TITLE: Prove that a subset in Hilbert space is open
QUESTION [5 upvotes]: I'm a undergraduate who enjoys math and follows his first course on real Analysis. I follow this course for fun and i'm stuck on the following problem:
Let $H^{\infty}$ be the Hilbert cube (the collection of all real sequences $x := (x_n)_{n \geq1}$ with $|x_n| \leq 1$ for $n = 1,2,...$).  For $x,y \in H^{\infty}$ let $d(x,y) = \sum_{n=1}^{\infty}{2^{-n}}|x_n-y_n|$
Prove that $\{x = (x_n)_{n \geq1} \in H^{\infty} : |x_k| < 1, k = 1,...2000\}$ is open in $H^{\infty}$
I already proved the following which may help:
$l \in \mathbb{N}$ and $ $ $M_l = \max(|x_1 - y_1|,...,|x_l-y_l|)$ then $2^{-l}M_l \leq d(x,y) \leq M_l + 2^{-l}$
I'm already stuck for a few days now so i hope you guys could give me a hint or guide me to the prove!

REPLY [3 votes]: Denote $\phi_k$ the $k$-th coordinate map $(x_n)_{n\geq 1}\longmapsto x_k$ and $B=\{\lambda\,|\,|\lambda|<1\}$ the open unit ball of the scalar field.
The set to prove open is $\displaystyle\bigcap_{k=1}^{2000}\phi_k^{-1}(B)$ so it suffices to show that every $\phi_k$ is continuous.
This follows from the following estimate 
$$|\phi_k(x)-\phi_k(y)|=|x_k-y_k|=2^k2^{-k}|x_k-y_k|\leq 2^kd(x,y)$$
which shows that $\phi_k$ is $2^k$-Lipschitz, hence continuous.