Norm Topology Induced by Metric Induced by Norm

Theorem

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

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.

Let $d$ be the metric induced on $X$ by $\norm {\, \cdot \,}$.

Then the open sets of $\struct {X, d}$ as a metric space are precisely the open sets of $\struct {X, \norm {\, \cdot \,} }$ as a normed vector space.

Proof

For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^d} x$ be the open ball in $\struct {X, d}$ with radius $\epsilon$ and center $x$.

For $\epsilon > 0$ and $x \in X$, let $\map {B_\epsilon^{\norm {\, \cdot \,} } } x$ be the open ball in $\struct {X, \norm {\, \cdot \,} }$ with radius $\epsilon$ and center $x$.

From the definition of an open set in $\struct {X, d}$, $U \subseteq X$ is open if and only if:

for each $x \in U$ there exists $\epsilon > 0$ such that $\map {B_\epsilon^d} x \subseteq U$.

From the definition of an open set in $\struct {X, \norm {\, \cdot \,} }$, $U \subseteq X$ is open if and only if:

for each $x \in U$ there exists $\epsilon > 0$ such that $\map {B_\epsilon^{\norm {\, \cdot \,} } } x \subseteq U$.

It is therefore sufficient to show that $\map {B_\epsilon^d} x = \map {B_\epsilon^{\norm {\, \cdot \,} } } x$.

Let $y \in X$.

We have $y \in \map {B_\epsilon^d} x$ if and only if:

$\map d {x, y} < \epsilon$

This is equivalent to:

$\norm {x - y} < \epsilon$

This is then equivalent to:

$y \in \map {B_\epsilon^{\norm {\, \cdot \,} } } x$

Hence we have:

$\map {B_\epsilon^d} x = \map {B_\epsilon^{\norm {\, \cdot \,} } } x$

$\blacksquare$