# Topology/Normed Vector Spaces

A normed vector space ${\displaystyle (V,|\cdot |)}$ is a vector space V with a function ${\displaystyle |\cdot |:V\times V\to \mathbb {R} }$ 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. ${\displaystyle |\cdot |}$  is a norm if these three conditions hold.

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

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

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

## Example

For a given ${\displaystyle n\in \mathbb {N} }$  we know that ${\displaystyle \mathbb {R} ^{n}}$  is a vector space and its norm can be defined to be ${\displaystyle |v-w|=d(v,w)}$  ie. ${\displaystyle |v|=d(v,0)}$ . 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 ${\displaystyle d(v,w)=|v-w|}$  satisfies the metric axioms. Let ${\displaystyle u,v,w\in V}$

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

2. ${\displaystyle d(v,w)=|v-w|=|-1||w-v|=|w-v|=d(w,v)}$

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

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

## Exercises

(under construction)