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)