Pseudometric Induced by Seminorm is Pseudometric
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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
Proof of Metric Space Axiom $(\text M 1)$
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:
- $\map {d_p} {x, x} = 0$
$\Box$
Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality
For $x, y, z \in X$ we have:
\(\ds \map {d_p} {x, z}\) | \(=\) | \(\ds \map p {x - z}\) | Definition of Pseudometric Induced by Seminorm | |||||||||||
\(\ds \) | \(=\) | \(\ds \map p {\paren {x - y} + \paren {y - z} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map p {x - y} + \map p {y - z}\) | Seminorm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {d_p} {x, y} + \map {d_p} {y, z}\) | Definition of Pseudometric Induced by Seminorm |
$\Box$
Proof of Metric Space Axiom $(\text M 3)$
For $x, y \in X$ we have:
\(\ds \map {d_p} {y, x}\) | \(=\) | \(\ds \map p {y - x}\) | Definition of Pseudometric Induced by Seminorm | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {-1} \map p {x - y}\) | Seminorm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
\(\ds \) | \(=\) | \(\ds \map p {x - y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {d_p} {x, y}\) | Definition of Pseudometric Induced by Seminorm |
$\blacksquare$