Absolute Value Functions Induce a Metric Theorem

Theorem

Absolute Values Induce a Metric

• $\mathcal{D}$ as an Integral Domain or Field
• Let $|\cdot |:\mathcal{D}\to \mathbb{R}$ be an absolute value
• Let $d(x,y)=|x-y|$
• For all $x,y\in \mathcal{D}$
• Then, $d$ is a metric for $\mathcal{D}$

Alternate Theorem

• Let $||\cdot ||:V\to \mathbb{R}$ be a Norm with $|\cdot |$
• Then, $d(x,y)=||x-y||$ is a Metric for $V$

Proof

Proving $d(x,y)\ge 0$

• Let $x,y\in \mathcal{D}$, then $d(x,y)=|x-y|\ge 0$ as $|\cdot |\ge 0$

Proving $d(x,y)=d(y,x)$

• $d(x,y)=|x-y|=|x+(-y)|-|(-1)(-1)(x)+y|$
• $=|(-1)\ast (y+(-1)(x)|=|(-1)\ast (y-x)|=|-1|\ast |y-x|=|y-x|$ as $|-1|=1$

Proving $d(x,y)=0⟺x=y$

• $d(x,y)=0$
• $⟺|x-y|=0$ by Absolute Function props
• $⟺x-y=0$ by defn Norm
• $⟺x=y$ by Integral Domain props

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

• $d(x,z)=|x-z|=|x-y+y-z|=|(x-y)+(y-z)|$
• $\le |x-y|+|y-z|=d(x,y)+d(y,z)$