Topology usually enters analysis as a language for nearness: open sets describe what can be detected locally, continuous maps preserve local information, and limits formalize the idea that points can approach one another. But some mathematical objects have no intended nearness relation. A finite state machine, a set of symbols, a basis indexed by a set, or the vertex set of a graph may come with equality but no geometry. If a topology is forced onto such a set from the outside, it may invent limit behavior that was never part of the problem. The discrete topology is the topology that refuses to invent any hidden nearness. Every subset is open, every point can be separated from every other point by a singleton neighbourhood, and every function out of the space is continuous. It is the finest possible topology on a fixed set, and because it makes continuity from the space automatic, it is the correct topology when points are meant to be individually observable. The first surprise is that the discrete topology can look harmless on finite sets but become rigid on infinite sets. A finite discrete space is compact, every sequence has the expected finite behavior, and every subset is both open and closed. An infinite discrete space is still locally simple, but globally it is not compact, not connected, and has no non-eventual convergent sequences. This contrast makes the discrete topology a useful test case throughout [Topology](/page/Topology), [Metric Space](/page/Metric%20Space), and [Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). [example: False Closeness from an Ambient Line] Let $X=\{1/n:n\in\mathbb{N}\}\subset\mathbb{R}$ and let $x_n=1/n$. In the ambient real line, the sequence approaches the missing point $0$: if $\varepsilon>0$, choose $N\in\mathbb{N}$ with $N>1/\varepsilon$. Then for every $n\ge N$, \begin{align*} 0<\frac{1}{n}\le \frac{1}{N}<\varepsilon, \end{align*} so $|x_n-0|<\varepsilon$. Thus $x_n\to 0$ in $\mathbb{R}$, but $0\notin X$. The same sequence has no limit in $X$ with the [subspace topology](/page/Subspace%20Topology). To see this, fix a candidate limit $1/m\in X$. The nearest neighbouring points of $1/m$ in $X$ are separated from it by positive real distances; for example, when $m\ge 2$, \begin{align*} \frac{1}{m-1}-\frac{1}{m}=\frac{1}{m(m-1)} \end{align*} and \begin{align*} \frac{1}{m}-\frac{1}{m+1}=\frac{1}{m(m+1)}. \end{align*} Choosing a positive radius smaller than both of these distances gives an open interval whose intersection with $X$ is exactly $\{1/m\}$. For $m=1$, the interval $(1/2,3/2)$ has intersection $\{1\}$ with $X$. Therefore $\{1/m\}$ is an open neighbourhood of $1/m$ in the subspace topology. But $x_n=1/m$ holds only when $n=m$, so for every $N$ there is some $n\ge N$ with $x_n\ne 1/m$. Hence $x_n$ does not converge to any point of $X$. If the intended object is merely the labelled set $X$, there is no intrinsic reason for $1/1000$ to be closer to $1/1001$ than to $1/2$. The discrete topology expresses exactly that choice: every singleton $\{1/m\}$ is open, and convergence to $1/m$ forces eventual membership in $\{1/m\}$, hence eventual equality to $1/m$. The ambient limit $0$ is therefore information coming from the embedding in $\mathbb{R}$, not from the labelled set itself. [/example] This example shows the guiding principle of the page. The discrete topology is not merely a list of open sets; it is a declaration that equality is the only local relation among points. The rest of the chapter studies how this declaration affects the standard topological notions of continuity, convergence, compactness, separation, and connectedness. ## Definition ### Open Sets and the Extreme Choice The page begins with the extreme choice itself. If the topology is supposed to record no nearness beyond equality, then no subset should be forbidden from being open. Each collection of points must be observable on its own, whether it is a singleton, a finite set, or an arbitrary infinite subset. [definition: Discrete Topology] Let $X$ be a set. The discrete topology on $X$ is the topology \begin{align*} \tau_{\mathrm{disc}} = \mathcal{P}(X). \end{align*} [/definition] The definition says that every subset of $X$ is open. Since complements of open sets are closed, every subset is also closed. This double status is not a contradiction; it means that the topology separates points so strongly that no subset has a forced boundary created by the topology alone. To use this extreme choice in the same language as other spaces, we need the general object of which it is an example. A topology is the rulebook that decides which subsets count as open, and its axioms are exactly the closure properties needed for local information to be restricted, combined finitely by intersections, and combined arbitrarily by unions. [definition: Topology] Let $X$ be a set. A topology on $X$ is a collection $\tau \subset \mathcal{P}(X)$ such that: 1. $\varnothing \in \tau$ and $X \in \tau$. 2. If $(U_i)_{i \in I}$ is a family of sets with $U_i \in \tau$ for all $i \in I$, then $\bigcup_{i \in I} U_i \in \tau$. 3. If $U_1, \ldots, U_n \in \tau$ for some $n \in \mathbb{N}$, then $\bigcap_{i=1}^n U_i \in \tau$. [/definition] The discrete topology satisfies these axioms because arbitrary unions and finite intersections of subsets of $X$ are still subsets of $X$. The axioms specify the collection, but mathematical arguments need a word for the members of that collection. The next definition lets us say that a set can be detected topologically, and it is the word that continuity, neighbourhoods, and compactness will use throughout the chapter. [definition: Open Set] Let $(X, \tau)$ be a [topological space](/page/Topological%20Space). A subset $U \subset X$ is open in $(X, \tau)$ if $U \in \tau$. [/definition] ### Isolated Points and Discrete Spaces A single point is the smallest possible local observation. In many spaces singleton sets need not be open, so it is useful to name the special case where a point can be detected by itself. This language lets us recognize discreteness without listing all subsets. [definition: Isolated Point] Let $(X, \tau)$ be a topological space and let $x \in X$. The point $x$ is isolated if $\{x\} \in \tau$. [/definition] If every point can be detected by itself, then every subset can be detected by taking a union of singleton observations. The obstruction to this happening in an arbitrary topology is that most subsets are not forced to be open merely because their points are present. A discrete space is the extreme case where there is no such obstruction: openness imposes no restriction beyond being a subset of the underlying set. [definition: Discrete Space] A topological space $(X, \tau)$ is a discrete space if $\tau = \mathcal{P}(X)$. [/definition] The two viewpoints match: a topological space is discrete precisely when every point is isolated. The forward direction comes from the definition of $\mathcal{P}(X)$. For the reverse direction, every subset $A \subset X$ is the union of the singleton sets $\{a\}$ with $a \in A$. Finite examples make the definition concrete because all open sets can be listed. The next example shows how quickly the topology becomes large even on a three-point set, and it also explains why singleton openness is enough to recover every [open set](/page/Open%20Set). [example: The Discrete Topology on Three Points] Let $X=\{a,b,c\}$. Its power set is obtained by choosing, independently for each of the three elements, whether that element is included in the subset. Hence the number of subsets is \begin{align*} |\mathcal{P}(X)|=2^3=8. \end{align*} Listing them explicitly, \begin{align*} \mathcal{P}(X)=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\}. \end{align*} Since the discrete topology on $X$ is $\mathcal{P}(X)$, these eight sets are exactly the open sets. Each subset is also a union of singleton sets: for example, \begin{align*} \{a,c\}=\{a\}\cup\{c\}. \end{align*} The empty set is the empty union of singleton sets, and $X=\{a\}\cup\{b\}\cup\{c\}$. Thus on a finite discrete space, knowing that all singletons are open already recovers every open set by taking unions. [/example] Finite examples can make the definition feel too small. On an infinite set, the same rule produces far more open sets than most familiar topologies. The next example emphasizes that no arithmetic or order property is required for openness. [example: A Discrete Topology on $\mathbb{N}$] Let $X=\mathbb{N}$ and equip $X$ with the discrete topology, so the open sets are exactly the subsets of $X$: \begin{align*} \tau_{\mathrm{disc}}=\mathcal{P}(\mathbb{N}). \end{align*} The set of even natural numbers is \begin{align*} E=\{2n:n\in\mathbb{N}\}. \end{align*} For every $n\in\mathbb{N}$, the number $2n$ is a natural number, so $E\subset\mathbb{N}$. Hence $E\in\mathcal{P}(\mathbb{N})$, and therefore $E$ is open in the discrete topology. The singleton $\{5\}$ is also a subset of $\mathbb{N}$, so \begin{align*} \{5\}\in\mathcal{P}(\mathbb{N}). \end{align*} Thus $\{5\}$ is open. Its complement inside $X$ is \begin{align*} \mathbb{N}\setminus\{5\}=\{n\in\mathbb{N}:n\ne 5\}. \end{align*} This complement is again a subset of $\mathbb{N}$, so \begin{align*} \mathbb{N}\setminus\{5\}\in\mathcal{P}(\mathbb{N}). \end{align*} Therefore $\mathbb{N}\setminus\{5\}$ is open, which means $\{5\}$ is closed as well as open. In the discrete topology, no arithmetic property such as being even, finite, or cofinite is needed for openness; being a subset of $\mathbb{N}$ is enough. [/example] These examples point to a useful mental model: a discrete space is a set whose points can be inspected one at a time. The topology does not impose continuity restrictions on functions leaving the space, but it imposes very strong restrictions on limits and compactness. ## Continuity Without Constraints ### Maps Out of a Discrete Space Continuity is usually a compatibility condition between two topologies. A function is continuous when the inverse image of every open set is open. If the domain is discrete, every inverse image is automatically open, so the domain topology places no restriction on the function. To state this property carefully, we first need the topological definition of continuity. The definition is formulated using inverse images because inverse images preserve arbitrary unions and finite intersections for every function, regardless of whether the function has any algebraic or metric structure. [definition: Continuous Map] Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be topological spaces. A function $f: X \to Y$ is continuous if for every open set $V \in \tau_Y$, the preimage \begin{align*} f^{-1}(V) = \{x \in X : f(x) \in V\} \end{align*} is open in $(X, \tau_X)$. [/definition] This sets up the theorem below: no preimage test can fail when the domain has the discrete topology. The result is needed because it is the main structural reason to choose the discrete topology for unconstrained input data. [quotetheorem:8519] The theorem explains why the discrete topology is the right topology for unconstrained data. If every point of the domain can be distinguished independently, then any assignment of values is compatible with the topology. [example: Arbitrary Functions from a Discrete Domain] Let $X=\mathbb{Z}$ with the discrete topology, and define $f:\mathbb{Z}\to\mathbb{R}$ by setting $f(n)=0$ when $n$ is even and $f(n)=1/n$ when $n$ is odd. We show that $f$ is continuous when $\mathbb{R}$ has its usual topology. Let $V\subset\mathbb{R}$ be open. Its preimage is \begin{align*} f^{-1}(V)=\{n\in\mathbb{Z}:f(n)\in V\}. \end{align*} This preimage is a subset of $\mathbb{Z}$, so \begin{align*} f^{-1}(V)\in\mathcal{P}(\mathbb{Z}). \end{align*} Because the topology on $X$ is the discrete topology, $\mathcal{P}(\mathbb{Z})$ is exactly the collection of open subsets of $X$. Hence $f^{-1}(V)$ is open in $X$ for every open set $V\subset\mathbb{R}$, so $f$ is continuous by the definition of continuity. The formula changes abruptly between even and odd integers, but continuity is being tested only through open subsets of the discrete domain $\mathbb{Z}$; since every subset of the domain is open, no jump in the assigned values can violate continuity. [/example] ### Maps Into a Discrete Space The direction of the map matters. A discrete codomain imposes strong conditions on continuity, because preimages of singleton sets must be open. This turns continuity into local constancy when the domain has connected pieces or intervals. The next definition names the resulting behavior. It is needed because ordinary formulas may jump between values, while a continuous map into a discrete target must hold one value throughout a neighbourhood of each point. [definition: Locally Constant Map] Let $(X, \tau_X)$ be a topological space and let $Y$ be a set. A function $f: X \to Y$ is locally constant if for every $x \in X$ there exists an open set $U \subset X$ such that $x \in U$ and $f|_U$ is constant. [/definition] Local constancy is exactly the continuity condition for maps into a discrete codomain. The reason this deserves a theorem is that it converts a preimage condition into a geometric description: near each point, the map must settle on a single value. [quotetheorem:8521] This theorem is a useful warning. A discrete topology in the codomain makes continuity harder, not easier. The codomain can test whether a function jumps between individual values. [example: A Noncontinuous Map into a Discrete Space] Let $\mathbb{R}$ have its usual topology, let $Y=\{0,1\}$ have the discrete topology, and define $f:\mathbb{R}\to Y$ by \begin{align*} f(x)=0\text{ for }x<0,\qquad f(x)=1\text{ for }x\ge 0. \end{align*} Since $Y$ has the discrete topology, every subset of $Y$ is open, so $\{1\}$ is open in $Y$. Its preimage is \begin{align*} f^{-1}(\{1\})=\{x\in\mathbb{R}:f(x)\in\{1\}\}. \end{align*} By the definition of $f$, the condition $f(x)\in\{1\}$ is equivalent to $f(x)=1$, and this holds exactly when $x\ge 0$. Hence \begin{align*} f^{-1}(\{1\})=\{x\in\mathbb{R}:x\ge 0\}=[0,\infty). \end{align*} This set is not open in the usual topology on $\mathbb{R}$. Indeed, if $[0,\infty)$ were open, then because $0\in[0,\infty)$ there would be some $\varepsilon>0$ with \begin{align*} (-\varepsilon,\varepsilon)\subset[0,\infty). \end{align*} But $-\varepsilon/2\in(-\varepsilon,\varepsilon)$ and $-\varepsilon/2<0$, so $-\varepsilon/2\notin[0,\infty)$. This contradicts the claimed containment. Therefore $f^{-1}(\{1\})$ is not open in $\mathbb{R}$, so $f$ is not continuous. [/example] The contrast between maps out of and into a discrete space is one of the fastest ways to understand the construction. Discreteness in the domain removes continuity constraints; discreteness in the codomain detects jumps. ## Convergence and Isolation ### Eventual Equality as Convergence Limits are where the discrete topology stops being merely permissive. If every singleton is open, then a sequence that converges to $x$ must eventually lie in the neighbourhood $\{x\}$. Thus convergence cannot mean gradual approach; it means eventual equality. The topological definition of sequence convergence uses neighbourhoods rather than distances. This is important because the discrete topology is not first introduced by a metric, even though it can be metrized later. [definition: Sequence Convergence in a Topological Space] Let $(X, \tau)$ be a topological space, let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $X$, and let $x \in X$. The sequence $(x_n)_{n \in \mathbb{N}}$ converges to $x$ if for every open set $U \subset X$ with $x \in U$, there exists $N \in \mathbb{N}$ such that $x_n \in U$ for all $n \ge N$. [/definition] Singleton neighbourhoods force a sequence converging to $x$ to enter the set $\{x\}$ and then never leave it. To separate this tail condition from the topological language of neighbourhoods, we name the purely set-theoretic behavior that remains: after some finite stage, every term is exactly the same chosen value. [definition: Eventually Constant Sequence] Let $X$ be a set and let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $X$. The sequence is eventually constant with value $x \in X$ if there exists $N \in \mathbb{N}$ such that $x_n = x$ for all $n \ge N$. [/definition] The previous definition is not merely convenient terminology. In a discrete space it is exactly what convergence means, because the singleton neighbourhood of the limit admits no other tail values. [quotetheorem:8523] This result is not a defect. It is the intended behavior. In a discrete data set, repeated equality is the only topological form of convergence. [example: The Alternating Sequence Does Not Converge] Let $X=\{0,1\}$ with the discrete topology, and define $x_n\in X$ by \begin{align*} x_n=0\text{ if }n\text{ is even},\qquad x_n=1\text{ if }n\text{ is odd}. \end{align*} We show directly that the sequence has no limit in $X$. Since $X$ has the discrete topology, every subset of $X$ is open, so $\{0\}$ is an open neighbourhood of $0$. If $x_n\to 0$, then there would be some $N\in\mathbb{N}$ such that $x_n\in\{0\}$ for all $n\ge N$. Choose an odd integer $n\ge N$. For this $n$, the definition gives $x_n=1$, so $x_n\notin\{0\}$, contradicting the required eventual membership in $\{0\}$. The same argument excludes convergence to $1$. The singleton $\{1\}$ is an open neighbourhood of $1$. If $x_n\to 1$, then there would be some $N\in\mathbb{N}$ such that $x_n\in\{1\}$ for all $n\ge N$. Choose an even integer $n\ge N$. Then $x_n=0$, so $x_n\notin\{1\}$, again a contradiction. Since the only points of $X$ are $0$ and $1$, the alternating sequence does not converge in the discrete space $X$. [/example] ### Ambient Limits Versus Discrete Limits The same reasoning separates discrete topology from the subspace topology on many familiar subsets of metric spaces. A subset of $\mathbb{R}$ may look pointwise separated while still carrying ambient limit behavior near a missing accumulation point. [example: A Sequence in $\{1/n : n \in \mathbb{N}\}$] Let $X=\{1/n:n\in\mathbb{N}\}$ and define $x_n=1/n$. First give $X$ the discrete topology. Fix a candidate limit $1/m\in X$. Since $X$ is discrete, the singleton $\{1/m\}$ is an open neighbourhood of $1/m$. If $x_n\to 1/m$, then there would be some $N\in\mathbb{N}$ such that $x_n\in\{1/m\}$ for every $n\ge N$. Choose $n\ge N$ with $n\ne m$, for instance $n=\max\{N,m+1\}$. Then \begin{align*} x_n=\frac{1}{n}\ne \frac{1}{m}, \end{align*} so $x_n\notin\{1/m\}$, contradicting eventual membership in the neighbourhood $\{1/m\}$. Since every point of $X$ has the form $1/m$, the sequence has no limit in the discrete space $X$. Now give $X$ the subspace topology inherited from $\mathbb{R}$. The same sequence approaches the missing ambient point $0$ in $\mathbb{R}$: if $\varepsilon>0$, choose $N\in\mathbb{N}$ with $N>1/\varepsilon$. For $n\ge N$, \begin{align*} 0<\frac{1}{n}\le\frac{1}{N}<\varepsilon, \end{align*} and therefore \begin{align*} |x_n-0|=\left|\frac{1}{n}\right|=\frac{1}{n}<\varepsilon. \end{align*} Thus $x_n\to 0$ in $\mathbb{R}$, while $0\notin X$ because $1/n>0$ for every $n\in\mathbb{N}$. However, this ambient convergence still does not give a limit inside $X$. Fix $1/m\in X$. If $m=1$, then \begin{align*} X\cap(1/2,3/2)=\{1\}. \end{align*} If $m\ge 2$, compute the distances to the adjacent points of the list: \begin{align*} \frac{1}{m-1}-\frac{1}{m}=\frac{m-(m-1)}{m(m-1)}=\frac{1}{m(m-1)}. \end{align*} \begin{align*} \frac{1}{m}-\frac{1}{m+1}=\frac{(m+1)-m}{m(m+1)}=\frac{1}{m(m+1)}. \end{align*} Choose \begin{align*} r=\frac{1}{2}\min\left\{\frac{1}{m(m-1)},\frac{1}{m(m+1)}\right\}. \end{align*} If $k<m$, then $k\le m-1$, so $1/k\ge 1/(m-1)$ and hence \begin{align*} \frac{1}{k}-\frac{1}{m}\ge \frac{1}{m-1}-\frac{1}{m}=\frac{1}{m(m-1)}>r. \end{align*} If $k>m$, then $k\ge m+1$, so $1/k\le 1/(m+1)$ and hence \begin{align*} \frac{1}{m}-\frac{1}{k}\ge \frac{1}{m}-\frac{1}{m+1}=\frac{1}{m(m+1)}>r. \end{align*} Therefore \begin{align*} X\cap\left(\frac{1}{m}-r,\frac{1}{m}+r\right)=\left\{\frac{1}{m}\right\}. \end{align*} So every singleton $\{1/m\}$ is open in the subspace topology, and the same eventual-membership argument shows that $x_n$ cannot converge to $1/m$. The ambient limit $0$ records information from the embedding $X\subset\mathbb{R}$; the discrete topology removes even that memory by making convergence inside $X$ mean eventual equality. [/example] This is why discrete spaces are good local models for isolated data but poor models for approximation by distinct points. Approximation requires a topology with meaningful nested neighbourhoods; discreteness collapses approximation into equality after finitely many terms. ## Compactness and Countability ### Open Covers by Singletons Compactness measures whether open covers admit finite subcovers. In a discrete space, every subset is open, so the cover by singletons is always available. This makes compactness equivalent to finiteness. We recall compactness in its open-cover form because this definition works in every topological space and exposes the role of singleton covers. The finite-subcover condition is the global test that an infinite discrete space fails. [definition: Compact Space] A topological space $(X, \tau)$ is compact if for every family $(U_i)_{i \in I}$ of open subsets of $X$ satisfying \begin{align*} X = \bigcup_{i \in I} U_i, \end{align*} there exists a finite subset $J \subset I$ such that \begin{align*} X = \bigcup_{j \in J} U_j. \end{align*} [/definition] For discrete spaces, compactness has no hidden analytic content. The key obstruction is the singleton cover: if every singleton is open, then an infinite set has an [open cover](/page/Open%20Cover) by singletons, and no finite subfamily can cover all of it. Thus compactness in a discrete space is exactly the question of whether the underlying set is finite. [quotetheorem:8524] The theorem is one of the standard tests for whether a proposed compactness argument is using topology or merely using finiteness. Infinite discrete spaces are locally simple but globally large. [example: The Singleton Cover of $\mathbb{N}$] Let $\mathbb{N}$ have the discrete topology. For each $n\in\mathbb{N}$, the singleton $\{n\}$ is a subset of $\mathbb{N}$, so $\{n\}$ is open. Hence \begin{align*} \mathcal{U}=\{\{n\}:n\in\mathbb{N}\} \end{align*} is a family of open subsets of $\mathbb{N}$. This family covers $\mathbb{N}$. Indeed, if $m\in\mathbb{N}$, then $m\in\{m\}$ and $\{m\}\in\mathcal{U}$, so $m\in\bigcup_{U\in\mathcal{U}}U$. Conversely, every set in $\mathcal{U}$ is a subset of $\mathbb{N}$, so its union is a subset of $\mathbb{N}$. Therefore \begin{align*} \bigcup_{U\in\mathcal{U}}U=\mathbb{N}. \end{align*} Now take any finite subcollection. If it is empty, its union is empty and cannot equal $\mathbb{N}$. Otherwise it has the form \begin{align*} \{\{n_1\},\{n_2\},\ldots,\{n_k\}\} \end{align*} for some $n_1,\ldots,n_k\in\mathbb{N}$. Its union is \begin{align*} \{n_1\}\cup\{n_2\}\cup\cdots\cup\{n_k\}=\{n_1,n_2,\ldots,n_k\}. \end{align*} Let \begin{align*} M=n_1+n_2+\cdots+n_k+1. \end{align*} Then $M\in\mathbb{N}$ and, for each $i\in\{1,\ldots,k\}$, \begin{align*} M=n_1+\cdots+n_k+1>n_i. \end{align*} Thus $M\notin\{n_1,n_2,\ldots,n_k\}$, so this finite subcollection does not cover $\mathbb{N}$. The open cover $\mathcal{U}$ has no finite subcover, and therefore $\mathbb{N}$ is not compact. [/example] ### Sequential Compactness Open covers are not the only compactness-like test used in analysis. In metric spaces, [sequential compactness](/page/Sequential%20Compactness) is often easier to check because it asks for convergent subsequences. In a discrete space, this test also detects finiteness. [definition: Sequentially Compact Space] A topological space $(X, \tau)$ is sequentially compact if every sequence $(x_n)_{n \in \mathbb{N}}$ in $X$ has a convergent subsequence. [/definition] This prepares the theorem below by translating the subsequence test into the language of eventual constancy. A finite discrete space forces repetition, while an infinite discrete space allows a sequence with no repeated tail value. We use the standard undergraduate convention that an infinite set admits a sequence of distinct elements; under this convention, the statement applies to arbitrary infinite discrete spaces. [quotetheorem:8527] The theorem also shows why compactness and sequential compactness agree for discrete spaces. This agreement should not be mistaken for a general topological principle; outside first-countable or metric settings, compactness and sequential compactness can diverge. [example: An Injective Sequence Has No Convergent Subsequence] Let $X=\mathbb{N}$ with the discrete topology, and define $x_n=n$. We show that no subsequence of $(x_n)$ converges. Let $(x_{n_j})_{j\in\mathbb{N}}$ be any subsequence, so $n_1<n_2<n_3<\cdots$. If $j<k$, then $n_j<n_k$, and therefore \begin{align*} x_{n_j}=n_j\ne n_k=x_{n_k}. \end{align*} Thus every two different terms of the subsequence are different. In particular, the subsequence cannot be eventually constant: if there were some $J\in\mathbb{N}$ and some $a\in\mathbb{N}$ such that $x_{n_j}=a$ for all $j\ge J$, then both $x_{n_J}=a$ and $x_{n_{J+1}}=a$, so $x_{n_J}=x_{n_{J+1}}$, contradicting the displayed inequality with $j=J$ and $k=J+1$. By *[Convergent Sequences in a Discrete Space](/theorems/8523)*, a sequence in a discrete space converges exactly when it is eventually constant with its limit value. Since no subsequence of $(x_n)$ is eventually constant, no subsequence of $(x_n)$ converges. [/example] Compactness therefore gives the first major global distinction between finite and infinite discrete spaces. Finite discrete spaces behave like compact zero-dimensional spaces; infinite discrete spaces behave like spaces with too many separated points to be captured by finitely many local observations. ## Separation and Metrization ### Hausdorff Separation The discrete topology separates points as strongly as a topology can. Given two distinct points, each singleton is an open neighbourhood containing one and not the other. This places discrete spaces at the strongest end of the usual separation hierarchy. The most familiar separation axiom is Hausdorffness. It asks that distinct points have disjoint open neighbourhoods, which is exactly the kind of test singleton neighbourhoods are designed to pass. [definition: Hausdorff Space] A topological space $(X, \tau)$ is Hausdorff if for every pair of distinct points $x,y \in X$, there exist open sets $U,V \subset X$ such that $x \in U$, $y \in V$, and $U \cap V = \varnothing$. [/definition] Once singleton neighbourhoods are available, the Hausdorff property follows. The result is worth stating because it connects the discrete topology to the general separation theory used throughout topology and analysis. [quotetheorem:8530] Hausdorffness has consequences for [uniqueness of limits](/theorems/625). In a discrete space, uniqueness is stronger than usual: not only can a sequence have at most one limit, but convergence forces eventual equality with that limit. [example: Separating Two Points] Let $X$ be a discrete space, and let $a,b\in X$ with $a\ne b$. Since $X$ is discrete, every subset of $X$ is open. In particular, \begin{align*} \{a\}\subset X \end{align*} so $\{a\}$ is open, and \begin{align*} \{b\}\subset X \end{align*} so $\{b\}$ is open. The set $\{a\}$ contains $a$, and the set $\{b\}$ contains $b$, so these are open neighbourhoods of $a$ and $b$ respectively. They are disjoint: if $x\in\{a\}\cap\{b\}$, then $x=a$ because $x\in\{a\}$, and $x=b$ because $x\in\{b\}$. Hence $a=b$, contradicting the assumption $a\ne b$. Therefore \begin{align*} \{a\}\cap\{b\}=\varnothing. \end{align*} Thus any two distinct points in a discrete space can be separated by disjoint open neighbourhoods, exactly as the Hausdorff condition requires. [/example] ### The Discrete Metric The discrete topology is also metrizable in a canonical way. The metric should encode no geometry beyond equality, so all distinct points are placed at the same distance. This supplies a useful bridge from abstract topology to metric-space intuition. [definition: Discrete Metric] Let $X$ be a set. The [discrete metric](/page/Discrete%20Metric) on $X$ is the function $d: X \times X \to \mathbb{R}$ defined by $d(x,x)=0$ for all $x \in X$ and $d(x,y)=1$ whenever $x,y \in X$ satisfy $x \ne y$. [/definition] Balls for this metric show exactly why the induced topology is discrete. The possible obstruction is that a metric topology might contain extra neighbourhood structure not present in the intended discrete topology. Here that obstruction disappears because every point has a ball, for example $B(x,1/2)$, containing only that point; once all singletons are open, every subset is open by taking unions. [quotetheorem:8359] This theorem is useful because it lets metric intuition be used without importing unwanted geometry. In the discrete metric, every point is isolated at radius $1/2$. [example: Balls in the Discrete Metric] Let $X$ be any set with at least two elements, and let $d$ be the discrete metric on $X$. Fix $x\in X$. The open ball of radius $1/2$ around $x$ is \begin{align*} B(x,1/2)=\{y\in X:d(x,y)<1/2\}. \end{align*} If $y=x$, then $d(x,y)=d(x,x)=0$, and since $0<1/2$, we have $y\in B(x,1/2)$. Thus $\{x\}\subset B(x,1/2)$. Conversely, if $y\in B(x,1/2)$ and $y\ne x$, then the definition of the discrete metric gives $d(x,y)=1$, but $1<1/2$ is false. Hence no point $y\ne x$ lies in $B(x,1/2)$, so \begin{align*} B(x,1/2)=\{x\}. \end{align*} Therefore every singleton is open in the metric topology. If $A\subset X$, then \begin{align*} A=\bigcup_{a\in A}\{a\}=\bigcup_{a\in A}B(a,1/2). \end{align*} Each $B(a,1/2)$ is open in the metric topology, and arbitrary unions of open sets are open by the topology axioms. Hence every subset $A\subset X$ is open, so the metric topology induced by the discrete metric is the discrete topology. [/example] The discrete metric also clarifies boundedness. Every discrete metric space has diameter at most $1$, but an infinite discrete metric space need not be compact. Boundedness alone is therefore far weaker than compactness outside Euclidean finite-dimensional settings. ## Connectedness and Components ### Connected Spaces Connectedness measures whether a space can be split into two nonempty open pieces. Since every subset of a discrete space is open, any subset and its complement form a separation unless one of them is empty. Thus a discrete space is connected only when it has at most one point. We recall the definition because connectedness is a global property: it is not about separating individual points, but about decomposing the entire space into open pieces that do not touch. [definition: Connected Space] A topological space $(X, \tau)$ is connected if there do not exist nonempty open sets $U,V \subset X$ such that $U \cap V = \varnothing$ and $X = U \cup V$. [/definition] The definition is formulated in terms of open decompositions, but it is often more useful to recognize connectedness by testing functions out of the space. A separation can be encoded by a function that takes two separated values, so connectedness should force every continuous map into a discrete two-valued target to be constant. The following equivalences turn that intuition into a flexible criterion. [quotetheorem:294] These characterisations explain why discrete spaces with two or more points fail to be connected: the characteristic function of a singleton gives a nonconstant continuous map to a discrete target such as $\mathbb{Z}$. They also prepare the component viewpoint, where connectedness is studied by looking for the largest pieces on which all such separation-detecting maps become constant. ### Components and Total Disconnectedness To state the previous idea precisely for arbitrary spaces, we need the notion of a connected component. It records the largest connected part containing a given point, and it lets us compare discrete spaces with spaces that are disconnected for subtler reasons. [definition: Connected Component] Let $(X, \tau)$ be a topological space and let $x \in X$. The connected component of $x$ is the union of all connected subsets of $X$ that contain $x$. [/definition] For a discrete space, the connected component of each point is just that point. This observation raises the next natural question: can we name spaces whose connected components are all point-sized, even if the points themselves are not isolated by open sets? The following definition records exactly this component-level absence of connected continua, separating it from the stronger requirement of discreteness. [definition: Totally Disconnected Space] A topological space $(X, \tau)$ is totally disconnected if every connected component of $X$ is a singleton. [/definition] The definition isolates a property that discrete spaces have, but that does not characterize them. The theorem below records the implication from discreteness to total disconnectedness before the next example shows that the converse fails. [quotetheorem:8533] This theorem separates two ideas that are often confused. A space may have no connected pieces with more than one point without having isolated points. Discreteness is stronger because it requires every singleton to be open. [example: The Rational Numbers Are Not Discrete] Let $\mathbb{Q}$ have the subspace topology inherited from $\mathbb{R}$. The space $\mathbb{Q}$ is totally disconnected: if a connected subset $C\subset\mathbb{Q}$ contained two points $a<b$, choose an irrational number $\alpha$ with $a<\alpha<b$. Then \begin{align*} C=(C\cap(-\infty,\alpha))\cup(C\cap(\alpha,\infty)). \end{align*} Both pieces are open in the subspace topology on $C$, they are disjoint, and they are nonempty because $a$ lies in the first and $b$ lies in the second. This contradicts connectedness, so every connected subset of $\mathbb{Q}$ has at most one point. However, $\mathbb{Q}$ is not discrete. Fix $q\in\mathbb{Q}$ and $\varepsilon>0$. Choose $n\in\mathbb{N}$ with $n>1/\varepsilon$. Then \begin{align*} 0<\frac{1}{n}<\varepsilon. \end{align*} Set $r=q+1/n$. Since $q\in\mathbb{Q}$ and $1/n\in\mathbb{Q}$, we have $r\in\mathbb{Q}$. Also $r\ne q$, because $1/n\ne 0$, and \begin{align*} q<q+\frac{1}{n}<q+\varepsilon. \end{align*} Thus $r\in(q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}$ and $r\ne q$. Therefore every subspace neighbourhood of $q$ in $\mathbb{Q}$ contains a rational point other than $q$, so no singleton $\{q\}$ is open. Hence $\mathbb{Q}$ is totally disconnected but not discrete. [/example] Discrete spaces therefore occupy an extreme position: they are not merely disconnected; their topology permits every point to be isolated by an open set. ## Products, Subspaces, and Quotients ### Subspaces A topology becomes more useful when we know how it behaves under standard constructions. Discreteness is stable under subspaces and finite products, but it can fail under infinite products. These facts explain why discrete objects can produce nondiscrete spaces after a construction. Subspaces are the first test. If a space already allows every subset to be open, then any subset inherits the same property through intersections with open sets in the ambient space. [definition: Subspace Topology] Let $(X, \tau)$ be a topological space and let $A \subset X$. The subspace topology on $A$ is \begin{align*} \tau_A = \{A \cap U : U \in \tau\}. \end{align*} [/definition] When passing to a subspace, openness is tested by intersecting with open sets from the ambient space. The possible obstruction is that the intersection process might remove too many open sets and create new limit behavior. In a discrete ambient space this cannot happen, because any subset of the subspace is already the intersection of the subspace with an open subset of the original space. [quotetheorem:8535] This permanence property is one reason discrete spaces are easy to work with locally. Passing to a subset never introduces new limit behavior. [example: A Subset of a Discrete Space] Let $X=\mathbb{Z}$ with the discrete topology, and let \begin{align*} A=2\mathbb{Z}=\{2k:k\in\mathbb{Z}\}. \end{align*} We compute the subspace topology on $A$. By definition, a subset of $A$ is open in the subspace topology if it has the form $A\cap U$ for some open set $U\subset X$. First, the singleton $\{4\}$ is open in $A$. Since $4=2\cdot 2$, we have $4\in A$, so \begin{align*} A\cap\{4\}=\{4\}. \end{align*} Because $X$ is discrete, every subset of $X$ is open, and in particular $\{4\}$ is open in $X$. Hence $\{4\}=A\cap\{4\}$ is open in the subspace topology on $A$. The same calculation works for every point of $A$. If $a\in A$, then $\{a\}\subset\mathbb{Z}$, so $\{a\}$ is open in the discrete space $X$. Also $a\in A$, hence \begin{align*} A\cap\{a\}=\{a\}. \end{align*} Thus every singleton in $A$ is open. If $B\subset A$, then \begin{align*} B=\bigcup_{b\in B}\{b\}. \end{align*} Since arbitrary unions of open sets are open, every subset $B\subset A$ is open in the subspace topology. Therefore the subspace topology on $A=2\mathbb{Z}$ is the discrete topology. [/example] ### Products Products behave differently depending on how many factors are used. A finite product of discrete spaces is discrete, because a product of singleton opens isolates each point. Infinite products usually lose discreteness because the [product topology](/page/Product%20Topology) only controls finitely many coordinates at a time. The definition of product topology is built to make all coordinate projections continuous. Its finite-coordinate condition is harmless for finite products, but it becomes the source of new limit behavior in infinite products. [definition: Product Topology] Let $((X_i,\tau_i))_{i \in I}$ be a family of topological spaces. The product topology on $\prod_{i \in I} X_i$ is the topology generated by all sets of the form \begin{align*} \prod_{i \in I} U_i, \end{align*} where $U_i \in \tau_i$ for all $i \in I$ and $U_i = X_i$ for all but finitely many $i \in I$. [/definition] The finite restriction in the definition is the key. In a finite product, controlling each coordinate is allowed. In an infinite product, a basic open set cannot prescribe infinitely many coordinates, so it cannot isolate a point in a non-singleton product. [quotetheorem:8537] The theorem is used whenever a finite tuple of discrete data is treated as discrete data again. Ordered pairs of symbols, finite words of fixed length, and finite state vectors inherit a discrete topology. [example: Infinite Product of Two-Point Discrete Spaces] Let \begin{align*} X=\prod_{n=1}^{\infty}\{0,1\}, \end{align*} where each factor $\{0,1\}$ has the discrete topology, and equip $X$ with the product topology. We show that $X$ is not discrete by proving that the singleton containing the all-zero sequence is not open. Let \begin{align*} z=(0,0,0,\ldots)\in X. \end{align*} A basic open set in the product topology has the form \begin{align*} B=\prod_{n=1}^{\infty}U_n, \end{align*} where each $U_n\subset\{0,1\}$ is open and $U_n=\{0,1\}$ for all but finitely many $n$. Suppose a basic open set $B$ contains $z$. Then $0\in U_n$ for every $n$, because the $n$th coordinate of $z$ is $0$. Let \begin{align*} F=\{n\in\mathbb{N}:U_n\ne\{0,1\}\}. \end{align*} By the definition of the product topology, $F$ is finite. Choose $k\in\mathbb{N}$ with $k\notin F$. Then \begin{align*} U_k=\{0,1\}. \end{align*} Define $y\in X$ by setting its coordinates as follows: \begin{align*} y_n=0\text{ for }n\ne k,\qquad y_k=1. \end{align*} For $n\ne k$, we have $y_n=0\in U_n$. For $n=k$, we have $y_k=1\in\{0,1\}=U_k$. Therefore $y_n\in U_n$ for every $n$, so \begin{align*} y\in\prod_{n=1}^{\infty}U_n=B. \end{align*} But $y\ne z$, since their $k$th coordinates are $1$ and $0$ respectively. Hence every basic open neighbourhood of $z$ contains a point other than $z$. If $\{z\}$ were open in the product topology, then some basic open neighbourhood $B$ of $z$ would satisfy $z\in B\subset\{z\}$. The previous paragraph shows that every such $B$ contains a point $y\ne z$, contradicting $B\subset\{z\}$. Thus $\{z\}$ is not open. Since a discrete topology makes every singleton open, the product space $\prod_{n=1}^{\infty}\{0,1\}$ is not discrete. [/example] ### Quotients Quotients are another construction where the source topology controls the result. Quotients identify points, and a quotient of a discrete space remains discrete because every proposed open set has an open preimage. The definition is designed so that the quotient map is continuous and so that the topology on the target is as large as possible subject to that requirement. This makes it the natural topology for spaces obtained by identifying points. [definition: Quotient Topology] Let $(X, \tau_X)$ be a topological space, let $Y$ be a set, and let $q: X \to Y$ be a surjective function. The [quotient topology](/page/Quotient%20Topology) on $Y$ induced by $q$ is \begin{align*} \tau_Y = \{V \subset Y : q^{-1}(V) \in \tau_X\}. \end{align*} [/definition] The theorem below applies the quotient definition to the special case of a discrete source. It is needed because identifications often look capable of creating new topology, but a discrete source makes every preimage test pass. [quotetheorem:8538] The theorem depends on starting with a discrete source. Quotients of nondiscrete spaces can identify points in ways that preserve or worsen nondiscreteness, so the source topology matters. [example: Parity Quotient of the Integers] Let $\mathbb{Z}$ have the discrete topology, and define $q:\mathbb{Z}\to\{0,1\}$ by \begin{align*} q(n)=0\text{ if }n\text{ is even},\qquad q(n)=1\text{ if }n\text{ is odd}. \end{align*} The map is surjective because $q(0)=0$ and $q(1)=1$. We compute the quotient topology on $\{0,1\}$ by checking every subset of $\{0,1\}$. There are four subsets: \begin{align*} \mathcal{P}(\{0,1\})=\{\varnothing,\{0\},\{1\},\{0,1\}\}. \end{align*} Their preimages under $q$ are \begin{align*} q^{-1}(\varnothing)=\varnothing. \end{align*} Also, \begin{align*} q^{-1}(\{0\})=\{n\in\mathbb{Z}:q(n)=0\}=\{n\in\mathbb{Z}:n\text{ is even}\}=2\mathbb{Z}. \end{align*} Similarly, \begin{align*} q^{-1}(\{1\})=\{n\in\mathbb{Z}:q(n)=1\}=\{n\in\mathbb{Z}:n\text{ is odd}\}=2\mathbb{Z}+1. \end{align*} Finally, \begin{align*} q^{-1}(\{0,1\})=\mathbb{Z}, \end{align*} because $q(n)$ is always either $0$ or $1$. Each of $\varnothing$, $2\mathbb{Z}$, $2\mathbb{Z}+1$, and $\mathbb{Z}$ is a subset of $\mathbb{Z}$. Since $\mathbb{Z}$ has the discrete topology, every subset of $\mathbb{Z}$ is open. Hence the preimage of every subset of $\{0,1\}$ is open in $\mathbb{Z}$. By the definition of the quotient topology, every subset of $\{0,1\}$ is open, so the quotient topology on $\{0,1\}$ is the discrete topology. [/example] Subspaces, finite products, and quotients show that discreteness is robust under many finite or pointwise constructions. Infinite products are the main warning: topological constructions can reintroduce limiting behavior even when each coordinate space is discrete. ## Beyond and Connected Topics The discrete topology is the simplest useful test object for [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). It sharpens the distinction between definitions that are local, such as continuity out of a space, and definitions that are global, such as compactness. Many early topology arguments can be checked against discrete spaces to see whether a claimed theorem has hidden hypotheses. In [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology), discrete spaces appear as spaces with no path information beyond constant paths. Their connected components and path components are singletons, and maps from discrete spaces encode indexed families of points in another space. This makes them useful for testing functorial constructions on the simplest possible domains. In [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology), infinite products of finite discrete spaces become important examples of compact totally disconnected spaces. The product $\prod_{n=1}^{\infty}\{0,1\}$ is not discrete, but it is a central model for Cantor-type behavior. This shows how nondiscrete spaces can be built from discrete pieces by changing the topology through an infinite construction. In [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis), discrete sets often appear through sequence spaces such as $\ell^p$. The underlying index set is discrete, but the function space carries a norm topology that is not the discrete topology except in degenerate finite cases. This distinction between a discrete domain and a nondiscrete function space is essential in analysis. The next natural topics are the indiscrete topology, which is the opposite extreme; metric topology, which explains how distances generate open sets; compactness, which separates finite and infinite discrete spaces; and product topology, where infinite products of discrete spaces produce rich examples. ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Androma, [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). Munkres, *Topology* (2000). Willard, *General Topology* (1970). Kelley, *General Topology* (1955).