# Topology/Normed Vector Spaces

A normed vector space $(V,|\cdot |)$ is a vector space V with a function $|\cdot |:V\times V\to \mathbb {R}$ that represents the length of a vector, called a norm.

## Definition

We know the vector space defintion, so we need to define the norm function. $|\cdot |$  is a norm if these three conditions hold.

1. Only the zero vector has zero length, with all others being positive. $|v|\geq 0,|v|=0\iff v=0$  for all $v\in V$ .

2. For $a\in \mathbb {R}$  and $v\in V$  we have $|av|=|a||v|$ .

3. The triangle inequality holds: $|v+w|\leq |v|+|w|$  for all $v,w\in V$ .

## Example

For a given $n\in \mathbb {N}$  we know that $\mathbb {R} ^{n}$  is a vector space and its norm can be defined to be $|v-w|=d(v,w)$  ie. $|v|=d(v,0)$ . This is not unusual, in fact we say that a norm induces a metric with the first equation. So normed vector spaces are always metric spaces. Let's prove this.

## Theorem

Normed vector spaces are metric spaces.

Proof

It suffices to show that $d(v,w)=|v-w|$  satisfies the metric axioms. Let $u,v,w\in V$

1. $d(v,w)=|v-w|\geq 0$  holds by definition and $d(v,w)=|v-w|=0\iff v-w=0\iff v=w$  as required.

2. $d(v,w)=|v-w|=|-1||w-v|=|w-v|=d(w,v)$

3. $d(u,v)+d(v,w)=|u-v|+|v-w|\geq |u-v+v-w|=d(u,w)$  so the triangle inequality translates correctly.

Since the axioms hold, we conclude that V is a metric space.

## Exercises

(under construction)