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!
Best Answer
Just a few hints: Let $A:=\{x\in H^\infty:\ |x_k|<1, k=1,\ldots, 2000\}$. Now fix $x\in A$. You need to find an $r>0$, such that the Ball $B_r(x):=\{y\in H^\infty:\ d(x,y)<r\}\subset A$. Let us pick an arbitrary $r>0$ and a $y\in B_r(x)$. Then $$d(x,y)=\sum_n 2^{-n}|x_n-y_n|<r.$$ Hence $|x_n-y_n| < C\cdot r$ for some constant $C>0$ and $n=1,\ldots,2000$. Now try to use $$|y_n|\leq |x_n-y_n|+|x_n|$$ to find an $r>0$ (you will have to choose it small), such that $|y_n| <1$ for $n=1,\ldots,2000$.