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.

We know the vector space definition, 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 ${\displaystyle v,w\in V}$.

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.

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.

(under construction)