Normed Vector Space is Hausdorff Topological Vector Space

Theorem
Let $\struct { K, +_K, \circ_K }$ be a valued field with norm $\norm {\,\cdot\,}_K$.

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

Let $\tau$ be the topology on $X$ that consists of all open sets in $X$.

That is, $U \in \tau$ :


 * $\forall x \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$

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

Then $\struct { X, \tau }$ is a topological vector space.

Proof
By its definition, a valued field is a normed division ring.

Let $\tau_K$ be the topology on $K$ induced by the metric induced by the norm $\norm {\,\cdot\,}_K$

From Normed Division Ring Operations are Continuous:Corollary, it follows that $\struct {K, \tau_K}$ is a topological field.

Let $d_X : X \times X \to \R_{\ge 0}$ be the metric induced by the inner product norm $\norm {\,\cdot\,}_X$ on $X$.

By definition of open set in normed vector space, it follows that $\tau$ is the topology on $K$ induced by the metric $d_X$.

To show that $\tau$ is a topology on $X$, we examine each of the open set axioms in turn:

Follows from Union of Open Sets of Normed Vector Space is Open.

Follows from Finite Intersection of Open Sets of Normed Vector Space is Open.

Follows from Normed Vector Space is Open in Itself.

Let $+_{\scriptscriptstyle X}$ denote the vector addition on $X$, and let $\circ_{\scriptscriptstyle X}$ denote the scalar multiplication on $X$.

To show that $\struct { X, \tau }$ is a topological vector space, we examine each of the axioms from the definition of topological vector space in turn:

$\tau$ is a Hausdorff topology
Follows from Normed Vector Space is Hausdorff.

$+_{\scriptscriptstyle X}: X \times X \to X$ is continuous with respect to $\tau$
Follows from Vector Addition on Normed Vector Space is Continuous.

$\circ_{\scriptscriptstyle X}: K \times X \to X$ is continuous with respect to $\tau$
Follows from Scalar Multiplication on Normed Vector Space is Continuous.