202009240421 Problem 2.1.1

For $x,\, y\in \mathbb{K}$ ( $\mathbb{R}$ or $\mathbb{C}$ ), define

$d(x,y)=\mathrm{min}\big\{ 1,\, |x-y| \big\}$ .

Prove that $d$ is a metric on $\mathbb{K}$ .

Motivation.

A ruler is marked by rules for the sake of measuring things. Not so much common to a ruler, but well worth the rule for the general, that a metric must measure its metric space, with the metric defined below:

Definition. (metric) Let $X$ be a non-empty set. A function $d: X\times X \rightarrow \mathbb{R}$ is said to be a metric on $X$ if it satisfies the following conditions:

i. $d(x,y)\ge 0\qquad \forall\, x,\, y\in X$ ;
ii. $d(x,y)=0 \Leftrightarrow x=y\qquad x,\, y\in X$ ;
iii. $d(x,y)=d(y,x) \qquad \forall\, x,\, y\in X$ ;
iv. $d(x,y)\le d(x,z)+d(z,y) \qquad \forall\, x,\, y,\, z\in X$ .

Remark.

i. As is known, distance should be either positive or zero (i.e., non-negative); ii. We are in one only by discrimination; iii. Fair and just from a symmetric point of view; and iv. The straighter the path, the shorter the distance (also known as the Triangle Inequality).

Proof.

Assume $x\, ,y\in\mathbb{C}$ for convenience. (Provided $\mathbb{R}\subset \mathbb{C}$ , the assumption is ready for reduction or extension.)

i.

If the minimum $\mathrm{min}\big\{ 1, |x-y|\big\} = 1$ , condition (i) $d(x,y)=1 \ge 0$ is seen. If the minimum $\mathrm{min}\big\{ 1,|x-y|\big\} = |x-y|$ , be it called the absolute value, the magnitude, the norm, the modulus, or whatsoever, a complex number is non-negative in norm.

Recall. (Norm of complex conjugate)

A complex number $z=x+\textrm{i}\, y$ contains two parts, the real part $x$ and the imaginary part $y$ . The norm $|z|$ (and the norm $|\bar{z}|$ of its complex conjugate $\bar{z}=x-\textrm{i}\, y$ )
is defined by the formula:

$|z|=|x+\textrm{i}\, y|=\sqrt{\big[\textrm{Re}(z)\big]^2+\big[\textrm{Im}(z)\big]^2}=\sqrt{x^2+y^2}$ .

$\dagger$ It turns out that $|z|=\sqrt{(x)^2+(y)^2}=\sqrt{(x)^2+(-y)^2}=|\bar{z}|$ (where $x,\, y\in\mathbb{R}$ ) is positive iff $x\enspace\textrm{\scriptsize OR}\enspace y\neq 0$ , and zero iff $x=y=0$ , but never negative.

$\ddagger$ The sum, the difference, the product, and the quotient of two complex numbers is one another complex number for the complex number field is algebraically closed.

ii.

(only-if) Giving a try straightforth:

$\begin{aligned} d(x,y) & = 0 \\ \textrm{min}\big\{ 1, |x-y| \big\} & = 0 \\ |x-y| & = 0\qquad \textrm{\scriptsize OR\qquad\quad } 1 = 0 \qquad \textrm{(rejected for\enspace }1\neq 0) \\ |x-y| & = 0 \\ x-y & = 0 \\ x & = y \end{aligned}$

(if) In reverse from backward:

$\begin{aligned} x & = y \\ x-y & = 0 \\ |x-y| & = 0 \\ \textrm{min}\big\{ 1, |x-y| \big\} & = 0 \\ d(x,y) & = 0 \end{aligned}$

If provided with appropriate explanation, the proof can be shortened by use of the two-way if-and-only-if.

iii.

Suffice it to check whether $d(x,y)=d(y,x)$ is true or not.

$\begin{aligned} \textrm{LHS} & = d(x,y) \\ & = \textrm{min}\big\{ 1,\, |x-y| \big\} \\ & = \textrm{min}\big\{ 1,\, |y-x| \big\} \\ & = d(y,x) \\ & = \textrm{RHS} \end{aligned}$

Roughwork.

$\forall\, x,\,y \in\mathbb{C}$ , let $x=a+\textrm{i}\, b$ and $y=c+\textrm{i}\, d$ where $a$ , $b$ , $c$ , and $d$ are real numbers. Then,
$\begin{aligned} x-y & = (a+\textrm{i}\, b)-(c+\textrm{i}\, d) \\ x-y & = (a-c) + \textrm{i}\, (b-d) \\ |x-y| & = \sqrt{(a-c)^2+(b-d)^2} \\ |x-y| & = \sqrt{(c-a)^2+(d-b)^2} \\ |x-y| & = |(c-a) + \textrm{i}\, (d-b)| \\ |x-y| & = |(c+\textrm{i}\, d) - (a+\textrm{i}\, b)| \\ |x-y| & = |y-x| \\ \end{aligned}$

iv.

Given here are some equations, I write out all them lest I might forget any:

$d(x,y)=\textrm{min}\big\{ 1, |x-y| \big\}$ ,
$d(x,z)=\textrm{min}\big\{ 1, |x-z| \big\}$ ,
$d(z,y)=\textrm{min}\big\{ 1, |z-y| \big\}$ .

RTP (i.e. required to prove):

$d(x,y)\le d(x,z) + d(z,y)$

$\begin{aligned} \textrm{RHS} & = d(x,z) + d(z,y) \\ & = \textrm{min}\big\{ 1,|x-z|\big\} + \textrm{min}\big\{ 1,|z-y|\big\} \\ & = \textrm{min}\big\{ 2,\, 1+|z-y|,\, 1+|x-z|,\, |x-z|+|z-y|\big\} \\ \end{aligned}$

If $d(x,y)=\textrm{min}\big\{ 1, |x-y|\big\} \stackrel{?}{=}1$ , so what have I done with?

WTS (i.e. wish to show):

$1\le \textrm{min}\big\{ 2, 1+|z-y|, 1+|x-z|, |x-z|+|z-y|\big\}$

The following inequalities hold evidently:

$\begin{aligned} 1 & \le 2 \\ 1 & \le 1+|z-y| \\ 1 & \le 1+|x-z| \\ \end{aligned}$

But $1\le |x-z|+|z-y|$ has not yet been ascertained.

The Argand diagram above replaces the usual $x$ – and $y$ -axes of the Cartesian plane with the real and the imaginary axes of Argand plane.

Owing to my giving too raw and rude maybe a proof, the problem should have otherwise been treated case-by-case. I.e., they are in either case:

$1\le |x-y| \qquad \qquad \textrm{\scriptsize OR}\qquad\qquad 1> |x-y|$

Just do it by rote:

$\begin{aligned} & \quad\, d(x,y) \\ & =\textrm{min}\big\{ 1, |x-y| \big\} \\ & =\textrm{min}\big\{ 1, |x-z+z-y| \big\} \\ \dots \, & \textrm{by the Triangle Inequality}\, \dots \\ & \leqslant \textrm{min} \big\{ 1, |x-z| + |z-y| \big\} \\ & \leqslant \textrm{min} \big\{ 1, |x-z| \big\} + \textrm{min} \big\{ 1, |z-y| \big\} \\ & \leqslant d(x,z) + d(z,y) \\ \end{aligned}$

$d(x,y) \le d(x,z) + d(z,y)$

QED

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