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.

Answer 2

Look at the definition of neighbourhoods for the weak topology, and show that any neighbourhood of any point contains a subspace (so, it is unbounded).

Answer 3

Let $B$ a unit ball of $E$. Clearly, $A=B^c$ is strongly closed in $E$. As $B$ is convex set, $A$ is not convex, then Mazur's Theorem says that $A$ is not weak closed. If $B$ is open in weak-topology $\sigma(E,E')$, $A$ would be weak closed, being a contradiction.