Why unit open ball is open in norm topology, but not open in weak topology? 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.
 A: 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. 
A: 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). 
A: 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.
