nets, and filters to make the same idea work in general topological spaces. In analysis, limit points govern where functions must respect limiting behaviour. A function defined on a [dense subset](/page/Dense%20Subset) may have at most one continuous extension to a Hausdorff target, because values near each limit point are forced. This principle lies behind completions, extension theorems, and the construction of [real numbers](/page/Real%20Numbers) from rational approximations. In algebraic geometry, closure and limit points appear through the Zariski topology, where open sets are large and closed sets are algebraic solution sets. The resulting topology is far from metric intuition: finite-looking algebraic data can have closures controlled by polynomial equations rather than by Euclidean distance. The natural course-level continuations are [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) for real sequences and limits, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) for metric and topological spaces, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis) for accumulation points of zeros and singularities, and [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry) for closure in the Zariski topology. ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). Androma, [Closure](/page/Closure). Androma, [Closed Set](/page/Closed%20Set). Androma, [Continuity](/page/Continuity). Androma, [Compact Space](/page/Compact%20Space). Munkres, *Topology* (2000). Kelley, *General Topology* (1955). Willard, *General Topology* (1970).