5
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ Why was this downvoted!? $\endgroup$
    – Shaun
    Commented Sep 14, 2018 at 12:29
  • $\begingroup$ Use $\{ X\}$ for $\{ X\}$. $\endgroup$
    – Shaun
    Commented Sep 14, 2018 at 12:29
  • $\begingroup$ Fixt! Thanks for the tip $\endgroup$
    – Henk
    Commented Sep 14, 2018 at 12:36

3 Answers 3

3
$\begingroup$

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 $n=1,\ldots,2000$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Before writing down a formal proof, try to understand concretely why the set (let's call it $A$) is open:

If $y \notin A$, then is must have a $|y_k| = 1$ for some $k \in \{1,...,2000\}$. So if we look at the distance between some $x \in A$ and this $y$, it cannot become arbitrarily small, right? There is always a positive distance because

$$d(x,y) = \sum_{n=1}^\infty 2^{-n} |x_n-y_n| \geq 2^{-k} |y_k - x_k| \geq 2^{-2000} |y_k-x_k|.$$

The last step works because $k$ can at most be 2000. Now you have a lower bound on the distance between some $y$ outside of $A$ and some $x \in A$. Can you go on from here to see why $A$ is thus an open set?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .