Why is the following set open in the dual space? Let $E$ be a normed vector space over $\mathbb{C}$, and consider $E^{\ast}$ the space of continuous linear functionals on $E$ with the topology induced by the norm $\|f\| = \sup_{\|x\| \leq 1 } |f(x)|$. Then the set $\{ f \in E^{\ast} |\ \forall x \in A: |f(x)| < \epsilon  \}$ where $A$ is bounded in norm is open in $E$, but I see no reason why it should be unless the bounded set $A$ were open or contained in it a disk centered at the origin.In other words I don't see why if $g \in E^{\ast}$ is such that $\|g-f\|$ is (sufficiently) small, then $|g(x)| < \epsilon$ for all $x \in A$.
 A: If the topology generated by $U_{\epsilon,A} = \{ f \in E^{\ast} : |f(x)| < \epsilon \forall x \in A \}$ where $A$ is a bounded subset of $E$ coincides with the topology induced by the norm $||f|| = \sup_{||x|| \leq 1 } |f(x)|$ when $E$ has its topology induced by a norm, then we would expect $U_{\epsilon,A}$ to be open in $E^{\ast}$ where $E^{\ast}$ has the topology induced by the above norm, and $A$ is bounded in norm. However I believe the following is a counterexample to this:
Let $E = \mathbb{R}^1$ with the usual norm, $A = (-\epsilon,\epsilon)$ for $\epsilon > 0$, and consider the set $U_{\epsilon,A}$. Clearly the function $f(x) = x$ is contained in $U_{\epsilon,A}$, and if $U_{\epsilon,A}$ is open in $E^{\ast}$ where $E^{\ast}$ has the topology induced by the norm on functionals, we require that there is some $\Delta > 0$ for which if $||g - f|| < \Delta$ and $g$ is a continuous linear functional on $E$, then $g \in U_{\epsilon,A} \iff |g(x)| < \epsilon \forall x \in A$. Suppose such a $\Delta > 0$ exists. Now define $g(x) = (1 + \delta)x$ for $0 < \delta < \Delta$, it is a continuous linear functional on $\mathbb{R}$. Furthermore we have $(g-f)(x) = \delta x$, and so $||g-f|| = \delta < \Delta$. Therefore we expect $g \in U_{\epsilon,A}$, but clearly $g(\epsilon) = \epsilon + \delta \epsilon > \epsilon > 0$ so given $a \in A$ sufficiently close to $\epsilon$, by continuity of $g$ we must have $|g(a)| > \epsilon$, hence $g \notin U_{\epsilon,A}$.
