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I have a doubt about weak topology in a Banach space.

Let $\mathcal{B}$ a infinite dimensional Banach space, I understood that the weak topology in $\mathcal{B}$, is the topology generated by $\Sigma = \lbrace \varphi^{-1}(U); \varphi \in \mathcal{B}^{*} \rbrace $, where $U \subset \mathbb{F} $ is open, it is, the coarsest topology in $\mathcal{B}$ such that each element of $X^{*}$ remains a continuous function. So, denoting the weak topology by $\tau_{w}$ and the norm topology by $\tau$, we have $\tau_{w} \subset \tau$.

But, let $x_{0} \in $$\mathcal{B}$ and $B(x_{0}; \epsilon)$ a open ball. For all $0 \neq f \in \mathcal{B}^{*}$, $f$ is surjetive, then by the open mapping theorem, $f(B(x_{0}; \epsilon))$ is open. Thus, for each $B(x_{0}; \epsilon)$ in $(\mathcal{B}, \tau)$, we have $B(x_{0}; \epsilon) = f^{-1}((B_{\mathbb{F}}(f(x_{0}); \epsilon')$. So, should I conclude that $\tau \subset \tau_{w}$?

I'm bit confused, any help will be appreciated.

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You are correct up until $B(x;\epsilon) = f^{-1}(B_\mathbb{F}(f(x_0);\epsilon'))$. All you can conclude from the statement before that is $B(x;\epsilon) \subseteq f^{-1}(B_\mathbb{F}(f(x_0);\epsilon'))$, which does not imply that $B(x;\epsilon)$ is weakly open.

In general, an open ball for the norm topology in an infinite-dimensional Banach space has empty weak interior.

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Yes, $f(B(x_{0}; \epsilon))$ is open. It does not follow that $B(x_{0}; \epsilon) = f^{-1}((B_{\mathbb{F}}(f(x_{0}); \epsilon')$.

It doesn't follow even in a finite-dimensional space. To get an idea what's happening here, say $X=\Bbb R^2$ with the euclidean norm. Say $B$ is the unit ball of $X$ and define $f(x,y)=x$. Then $f(B)$ is the open interval $(-1,1)$. But for any $\delta>0$ the set $f^{-1}((-\delta,\delta))$ is a vertical strip in $\Bbb R^2$, not contained in $B$.

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  • $\begingroup$ @DavidC.Ulrich Thanks for help me again, it's very clear for me now. $\endgroup$
    – BBVM
    Feb 3 '16 at 19:06

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