A metric is supposed to measure closeness, but there is a strange way for closeness to disappear. Imagine a space in which every point can be recognized perfectly: two distinct points are always exactly one unit apart, no matter how similar they look or how they were named. In such a space, a sequence cannot creep toward a new point by making smaller and smaller moves. It must eventually land on the point and stay there. This is the world of the discrete metric. The discrete metric is often the first example showing that the axioms of a [metric space](/page/Metric%20Space) allow more than geometric distance in Euclidean space. It converts any set into a metric space, but the resulting geometry is rigid: balls of radius less than $1$ isolate single points, every subset is open, compactness becomes finiteness in many cases, and continuous maps out of the space lose all local restrictions. [example: A Sequence That Cannot Approach Without Arriving] Let $X$ be a set with at least two points, fix $x_0\in X$, and equip $X$ with the discrete metric: $d(x,x)=0$ and $d(x,y)=1$ when $x\ne y$. We show that a sequence $(x_n)_{n=1}^{\infty}$ cannot converge to $x_0$ unless it is eventually equal to $x_0$. Assume $x_n\to x_0$. By the definition of metric convergence with $\varepsilon=1/2$, there exists $N\in\mathbb{N}$ such that for every $n\ge N$, \begin{align*} d(x_n,x_0)<\frac{1}{2}. \end{align*} Fix such an $n\ge N$. In the discrete metric, exactly one of the following two cases holds. If $x_n=x_0$, then $d(x_n,x_0)=0$. If $x_n\ne x_0$, then $d(x_n,x_0)=1$. The second case is impossible, because \begin{align*} 1<\frac{1}{2} \end{align*} is false. Therefore the only possible case is $d(x_n,x_0)=0$, and by the defining property of the discrete metric this means $x_n=x_0$. Since this holds for every $n\ge N$, the sequence is eventually constant with value $x_0$. Thus, in the discrete metric, convergence to a point means that the sequence eventually arrives at that point and then stays there. [/example] This example is the guiding phenomenon for the chapter. The discrete metric is not hard because its formula is complicated; it is important because it shows how much topology can be forced by a metric with only two distance values. It is the metric counterpart of the discrete topology, and it acts as a testing ground for definitions involving continuity, compactness, connectedness, and completeness. ## Definition The guiding question is whether every set can be made into a metric space without imposing coordinates, order, algebra, or an ambient Euclidean space. The construction we need should distinguish equality from non-equality and refuse to measure any finer relationship between distinct points. [definition: Discrete Metric] Let $X$ be a set. The discrete metric on $X$ is the function $d:X\times X\to\mathbb{R}$ defined by $d(x,x)=0$ for every $x\in X$ and $d(x,y)=1$ whenever $x,y\in X$ satisfy $x\ne y$. [/definition] The formula is only useful if it really behaves like a distance. Positivity and symmetry follow directly from the two cases in the definition, but the triangle inequality requires checking that a distance equal to $1$ cannot be forced to be less than the sum of two zero distances. That is the only possible obstruction to turning this equality test into a metric. [quotetheorem:8358] ## Scaling and Balls This theorem gives a metric space structure on every set. The construction has a visible scale equal to $1$, so the next definition records the same idea with an arbitrary positive separation size. This is useful when numerical diameters or Lipschitz constants are being compared. [definition: Scaled Discrete Metric] Let $X$ be a set and let $a>0$. The scaled discrete metric with scale $a$ is the function $d_a:X\times X\to\mathbb{R}$ defined by $d_a(x,x)=0$ for every $x\in X$ and $d_a(x,y)=a$ whenever $x,y\in X$ satisfy $x\ne y$. [/definition] All scaled discrete metrics determine the same open sets, but they do not have the same numerical diameter when $X$ has at least two points. The scale records a chosen unit of separation, while the topology records only that distinct points are separated. The metric formula is so short that its topology can be read directly from the balls. Before discussing open sets, we need the concrete ball computation because it is the mechanism behind almost every later theorem on convergence, compactness, and continuity. [example: Balls in a Discrete Metric Space] Let $(X,d)$ be a set with the discrete metric, fix $x\in X$, and let $r>0$. We compute the open ball \begin{align*} B(x,r)=\{y\in X:d(x,y)<r\}. \end{align*} First suppose $0<r\le 1$. Since $d(x,x)=0$ and $0<r$, we have \begin{align*} d(x,x)=0<r, \end{align*} so $x\in B(x,r)$. If $y\ne x$, then the definition of the discrete metric gives $d(x,y)=1$. Because $r\le 1$, the inequality $1<r$ is false, so $d(x,y)<r$ is false. Hence no point $y\ne x$ belongs to $B(x,r)$, and therefore \begin{align*} B(x,r)=\{x\}. \end{align*} Now suppose $r>1$. For $y\in X$, either $y=x$ or $y\ne x$. If $y=x$, then \begin{align*} d(x,y)=d(x,x)=0<r. \end{align*} If $y\ne x$, then \begin{align*} d(x,y)=1<r. \end{align*} In both cases $y\in B(x,r)$, so every point of $X$ lies in the ball. Thus \begin{align*} B(x,r)=X. \end{align*} The radius $1$ is the threshold: balls of radius at most $1$ isolate the centre, while balls of radius greater than $1$ contain the whole space. [/example] ## The Discrete Topology Generated by the Metric A metric space generates a topology by declaring unions of open balls to be open. For the Euclidean metric, this produces a topology with local shape and boundary behavior. For the discrete metric, the balls of radius $1/2$ already isolate individual points, so the resulting topology is as fine as possible. To make that comparison precise, we first need the general definition of the topology associated with a metric. This definition is the bridge from numerical distances to topological language such as [open set](/page/Open%20Set), closure, compactness, and connectedness. [definition: Metric Topology] Let $(X,d)$ be a metric space. The metric topology induced by $d$ is the collection $\tau_d$ of subsets $U\subset X$ such that for every $x\in U$ there exists $r>0$ with $B(x,r) \subset U$. [/definition] The metric topology records which subsets are visible through open balls. In the discrete metric, every singleton is visible at scale $1/2$. We therefore need a name for the topology in which every possible subset is already open. [definition: Discrete Topology] Let $X$ be a set. The discrete topology on $X$ is the topology $\tau = \{U : U \subset X\}$. [/definition] The two definitions now have to be connected at the level of arbitrary subsets, not just singletons. Since every point of a subset has a radius-$1/2$ ball that stays inside the subset, there is no remaining local condition for openness to impose. The issue is whether this singleton-ball computation really promotes every subset to an open set in the induced topology. [quotetheorem:8359] The theorem explains why the phrase discrete metric is not only about distances: every subset is topologically open, so there are no forced limit points and no local constraints inside the domain. It also raises the converse problem: if the topology has already declared every subset open, the next theorem is needed to show that this topological structure really comes from a metric. [quotetheorem:8360] This result is often the first metrization theorem a student meets: there is no separation axiom to check, no countability condition, and no construction beyond the equality test. ## Convergence and Cauchy Behavior ### Eventual Equality Metric spaces are built to discuss limiting processes, and the discrete metric turns limiting processes into stabilization. This makes it a useful boundary case: whenever a theorem about metric spaces has a conclusion involving convergent subsequences, Cauchy sequences, or closure, the discrete metric often translates that conclusion into a finite or eventual statement. We need a term for sequences that have stopped changing after a finite time. This definition is not special to metric spaces, but in the discrete metric it becomes exactly the metric notion of convergence. [definition: Eventually Constant Sequence] Let $X$ be a set. A sequence $(x_n)_{n=1}^{\infty}$ in $X$ 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 definition isolates the exact behavior suggested by the opening example, but convergence is still phrased in terms of every positive error tolerance. In the discrete metric the decisive tolerance is $1/2$: once all late terms are within that distance of the proposed limit, they must actually equal it. The question is whether this single threshold captures the full convergence definition. [quotetheorem:7790] This theorem is the main computational test for limits. To understand completeness, however, we must look not only at sequences with identified limits but also at sequences whose late terms are mutually close. ### Cauchy Sequences and Completeness Cauchy sequences are designed to detect potential convergence before a candidate limit has been named. In the discrete metric, this idea becomes especially rigid because mutual closeness at radius $1/2$ forces equality among all sufficiently late terms. [definition: Cauchy Sequence in a Metric Space] Let $(X,d)$ be a metric space. A sequence $(x_n)_{n=1}^{\infty}$ in $X$ is Cauchy if for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that $d(x_m,x_n)<\varepsilon$ for all $m,n\ge N$. [/definition] The definition is usually weaker than convergence, since the limit may lie outside the space. In a discrete metric space, there is no such gap because the Cauchy condition itself forces the sequence to choose a permanent value. [quotetheorem:8361] The previous theorem supplies exactly the missing ingredient for completeness, because an eventually constant sequence converges to its eventual value. This motivates the completeness statement: no completion adds new points to a discrete metric space, since every Cauchy process has already stabilized inside the original set. [quotetheorem:8362] Completeness alone should not be mistaken for compactness. The following example is needed because it shows that an infinite discrete metric space can be complete while still having sequences with no convergent subsequence. [example: A Complete Space With No Convergent Subsequence] Let $X=\mathbb{N}$ with the discrete metric, and define $x_n=n$. We show that the sequence has no convergent subsequence, although the metric space itself is complete. Let $(x_{n_j})_{j=1}^{\infty}$ be any subsequence. Since $n_1<n_2<\cdots$, for any $i\ne j$ we have $n_i\ne n_j$, and hence \begin{align*} x_{n_i}=n_i\ne n_j=x_{n_j}. \end{align*} Therefore the discrete metric gives \begin{align*} d(x_{n_i},x_{n_j})=1 \end{align*} whenever $i\ne j$. In particular, the subsequence is not eventually constant: after any index $J$, the two later terms $x_{n_J}$ and $x_{n_{J+1}}$ are distinct. If the subsequence converged to some $a\in\mathbb{N}$, then applying the definition of convergence with $\varepsilon=1/2$ would give an index $J$ such that \begin{align*} d(x_{n_j},a)<\frac{1}{2} \end{align*} for every $j\ge J$. In the discrete metric this inequality forces $x_{n_j}=a$ for every $j\ge J$, because the only other possible distance is $1$, and $1<1/2$ is false. That would make the subsequence eventually constant, contradicting the previous paragraph. Hence no subsequence of $(x_n)$ converges. On the other hand, $X$ is complete. Indeed, let $(y_n)_{n=1}^{\infty}$ be a [Cauchy sequence](/page/Cauchy%20Sequence) in $X$. Using $\varepsilon=1/2$ in the Cauchy condition, there is $N\in\mathbb{N}$ such that \begin{align*} d(y_m,y_n)<\frac{1}{2} \end{align*} for all $m,n\ge N$. Fixing $m=N$, we get \begin{align*} d(y_N,y_n)<\frac{1}{2} \end{align*} for every $n\ge N$. The discrete metric again forces $y_n=y_N$ for all $n\ge N$, so every Cauchy sequence is eventually constant and therefore converges to its eventual value. Thus $\mathbb{N}$ with the discrete metric is complete, even though the sequence $x_n=n$ has no convergent subsequence. [/example] ## Continuity and Quantitative Control ### Maps Out of Discrete Spaces Continuity is local: a function is continuous if small changes in the input force small changes in the output. The discrete metric removes the input side of this restriction, because sufficiently small balls contain only the base point. Therefore every function from a discrete metric space is continuous, whatever the codomain is. We first recall the pointwise metric definition of continuity so that the role of the radius $1/2$ is visible. The definition is needed before the theorem because the theorem works by choosing a universal small ball in the domain. [definition: Continuous Map Between Metric Spaces] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A map $f:X\to Y$ is continuous at $x\in X$ if for every $\varepsilon>0$ there exists $\delta>0$ such that $d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon$ for all $y\in X$. The map $f$ is continuous if it is continuous at every point $x\in X$. [/definition] The definition becomes automatic on a discrete domain because a small enough input ball contains no point other than its centre. Thus the usual burden of controlling nearby but distinct inputs disappears: after choosing $\delta<1$, the implication only tests $y=x$, where the output distance is zero. This removes the only local obstruction to continuity. [quotetheorem:4952] The theorem is asymmetric. Maps out of a discrete space are unconstrained, but maps into a discrete space must keep each output value stable on an open neighbourhood of every point mapping to it. Equivalently, the possible obstruction is a fibre whose points cannot be protected from nearby points with different values. [quotetheorem:8363] This characterization explains why integer-valued continuous functions on connected domains are forced to be constant. The example below shows how a discrete codomain turns continuity into a statement about a partition by open fibres. [example: Integer-Valued Continuous Functions] Let $U\subset\mathbb{R}$ be a nonempty interval with the Euclidean metric, equip $\mathbb{Z}$ with the discrete metric, and let $f:U\to\mathbb{Z}$ be continuous. We show that only one integer occurs as a value of $f$. Fix $k\in\mathbb{Z}$ and consider the fibre \begin{align*} F_k=f^{-1}(\{k\})=\{x\in U:f(x)=k\}. \end{align*} If $x\in F_k$, then $f(x)=k$. By continuity of $f$ at $x$ with $\varepsilon=1/2$, there exists $\delta>0$ such that for every $y\in U$, \begin{align*} |x-y|<\delta \implies d_{\mathbb{Z}}(f(x),f(y))<\frac{1}{2}. \end{align*} Since $f(x)=k$, this becomes \begin{align*} |x-y|<\delta \implies d_{\mathbb{Z}}(k,f(y))<\frac{1}{2}. \end{align*} In the discrete metric on $\mathbb{Z}$, either $f(y)=k$ and $d_{\mathbb{Z}}(k,f(y))=0$, or $f(y)\ne k$ and $d_{\mathbb{Z}}(k,f(y))=1$. The second case is impossible under the displayed inequality, because $1<1/2$ is false. Hence $f(y)=k$, so $y\in F_k$. Therefore every point of $F_k$ has a relative open neighbourhood contained in $F_k$, and $F_k$ is open in $U$. The same argument applies to every fibre $F_j$. For fixed $k$, the complement of $F_k$ is the union of all fibres with labels different from $k$: \begin{align*} U\setminus F_k=\bigcup_{\{j\in\mathbb{Z}:j\ne k\}} F_j. \end{align*} This is a union of open subsets of $U$, so $U\setminus F_k$ is open in $U$ and $F_k$ is closed in $U$. The fibres are pairwise disjoint, since one point cannot have two distinct integer values, and they cover $U$ because every $x\in U$ has a value $f(x)\in\mathbb{Z}$. Because $U$ is a nonempty interval, it is connected by *Connectedness of Intervals*. If two different fibres were nonempty, then one nonempty fibre $F_k$ would be a nonempty proper subset of $U$ that is both open and closed, separating $U$ from its open complement. Hence exactly one fibre is nonempty, so there is some $k_0\in\mathbb{Z}$ such that $f(x)=k_0$ for every $x\in U$. Thus every continuous integer-valued function on an interval is constant. [/example] ### Lipschitz and Uniform Continuity Continuity only tests sufficiently close pairs of points, which makes it automatic on a discrete domain. Lipschitz continuity is different because it compares all pairs quantitatively, so it can still detect whether the image has bounded diameter. [definition: Lipschitz Map] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A map $f:X\to Y$ is Lipschitz if there exists $L\ge 0$ such that $d_Y(f(x),f(y)) \le L d_X(x,y)$ for all $x,y\in X$. [/definition] The discrete metric has $d_X(x,y)=1$ for distinct points, so the Lipschitz inequality no longer asks about small neighbourhoods; it asks whether all distances between image points are bounded by one constant. The obstruction is therefore global: a function from a discrete space can be continuous while its image spreads out without bound. [quotetheorem:8364] This distinction matters on infinite discrete domains. Continuity alone sees only local behavior, while Lipschitz continuity still sees the global size of the image. [example: Continuous But Not Lipschitz] Let $X=\mathbb{N}$ with the discrete metric, and define $f:\mathbb{N}\to\mathbb{R}$ by $f(n)=n$, where $\mathbb{R}$ has its usual metric. The map is continuous by *Every Map From a Discrete Metric Space Is Continuous*, since its domain is equipped with the discrete metric. We show that $f$ is not Lipschitz. Suppose, for contradiction, that there is some $L\ge 0$ such that for all $m,n\in\mathbb{N}$, \begin{align*} |f(m)-f(n)|\le Ld(m,n). \end{align*} Take $n=1$ and $m=L+2$ if $L+2\in\mathbb{N}$. If $L+2$ is not an integer, take instead any integer $m>L+1$, which exists by the [Archimedean property](/theorems/737) of $\mathbb{N}$. Then $m\ne 1$, so the discrete metric gives \begin{align*} d(m,1)=1. \end{align*} Using $f(k)=k$, the Lipschitz inequality with the pair $(m,1)$ becomes \begin{align*} |f(m)-f(1)|\le Ld(m,1). \end{align*} Substituting $f(m)=m$, $f(1)=1$, and $d(m,1)=1$, this is \begin{align*} |m-1|\le L. \end{align*} Since $m>1$, we have $|m-1|=m-1$. Since $m>L+1$, subtracting $1$ from both sides gives \begin{align*} m-1>L. \end{align*} Thus the same number $m-1$ is both at most $L$ and greater than $L$, a contradiction. Hence no Lipschitz constant exists. This example shows that discreteness of the domain makes continuity automatic, but Lipschitz continuity still detects the unbounded growth of the image values $f(n)=n$. [/example] [Uniform continuity](/page/Uniform%20Continuity) still asks only about sufficiently close pairs, but it asks for one radius that works over the whole domain. In the discrete metric, the same small radius works everywhere, so uniform continuity returns to being automatic. [definition: Uniformly Continuous Map] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A map $f:X\to Y$ is uniformly continuous if for every $\varepsilon>0$ there exists $\delta>0$ such that $d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon$ for all $x,y\in X$. [/definition] This definition is stronger than pointwise continuity on many domains because the same $\delta$ must work everywhere. In a discrete domain, however, the radius $1/2$ isolates every point at once, so the usual obstruction to uniformity has nowhere to appear. The only sufficiently close pairs are equal pairs, independently of their location in the space. [quotetheorem:8365] This theorem helps separate uniform continuity from Lipschitz continuity. Uniform continuity is still local in the sense that it only asks for sufficiently close input pairs, and the only sufficiently close pairs are equal pairs. ## Compactness, Total Boundedness, and Separability ### Finite Covers by Small Balls Compactness asks whether every open cover has a finite subcover, or equivalently in many metric settings whether every sequence has a convergent subsequence. In the discrete metric, both viewpoints reveal the same obstruction: an infinite discrete metric space has infinitely many isolated points that cannot be grouped by small balls. The right metric notion before compactness is [total boundedness](/page/Total%20Boundedness). We need it because boundedness alone is too weak: every discrete metric space has diameter at most $1$, but an infinite one cannot be covered by finitely many balls of radius $1/2$. [definition: Totally Bounded Metric Space] A metric space $(X,d)$ is totally bounded if for every $\varepsilon>0$ there exists a finite set $F\subset X$ such that $X \subset \bigcup_{x\in F} B(x,\varepsilon)$. [/definition] Total boundedness measures how many small balls are required to cover the whole space. In a discrete metric space, the radius-$1/2$ balls are singletons, so a finite cover by such balls can contain only finitely many points. The obstruction to total boundedness is therefore exactly the presence of infinitely many isolated points. [quotetheorem:1088] This theorem captures the main compactness obstruction. Since every discrete metric space is complete, compactness can hold only when the small-ball obstruction disappears, which leads to the finite classification. [quotetheorem:316] The open-cover proof is especially transparent: cover $X$ by singleton open sets. The example below is the failure mode that prevents compactness in every infinite discrete metric space. [example: The Singleton Cover of an Infinite Discrete Space] Let $X$ be an infinite set with the discrete metric. For each $x\in X$, the singleton $\{x\}$ is open: if $y\in\{x\}$, then $y=x$, and the open ball of radius $1/2$ around $y$ satisfies \begin{align*} B(y,1/2)=\{z\in X:d(y,z)<1/2\}=\{y\}\subset\{x\}. \end{align*} Indeed, $d(y,y)=0<1/2$, while for $z\ne y$ the discrete metric gives $d(y,z)=1$, and $1<1/2$ is false. Now let \begin{align*} \mathcal{U}=\{\{x\}:x\in X\}. \end{align*} This is an open cover of $X$, because every member of $\mathcal{U}$ is open and, for every $p\in X$, the set $\{p\}\in\mathcal{U}$ contains $p$. Suppose a finite subfamily is chosen: \begin{align*} \{\{x_1\},\{x_2\},\ldots,\{x_n\}\}\subset\mathcal{U}. \end{align*} Its union is \begin{align*} \{x_1\}\cup\{x_2\}\cup\cdots\cup\{x_n\}=\{x_1,x_2,\ldots,x_n\}. \end{align*} This set has at most $n$ elements, so it is finite. Since $X$ is infinite, there exists some $p\in X$ with \begin{align*} p\notin \{x_1,x_2,\ldots,x_n\}. \end{align*} Hence the finite subfamily does not cover $X$. Therefore the open cover $\mathcal{U}$ has no finite subcover, so $X$ is not compact. [/example] ### Dense Sets Separability measures whether a metric space has a countable [dense subset](/page/Dense%20Subset). In a discrete metric space, density is demanding because every singleton is open: a dense set must meet every singleton, hence must contain every point. We need the definition of separability to make the size distinction precise. It will show that countability is the exact threshold for having a countable dense subset in the discrete metric. [definition: Separable Metric Space] A metric space $(X,d)$ is separable if there exists a countable subset $D\subset X$ such that $\overline{D}=X$. [/definition] In the discrete topology, closures do not add new points: if a point is missing from $D$, the open singleton around that point misses $D$ entirely. Thus density forces $D=X$. The only remaining question for separability is whether the whole underlying set can itself be chosen countable. [quotetheorem:8366] Thus the discrete metric sharply distinguishes three size conditions: finite means compact, countable means separable, and arbitrary size still gives completeness. ## Connectedness and Path Structure ### Separation by Singletons Connectedness asks whether a space can be split into two separated open pieces. Since every subset of a discrete metric space is open, any subset is also closed. This makes nontrivial connectedness impossible once the space has more than one point. To state the classification, we first need the topological definition of connectedness. The definition matters here because discreteness gives many open-and-closed subsets, and connectedness is exactly the condition that forbids such separations. [definition: Connected Topological Space] A [topological space](/page/Topological%20Space) $X$ is connected if there do not exist nonempty open sets $U,V\subset X$ such that $X = U \cup V$ and $U\cap V=\varnothing$. [/definition] The definition detects whether the space can be decomposed into separated pieces. In a discrete metric space with two distinct points, choosing one point and its complement immediately produces two disjoint open pieces. The obstruction to connectedness is therefore not subtle topology, but simply the existence of more than one point. [quotetheorem:8367] The same rigidity appears for paths. To explain it, we need the definition of a path as a continuous map from an interval, because intervals carry connectedness from the domain into the image. ### Paths Cannot Move [Path connectedness](/page/Path%20Connectedness) is a stronger way of asking whether points can be joined by continuous motion. In a discrete codomain, continuous motion has no room to jump, because each point is separated from every other by a fixed positive distance. [definition: Path in a Metric Space] Let $(X,d)$ be a metric space. A path in $X$ is a continuous map $\gamma:[0,1] \to X$. [/definition] This definition turns motion into continuity. For a path in a discrete metric space, each fibre of the path is open, while the interval domain cannot be split into separated nonempty open pieces. The obstruction to a nonconstant path is that changing value would create such a separation of $[0,1]$. [quotetheorem:8368] This is why the discrete metric is not a geometry of movement. It is a geometry of exact identification: points can be named, separated, and selected, but not approached along a nonconstant continuous curve. [example: No Continuous Path Between Two Integers] Equip $\mathbb{Z}$ with the discrete metric. We show that there is no continuous path $\gamma:[0,1]\to\mathbb{Z}$ satisfying $\gamma(0)=0$ and $\gamma(1)=1$. Suppose such a path existed, and set \begin{align*} A=\gamma^{-1}(\{0\})=\{t\in[0,1]:\gamma(t)=0\}. \end{align*} The set $A$ is nonempty because $\gamma(0)=0$, so $0\in A$. It is also proper because $\gamma(1)=1\ne 0$, so $1\notin A$. We first show that $A$ is open in $[0,1]$. Let $t\in A$. Then $\gamma(t)=0$. Since $\gamma$ is continuous at $t$, applying continuity with $\varepsilon=1/2$ gives some $\delta>0$ such that for every $s\in[0,1]$, \begin{align*} |s-t|<\delta \implies d_{\mathbb{Z}}(\gamma(s),\gamma(t))<\frac{1}{2}. \end{align*} Because $\gamma(t)=0$, this says \begin{align*} |s-t|<\delta \implies d_{\mathbb{Z}}(\gamma(s),0)<\frac{1}{2}. \end{align*} In the discrete metric, if $\gamma(s)\ne 0$, then $d_{\mathbb{Z}}(\gamma(s),0)=1$, and the inequality $1<1/2$ is false. Hence $\gamma(s)=0$ whenever $|s-t|<\delta$, so a relative neighbourhood of $t$ lies in $A$. Thus $A$ is open in $[0,1]$. We next show that $[0,1]\setminus A$ is open. Let $t\in[0,1]\setminus A$, and write $\gamma(t)=k$. Then $k\ne 0$. By continuity at $t$ with $\varepsilon=1/2$, there exists $\delta>0$ such that for every $s\in[0,1]$, \begin{align*} |s-t|<\delta \implies d_{\mathbb{Z}}(\gamma(s),k)<\frac{1}{2}. \end{align*} If $\gamma(s)\ne k$, then the discrete metric gives $d_{\mathbb{Z}}(\gamma(s),k)=1$, again impossible because $1<1/2$ is false. Therefore $\gamma(s)=k\ne 0$ whenever $|s-t|<\delta$, so this relative neighbourhood of $t$ lies in $[0,1]\setminus A$. Hence $[0,1]\setminus A$ is open in $[0,1]$. Thus $A$ is a nonempty proper subset of $[0,1]$ whose complement is also open, so $A$ separates $[0,1]$. This contradicts *Connectedness of Intervals*. Therefore no continuous path in the discrete metric on $\mathbb{Z}$ can start at $0$ and end at $1$. [/example] ## Function Spaces on Discrete Domains Another useful viewpoint comes from function spaces. If $X$ is discrete, then functions on $X$ are just indexed families of values. Topological restrictions disappear, and analytic restrictions come only from boundedness, summability, or other imposed norms. For countable discrete domains, the basic analytic examples are sequence spaces. The exponent $p$ measures summability rather than continuity: once the domain is discrete, the only remaining question is how large the indexed family is allowed to be. [definition: $\ell^p$ Space] For $1\le p<\infty$, the space $\ell^p$ is the set of all sequences $a=(a_n)_{n=1}^{\infty}$ of [real numbers](/page/Real%20Numbers) such that \begin{align*} \|a\|_{\ell^p} := \left(\sum_{n=1}^{\infty} |a_n|^p\right)^{1/p} < \infty. \end{align*} The norm functional is \begin{align*} \|\cdot\|_{\ell^p}:\ell^p &\to [0,\infty). \end{align*} [/definition] The case $p=\infty$ asks for boundedness rather than summability. This is the right space when continuity has become automatic but global size still matters. [definition: Bounded Functions on a Discrete Set] Let $X$ be a set. The space $\ell^\infty(X)$ is the set of functions $f:X\to\mathbb{R}$ such that \begin{align*} \|f\|_{\ell^\infty(X)} := \sup_{x\in X} |f(x)| < \infty. \end{align*} The norm functional is \begin{align*} \|\cdot\|_{\ell^\infty(X)}:\ell^\infty(X) &\to [0,\infty). \end{align*} [/definition] This definition is not part of the metric structure itself, but it is a natural analytic layer placed on a discrete domain. The following example is needed because it separates continuity, uniform continuity, and boundedness in a single familiar space. [example: Continuous Functions Versus Bounded Functions] Let $X=\mathbb{N}$ with the discrete metric, and define $f(n)=n$ and $g(n)=(-1)^n$. First observe that every function $h:\mathbb{N}\to\mathbb{R}$ is uniformly continuous in this metric. Let $\varepsilon>0$ and choose $\delta=1/2$. If $m,n\in\mathbb{N}$ satisfy $d(m,n)<\delta$, then $m\ne n$ would imply $d(m,n)=1$, which would give $1<1/2$, impossible. Hence $m=n$, and therefore \begin{align*} |h(m)-h(n)|=|h(n)-h(n)|=0<\varepsilon. \end{align*} Thus $h$ is uniformly continuous. Since the same $\delta$ works at each fixed base point, $h$ is also continuous. Applying this to $h=f$ and $h=g$, both $f$ and $g$ are continuous and uniformly continuous. The two functions differ in boundedness. For $f(n)=n$, suppose there were a finite number $M$ such that $|f(n)|\le M$ for every $n\in\mathbb{N}$. By the Archimedean property, choose $n\in\mathbb{N}$ with $n>M$. Then \begin{align*} |f(n)|=|n|=n>M, \end{align*} contradicting $|f(n)|\le M$. Hence $f$ has no finite uniform bound, so \begin{align*} \sup_{n\in\mathbb{N}} |f(n)|=\infty, \end{align*} and $f\notin \ell^\infty(\mathbb{N})$. For $g(n)=(-1)^n$, every value is either $1$ or $-1$, so for every $n\in\mathbb{N}$, \begin{align*} |g(n)|=|(-1)^n|=1. \end{align*} Thus $1$ is an upper bound for the set $\{|g(n)|:n\in\mathbb{N}\}$, and it is also attained at every index. Therefore \begin{align*} \|g\|_{\ell^\infty(\mathbb{N})}=\sup_{n\in\mathbb{N}} |g(n)|=1. \end{align*} This example shows that discreteness of the domain removes local continuity restrictions, but membership in $\ell^\infty(\mathbb{N})$ still requires a global bound on the function values. [/example] The lesson is that discreteness removes local analytic constraints but does not remove global size constraints. This distinction appears throughout analysis on graphs, sequences, counting measure spaces, and Banach spaces of functions on discrete sets. ## Beyond and Connected Topics The discrete metric is the metric-space version of the [discrete topology](/page/Topology): every point is isolated and every subset is open. It is the cleanest example to keep in mind when studying [metric spaces](/page/Metric%20Space), because it tests which statements are genuinely metric and which depend on Euclidean intuition. The comparison with the indiscrete topology also explains why metrizability carries separation information: a metric can produce the discrete topology on any set, but it cannot produce the indiscrete topology on a set with more than one point. In analysis, infinite discrete metric spaces separate completeness from compactness. They are complete for a sequence-level reason, yet fail compactness because singleton covers have no finite subcover. This makes them standard test examples for the compactness and sequence arguments developed in [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) and for the foundational limiting arguments in [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Discrete domains also lead naturally to sequence spaces such as $\ell^p$ and $\ell^\infty$. Once the domain is a countable discrete set, a function becomes a sequence, and analytic questions become questions about summability, boundedness, operators, and convergence. This viewpoint connects the topic to function spaces and normed-space methods in [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Finally, discrete codomains appear whenever a continuous object is forced to take separated values. Integer-valued continuous functions, locally constant functions, covering space fibres, and winding-number type invariants all use the same principle: a continuous map from a [connected space](/page/Connected%20Space) into a discrete space must be constant. This principle also appears indirectly in complex analysis, where integer-valued invariants remain constant under continuous deformation, as in themes from [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). ## 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 Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Munkres, *Topology* (2000). Willard, *General Topology* (1970). Rudin, *Principles of Mathematical Analysis* (1976).