1-Metric

Definition

• Let $x,y\in {\mathbb{F}}^n$
• Let $d(x,y)$ be a Metric for $\mathbb{F}$
• Let $f(x,y)={\sum }_{i=1}^{n}d(x_i,y_i)$
• Then, $f$ is a metric for ${\mathbb{F}}^n$

Proof

$f(x,y)\ge 0$

• As $d(x_i,y_i)\ge 0$
• Then, $\forall i,{\sum }_{i=1}^{n}d(x_i,y_i)\ge 0$
• Then, $\forall i,f(x,y)\ge 0$

$f(x,y)=f(y,x)$

• Let $x,y\in {\mathbb{F}}^n$
• Then, $x=(x_1,\dots ,x_n)$
• And $y=(y_1,\dots ,y_n)$
• $f(x,y)={\sum }_{i=1}^{n}d(x_i,y_i)$ by defn of $f$
• $={\sum }_{i=1}^{n}d(y_i,x_i)$ as $d$ is a metric
• $=f(y,x)$ by defn of $f$

$f(x,y)=0⟺x=y$

• $f(x,y)=0$
• $⟺{\sum }_{i=1}^{n}d(x_i,y_i)=0$
• $⟺d(x_i,y_i)=0$
• $⟺x_i=y_i\forall i$
• $⟺x=y$

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

• Let $x,y,z\in {\mathbb{F}}^n$
• Then, $x=(x_1,\dots ,x_n)$
• Then, $y=(y_1,\dots ,y_n)$
• Then, $z=(z_1,\dots ,z_n)$
• $f(x,z)={\sum }_{i=1}^{n}d(x_i,z_i)$
• $\le {\sum }_{i=1}^n[d(x_i,y_i)+d(y_i,z_i])$ as $d$ is a Metric
• $={\sum }_{i=1}^{n}d(x_i,y_i)+{\sum }_{i=1}^{n}d(y_i,z_i)$
• $=f(x,y)+f(y,z)$