# 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.

• You can sohw that its interior is empty in weak topology Jun 14, 2014 at 21:37
• What do weak neighborhoods of zero look like?
– user147263
Jun 14, 2014 at 21:41
• $B_{\epsilon,\{f_1,\dots,f_k\}}=\{ x \in X : |f_i(x)|<\epsilon \:\:\: \forall i\in \{1,\dots, k \}\}$ where $f_i$ are continuous linear fucntionals on $X$ Jun 14, 2014 at 21:50

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.

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).

• I added comment above. But I dont get it which subspace it contains. Jun 14, 2014 at 21:51
• See DiegoMath's answer for details. Jun 15, 2014 at 2:44

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.