Weak Convergence Confusion Let $X$ a Hilbert space. Suppose there is a sequence $(x_n)$ such that for any weak neighbourhood $U$ of $0$,there is some $N$ such that for $n,m \geq N$, $x_n - x_m \in U$.
I have two questions:
(a) Is $(x_n)$ a cauchy sequence? Here is my attempt at a proof:
For any open ball around $0$, $B(\varepsilon)$,there is some weak open neighbourhood of $0$, $U$, depending on $\varepsilon$, contained in $B(\varepsilon)$. Then there is some $N$ such that for any $n,m \geq N$,
\begin{equation}
x_n - x_m \in U \subseteq B(\varepsilon).
\end{equation}
So $\|x_n - x_m\| \leq \varepsilon$, and so $(x_n)$ is a cauchy sequence (in norm topology).
(b) For such a sequence $(x_n)$, I think I should be able to show that given $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ so that for any $n,m \geq N$ and $y\in H$, $\langle x_n - x_m,y \rangle < \varepsilon$. It seems reasonable, but I have not managed to do it rigourously. 
Any help for either (or both) questions would be very appreciated. 
 A: The condition you're given is otherwise known as weakly Cauchy.

For any open ball around $0$, $B(\varepsilon)$,there is some weak open neighbourhood of $0$, $U$, depending on $\varepsilon$, contained in $B(\varepsilon)$.

That's not true.  In fact, given a weakly open neighborhood, there is no ball that contains it.
To get a counterexample, you might try to show that any weakly convergent sequence is weakly Cauchy.  You hopefully know an example of a weakly convergent sequence that does not converge in norm.  In particular, it cannot be Cauchy in norm (by completeness).

For such a sequence $(x_n)$, I think I should be able to show that given $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ so that for any $n,m \geq N$ and $y\in H$, $\langle x_n - x_m,y \rangle < \varepsilon$. It seems reasonable, but I have not managed to do it rigourously. 

The set $\{x : \langle x ,y \rangle < \varepsilon\}$ is itself weakly open, i.e. it's a weak open neighborhood of $y$.  That is pretty much the very definition of the weak topology.  So that should help a lot.
