Normed Vector Space is Locally Convex Space

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

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

Let $\tau$ be the standard topology on the locally convex space $\struct {X, \set {\norm {\, \cdot \,} } }$.

Then $\tau$ consists precisely of the open sets of $\struct {X, \norm {\, \cdot \,} }$.

Proof
Let $U \subseteq X$.

From Open Sets in Standard Topology of Locally Convex Space, we have $U \in \tau$ :


 * for each $x \in U$ there exists $\epsilon > 0$ such that:


 * $\set {y \in X : \norm {y - x} < \epsilon} \subseteq U$

That is for each $x \in U$ there exists $\epsilon > 0$ such that:


 * $\map {B_\epsilon} x \subseteq U$

where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.

So from the definition of an open set of $\struct {X, \norm {\, \cdot \,} }$, this is equivalent to $U$ being an open set of $\struct {X, \norm {\, \cdot \,} }$.