Every inner product space is a metric space.

Question

Show that every inner product space is a metric space.

To show this should I set the distance metric as $d(x,y) = \langle x-y,x-y\rangle$, then show properties of being metric space such as $d(x,y) = d(y,x)$ etc.? If so the point I do not understand is why we set metric as $d(x,y) = \langle x-y,\ x-y \rangle$ (this metric is mentioned in wolfram)

Answer 1

That is wrong. It should be $d(x,y)=\sqrt{\langle x-y,x-y\rangle}$ because the map $x\mapsto\sqrt{\langle x,x\rangle}$ is a norm. And, whenever you have a norm $\lVert\cdot\rVert$, the map $(x,y)\mapsto\lVert x-y\rVert$ is a distance.

Answer 2

The inner product induces a norm via $$||x||:=\sqrt{\langle x,x\rangle}$$ and then, the norm induces a metric via $$d(x,y):=||x-y||$$ which is precisely what is mentioned in the comment above. In general, one has the relation $\text{Inner Product Spaces}\subset\text{Normed Vector Spaces}\subset\text{Metric Spaces}$.

Answer 3

You are setting $d(x,y)=||x-y||$. So your metric here comes from the norm. You can check that it satisfies the required metric conditions.

We are not sating that this is the only way to choose a metric on the space, but it is one way.