202010020718 Problem 2.1.2

Let $(X,d)$ be a metric space and let $k$ be a fixed positive real number. For $x,\, y\in X$ , define

$d^{*}=kd(x,y)$ .

Prove that $d^{*}$ is a metric on $X$ .

Recall.

Definition. (metric) Let $X$ be a non-empty set. A metric on $X$ is a real-valued function $d:\enspace X\times X\rightarrow \mathbb{R}$ satisfying the following conditions i– iv:

i. $d(x,y)\ge 0$ ;
ii. $d(x,y)=0\Leftrightarrow x=y$ ;
iii. (Symmetry) $d(x,y)=d(y,x)$ ;
iv. (Triangle Inequality) $d(x,y)\le d(x,z)+d(z,y)$ for any $x,\, y,\, z\in X$ .

N.b. Given $x,\,y\in X$ , $d(x,y)$ is sometimes called the distance between $x$ and $y$ with respect to $d$ .

Proof.

i.

WTS (wish to show)

$d^{*}(x,y)\ge 0$

By definition $d^{*}(x,y)=kd(x,y)$ and in that the metric $d$ is let clear ( $\therefore d(x,y)\ge 0$ ) and $k$ a fixed positive real number ( $\therefore k>0$ ),

one can see

$\begin{aligned} d^{*}(x,y) & = kd(x,y) \\ \textrm{\dots because\enspace} & k>0 \enspace \textrm{and}\enspace d(x,y)\ge 0\textrm{\enspace \dots}\\ d^{*}(x,y) & \geqslant 0 \\ \end{aligned}$

$\therefore$ Condition i. is made.

ii.

$\begin{aligned} d^{*}(x,y) & = 0 \\ \Leftrightarrow kd(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace k>0 \enspace & \textrm{so}\enspace k\neq 0\enspace \dots \\ \Leftrightarrow d(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace d \enspace \textrm{was} &\enspace\textrm{foretold to be a metric}\enspace \dots \\ \Leftrightarrow x & = y \\ \end{aligned}$

$\therefore$ Condition ii. is made.

iii.

NTS (need to show)

$d^{*}(x,y)=d^{*}(y,x)$

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \enspace & \textrm{by the symmetric property of }d\enspace \dots \\ & =kd(y,x) \\ & = d^{*}(y,x) \\ & = \textrm{RHS} \\ \end{aligned}$

$\therefore$ Condition iii. is made.

iv.

RTP (required to prove)

$d^{*}(x,y)\leqslant d^{*}(x,z)+d^{*}(z,y)$

One starts with the left hand side,

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \textrm{as does}\enspace & d(x,y)\le d(x,z)+d(z,y)\enspace\textrm{the metric}\enspace d\enspace \textrm{do} \dots \\ & \le k\Big( d(x,z) + d(z,y) \Big) \\ & = kd(x,z) + kd(z,y) \\ & = d^{*}(x,z) + d^{*}(z,y) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition iv. is made.

In conclusion, $(X,d^{*})$ is a metric space metered by a well-defined metric $d^{*}$ . This metric space shall simply be called $X$ hence.

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