[Math] Why unit open ball is open in norm topology, but not open in weak topology

Question

Why unit open ball is open in norm topology, but not open in weak topology? I will be grateful for any explanation.

Edit: Obviously in infnite dimensional spaces.

Answer 1

Consider $E$ a normed space. Let $U$ be the unitary ball. By contradiction, suppose that the interior of the unitary ball is not empty. Note that $0$ is in $U$. Thus we can construct a neighborhood around $0$. $$N_{\varepsilon;0}=\{x\in E; |f_i(x)|<\varepsilon;f_1,\ldots,f_n\in E^*,\varepsilon>0 \},$$ with $0\in N\subset U$.

$\textbf{Statement}: $ There exists $y\neq0$ such that $y\in\displaystyle\bigcap_{i=1}^n Ker(f_i)$.

If this don't occurs we have that the linear map $$T:E\to\mathbb{R}^n$$ $$T(x)=(f_1(x),\ldots,f_n(x))$$ is injective, thus $\dim(E)<\infty$, contradiction.

Hence, we have that the line $ty\in N,\forall t\in\mathbb{R}$, because $$|f(ty)|=|tf(y)|=0<\varepsilon.$$ But if we make $t$ sufficiently big then $ty\notin U$, thus the interior should be empty.