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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.

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    $\begingroup$ You can sohw that its interior is empty in weak topology $\endgroup$
    – DiegoMath
    Jun 14, 2014 at 21:37
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    $\begingroup$ What do weak neighborhoods of zero look like? $\endgroup$
    – user147263
    Jun 14, 2014 at 21:41
  • $\begingroup$ $B_{\epsilon,\{f_1,\dots,f_k\}}=\{ x \in X : |f_i(x)|<\epsilon \:\:\: \forall i\in \{1,\dots, k \}\}$ where $f_i$ are continuous linear fucntionals on $X$ $\endgroup$
    – luka5z
    Jun 14, 2014 at 21:50

3 Answers 3

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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.

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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).

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  • $\begingroup$ I added comment above. But I dont get it which subspace it contains. $\endgroup$
    – luka5z
    Jun 14, 2014 at 21:51
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    $\begingroup$ See DiegoMath's answer for details. $\endgroup$ Jun 15, 2014 at 2:44
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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.

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