Open Ball in Infinite-Dimensional Normed Vector Space is not Weakly Open

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

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

Suppose that $\dim X = \infty$.

Let $\map {B_r} x$ be the open ball in $X$ with radius $r > 0$ and center $x \in X$.

Then $\map {B_r} x$ is not weakly open.

Proof
From Translation of Open Set in Normed Vector Space is Open and Dilation of Open Set in Normed Vector Space is Open, it suffices to show that $\map {B_1} 0$ is not weakly open.

that $\map {B_1} 0$ is weakly open.

From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $\struct {X, w}$ is a locally convex space induced by:


 * $\set {p_f : f \in X^\ast}$

with $p_f : X \to \hointr 0 \infty$ defined by:


 * $\map {p_f} x = \cmod {\map f x}$

for each $x \in X$.

From Open Sets in Standard Topology of Locally Convex Space, there exists $f_1, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that:


 * $\set {y \in X : \cmod {\map {f_k} y} < \epsilon \text { for each } 1 \le k \le n} \subseteq \map {B_1} 0$

Define $T : X \to \GF^n$ by:


 * $T x = \tuple {\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x}$

for each $x \in X$.

Suppose that:


 * $\ds \bigcap_{i \mathop = 1}^n \ker f_i = \set {\mathbf 0_X}$

Then $\ker T = \set {\mathbf 0_X}$.

So $T$ is injective, and is therefore a bijection onto $T \sqbrk X$.

So $T$ is a vector space isomorphism between $X$ and $T \sqbrk X$.

So we have:


 * $\dim X = \dim T \sqbrk X$

From Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space, we have:


 * $\dim T \sqbrk X \le n$

and so $\dim X \le n < \infty$, contrary to our assumption that $\dim X = \infty$.

So we must have:


 * $\ds \bigcap_{i \mathop = 1}^n \ker f_i \ne \set {\mathbf 0_X}$

Pick $x \ne \mathbf 0_X$ with:


 * $\ds x \in \bigcap_{i \mathop = 1}^n \ker f_i$

Then for each $\lambda \in \GF$, we have $\map {f_i} {\lambda x} = \lambda \map {f_i} x = 0 < \epsilon$ for each $1 \le i \le n$.

So $\lambda x \in \map {B_1} 0$ for each $\lambda \in \GF$.

But then, for each $\lambda \in \GF \setminus \set 0$, we have:


 * $\ds \norm x < \frac 1 {\cmod \lambda}$

Taking $\lambda \to \infty$ we obtain $\norm x = 0$.

Hence $x = \mathbf 0_X$ from, contrary to our assumption that $x \ne \mathbf 0_X$.