About Hilbert spaces How can I prove this fact:
We're working in a Hilbert space $$ \mathcal{H} := \left\{ (x_n)_{n \in \mathbb{N}} \in {\mathbb{R}}^{\mathbb{N}} \mid \sum_{n=1}^{\infty}\,(x_n)^2 < \infty \right\} $$ and we're taking this metric, which we can proof is an ultrametric $$ d(\mathbf{x},\mathbf{y}) = \dfrac{1}{2^k} $$ where $k$ is the first different component in both successions. What I'm trying to proof is:
Let be $\mathbf{x} \in \mathcal{H}$, with $\mathcal{H}$ Hilbert space and $\varepsilon > 0$. Show that if $\mathbf{y} \in B_{\varepsilon}\,(\mathbf{x})$, then $B_{\varepsilon}\,(\mathbf{x}) = B_{\varepsilon}\,(\mathbf{y})$.
Thank you so much
 A: I can't post comments, but i guess you got a typo there. If not, this is obviously wrong. Consider any ball in $\Bbb R^2$ with the standard scalar product, and think about it geometrically.
(Please feel free to delete the answer again after an update from the OP)
A: This is a general fact for any ultrametric space $(X,d)$: this is a metric space where the triangle inequality is strengthened to: for all $x,y,z \in X$ we have $d(x,z) \le \max(d(x,y), d(y,z))$. In that case, any two $r$-balls either are disjoint, or are equal.
For suppose that $z \in B_r(x) \cap B_r(y)$. So $d(x,z) < r$ and $d(y,z) < r$. It follows that $d(x,y) \le \max(d(x,z),d(z,y)) < r$ as well (the maximum of two numbers less than $r$ is also less than $r$). 
Now suppose $u \in B_r(x)$, we want to show it is in $B_r(y)$, to get one inclusion. So we know that $d(u,x) < r$ and so $d(u,y) \le \max(d(u,x), d(x,y)) < r$, as both numbers are $<r$ again. Symmetrically, if $u \in B_r(y)$, then $u \in B_r(x)$, so we have equality of the sets $B_r(x)$ and $B_r(y)$.  
