Pseudometric Induced by Seminorm is Pseudometric

Definition
Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $p$ be a seminorm on $X$.

Let $d_p$ be the pseudometric induced by $p$.

Then $d_p$ is a pseudometric.

Proof of
For each $x \in X$ we have:


 * $\map {d_p} {x, x} = \map p {x - x} = \map p { {\mathbf 0}_X}$

From Seminorm Maps Zero Vector to Zero, we therefore have:


 * $\map {d_p} {x, x} = 0$

Proof of
For $x, y, z \in X$ we have:

Proof of
For $x, y \in X$ we have: