Topology/Normed Vector Spaces

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

Definition

We know the vector space definition, 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. $|v|\ge 0,|v|=0⟺v=0$ for all $v\in V$.

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

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

Example

For a given $n\in \mathbb{N}$ we know that ${ℝ}^n$ 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. $d(v,w)=|v-w|\ge 0$ holds by definition and $d(v,w)=|v-w|=0⟺v-w=0⟺v=w$ as required.

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

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

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

Exercises

(under construction)