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In a metric space $(X,d)$, an open ball of radius $r>0$ about $a$ is $B_d(a,r)=\{x\in X:d(a,x)<r\}$. A closed ball instead uses $d(a,x)\le r$. The strict inequality matters: a point at distance exactly $r$ belongs to the closed ball and is excluded from the open ball. In particular, $(1,0)$ is excluded from all three open unit balls above, and included in all three closed unit balls.

The three closed unit balls centred at zero are $|x_1|+|x_2|\le 1$, $x_1^2+x_2^2\le 1$, and $\max \{|x_1|,|x_2|\}\le 1$. Their shapes are a diamond (a rotated square), a disk with a circle as its boundary, and an axis-aligned square. A ball need not look round! The corresponding open balls are the interiors of these shapes. At $(0.8,0.8)$, Manhattan distance from zero is $1.6$ and Euclidean distance is $\sqrt{1.28}\approx 1.131$, whereas Chebyshev distance is $0.8$: only the square's open unit ball contains this point.

These shapes illustrate $B_1(0,1)\subseteq B_2(0,1)\subseteq B_{\infty }(0,1)$, with either the open ball or closed ball convention used consistently. Their outlines are included in the displayed closed balls. In an arbitrary metric space, however, points at distance exactly $r$ need not form the topological boundary; the discrete metric below shows why this distinction matters.

The same self-map can be a contraction mapping for one metric but not for a bilipschitz equivalent metric. On three points with distances $d^{\prime }(x,y)=d^{\prime }(x,z)=2$, $d^{\prime }(y,z)=1$, the map $f(x)=y$, $f(y)=f(z)=z$ has Lipschitz constant $1/2$. For the discrete metric, its Lipschitz constant is one. The metrics satisfy $d\le d^{\prime }\le 2d$, so they preserve topology and completeness but not the strict numerical contraction condition. A contraction mapping theorem fixed-point proof therefore needs the specified metric and its contraction constant.

The discrete metric sets $d(x,x)=0$ and $d(x,y)=1$ for distinct points. It induces the discrete topology, and its Cauchy sequences are exactly the eventually constant sequences.

Every nonempty set equipped with the discrete metric is a bounded set, even if it is an infinite set. For $0<r<1$, both the open ball and closed ball about $a$ are the singleton $\{a\}$. At $r=1$, the open ball is still $\{a\}$, but the closed ball is the whole space; for $r>1$, both are the whole space. Every subset is both an open set and a closed set. Consequently the boundary of the radius-one closed ball is empty, even though every point other than $a$ is at distance exactly one. Also, the closure of $B_d(a,1)$ is $\{a\}$, rather than the radius-one closed ball.

The contraction mapping theorem states that a self-map $f$ of a nonempty complete metric space $(X,d)$ satisfying $d(fu,fv)\le qd(u,v)$ for a fixed $0\le q<1$ has a unique fixed point. Its iterates converge to that point from every starting point.

To prove it, set $u_{n+1}=f(u_n)$. Induction gives $d(u_{n+1},u_n)\le q^{n}d(u_1,u_0)$. Hence for $m>n$, the triangle inequality yieldsThus the iterates are a Cauchy sequence, with limit $u$ by completeness. The contraction inequality gives $d(fu,u)\le qd(u,u_n)+d(u_{n+1},u)\to 0$, so $f(u)=u$. For any two fixed points $u,v$, $d(u,v)\le qd(u,v)$ forces equality of the points. This proves existence, uniqueness and convergence; passing $m\to \infty$ also gives the displayed error bound.

For the three-point example, positivity and symmetry of $d^{\prime }$ are immediate. For three distinct points, the largest side has length two, no more than the sum of the other two; repeated-point triangle inequalities are trivial. Thus $d^{\prime }$ is a metric. The discrete metric satisfieswhich proves bilipschitz equivalence of the two metrics. DefineThe ratios of output to input distances for the three distinct pairs are $1/2,1/2,0$ in $d^{\prime }$, so it is a contraction with $q=1/2$. In $d$, the pair $(x,y)$ has input and output distance one, precluding any $q<1$. This illustrates the contraction property under equivalent metrics.

False. We construct a compact Hausdorff space with a countable dense discrete subspace but no compatible metric. Letfor each subset $S\subseteq \mathbb{N}$. Give $P$ its product topology, and let $X=\overline{j(\mathbb{N})}$ in $P$. The Tychonoff theorem makes $P$ compact, and the product is Hausdorff; hence $X$ is a compact Hausdorff space. By construction, $Y=j(\mathbb{N})$ is countable and dense in $X$, so $X$ is a separable topological space. The coordinate $S=\{n\}$ isolates $j(n)$ within $Y$. Thus the subspace topology on $Y$ is discrete and is induced by the discrete metric.

Suppose $X$ were metrizable. An infinite sequence of distinct points of $Y$ would have a convergent subsequence, by sequential compactness of a compact metric space. Write this subsequence as $j(n_k)$ with distinct $n_k$. For $S=\{n_2,n_4,n_6,\dots \}$, its $S$ coordinate alternates between zero and one and therefore does not converge. But coordinate projections are continuous in the product topology, so every coordinate of a convergent sequence must converge. This contradiction proves that $X$ is not metrizable. This is a separable compactification need not be metrizable counterexample.

Take $X$ to be the set $\mathbb{Q}$, equipped with the discrete metricEvery Cauchy sequence is eventually constant: its terms must eventually have mutual distance in this metric less than $1/2$. Consequently $X$ is a complete metric space. Its topology is the discrete topology, since every singleton is a ball of radius $1/2$ and every subset is a union of such balls.

For the codomain use the usual subspace topology on $\mathbb{Q}$. The inverse image of any open set there is a subset of the discrete $X$, hence is open. Therefore $f$ is continuous and plainly surjective. A continuous image of a complete metric space need not be complete. The two occurrences of $\mathbb{Q}$ have different domain and codomain metrics.

Let $X=\mathbb{R}$ with the usual metric and let $Y=\mathbb{R}$ with the discrete metric, and take $f(x)=x$. Every subset of $Y$ is an open set, so the image of every open subset of $X$ is open: $f$ is an open map.

It is not continuous at any point. For $ϵ=1/2$, every $\delta >0$ allows a distinct $y$ with $|y-x|<\delta$, but the distance between $f(y)$ and $f(x)$ in the discrete metric is 1. Equivalently, the preimage of the open singleton $\{x\}\subset Y$ is not open in $X$. Thus $\boxed{\text{an open map need not be continuous}}$ even between metric spaces.

No. Give an infinite set $X$ the discrete metricThis space is bounded. Every Cauchy sequence is eventually constant, so it is complete. However, the open cover by the singleton balls $B_{1/2}(x)=\{x\}$ has no finite subcover. Hence $X$ is not compact.

For a locally finite discrete metric space $(X,d)$ with basepoint $x_0$, the growth function of a discrete metric space isIn a Cayley graph, left multiplication by $g{x}_0^{-1}$ is a graph isometry carrying $x_0$ to any other vertex $g$. It therefore gives a bijection $B(x_0,r)\to B(g,r)$ for every $r$. Thus the growth function of a fixed Cayley graph is exactly independent of the basepoint.