Many metrics can be converted into bounded metrics while preserving the same zero-distance relation and triangle behavior. This is useful when large distances must be compressed without changing the underlying metric structure. Two standard constructions are the minimum metric $\min\{1,d\}$ and the fractional transform $d/(1+d)$. :::definition[Minimum Bounded Metric] Let $(X,d)$ be a metric space. Define $$\begin{equation*} e(x,y)=\min\{1,d(x,y)\},\qquad x,y\in X. \end{equation*}$$ Then $e$ is called the $minimum bounded metric$ generated by $d$. ::: :::theorem[Metric Property of the Minimum Construction] Let $(X,d)$ be a metric space and define $e(x,y)=\min\{1,d(x,y)\}$. Then $e$ is a metric on $X$. ::: :::proof Given that $(X,d)$ is a metric space and $e(x,y)=\min\{1,d(x,y)\}$. To prove that $e$ is a metric on $X$. Non-negativity and symmetry follow from $d$. Also, $$\begin{align*} e(x,y)=0 &\iff \min\{1,d(x,y)\}=0\\ &\iff d(x,y)=0\\ &\iff x=y. \end{align*}$$ For $a,b\geq 0$, $$\begin{equation*} \min\{1,a+b\}\leq \min\{1,a\}+\min\{1,b\}. \end{equation*}$$ Therefore, using the triangle inequality for $d$, $$\begin{align*} e(x,y)&=\min\{1,d(x,y)\}\\ &\leq \min\{1,d(x,z)+d(z,y)\}\\ &\leq e(x,z)+e(z,y). \end{align*}$$ Hence $e$ is a metric. $\square$ ::: :::definition[Fractional Bounded Metric] Let $(X,d)$ be a metric space. Define $$\begin{equation*} d'(x,y)=\frac{d(x,y)}{1+d(x,y)},\qquad x,y\in X. \end{equation*}$$ Then $d'$ is called the $fractional bounded metric$ generated by $d$. ::: :::theorem[Bounded Metric from a Fractional Transform] Let $(X,d)$ be a metric space. Then $d'(x,y)=d(x,y)/(1+d(x,y))$ is a metric on $X$ and $d'(x,y)<1$. ::: :::proof Given that $d'(x,y)=d(x,y)/(1+d(x,y))$. To prove that $d'$ is a metric and bounded by $1$. The first three axioms follow from $d$ and positivity of $1+d(x,y)$. Let $\phi(t)=t/(1+t)$ on $[0,\infty)$. The function $\phi$ is increasing and satisfies $\phi(a+b)\leq \phi(a)+\phi(b)$ for $a,b\geq 0$. Hence $$\begin{align*} d'(x,y)&=\phi(d(x,y))\\ &\leq \phi(d(x,z)+d(z,y))\\ &\leq \phi(d(x,z))+\phi(d(z,y))\\ &=d'(x,z)+d'(z,y). \end{align*}$$ Also $0\leq d(x,y)/(1+d(x,y))<1$. Hence $d'$ is a bounded metric. $\square$ ::: :::example On $\mathbb{R}$, the formula $$\begin{equation*} d'(x,y)=\frac{|x-y|}{1+|x-y|} \end{equation*}$$ defines a bounded metric. ::: :::definition[Sequence Transform Metric] Let $X$ be the space of all scalar sequences. For $x=\{x_i\}_{i\geq 1}$ and $y=\{y_i\}_{i\geq 1}$, define $$\begin{equation*} d(x,y)=\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|}. \end{equation*}$$ This formula defines a metric on $X$. ::: :::proof Given that $X$ is the set of all scalar sequences and $d$ is defined by the stated series. To prove that $d$ is a metric. Each term is non-negative and at most $2^{-i}$, so the series converges. Non-negativity and symmetry follow termwise. If $d(x,y)=0$, every term is zero, so $x_i=y_i$ for every $i$ and $x=y$. The triangle inequality follows by applying the fractional metric inequality in each coordinate, multiplying by $2^{-i}$, and summing. Hence $d$ is a metric. $\square$ ::: :::exercise [1] Why is $\min\{1,d(x,y)\}$ bounded? [2] Show that $d(x,y)/(1+d(x,y))<1$. [3] Why is the sequence-transform series finite? ::: :::answer [1] It is always at most $1$. [2] Since $d(x,y)\geq 0$, the numerator is strictly smaller than the denominator unless $d(x,y)=0$, and then the value is $0$. [3] Each term is at most $2^{-i}$, and $\sum_{i=1}^{\infty}2^{-i}=1$. ::: :::call-to-action[Continue to Pseudometrics] subtitle: Study what happens when zero distance is allowed between distinct points. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/pseudometrics-and-quotient-metrics button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/standard-metrics-on-euclidean-spaces :::