Compactness of unit ball in $\ell_\infty$ with a different norm

Consider the normed spaces (over the field of real numbers) $X=(\ell_\infty,\|\cdot\|_\infty)$ and $Y=(\ell_\infty,\|\cdot\|)$ where $$\|x\|=\sup_{n\in\mathbf{N}}\frac{|x_n|}{2^n}.$$

How can I show that the closed unit ball in $X$ is compact in $Y$?

-
Hint: $Y$ is also known as the Hilbert cube. Show that the topology induced by $\Vert \cdot\Vert$ on $[-1,1]^{\mathbb N}$ is the same as the product topology on this set. Then use Tychonoff. – Sam Aug 31 '12 at 22:20
@Sam Is the whole $Y$ a Hilbert cube? Or just that closed ball? – Godot Sep 1 '12 at 0:42
@Godot: You are right. It is just the closed ball, of course. Thanks for the correction. =) – Sam Sep 1 '12 at 10:29
Hum. How can the unit ball be precompact in a infinite-dimensional normed space? I guess that $Y$ is just a metric space, not a normed space. – Giuseppe Negro Sep 7 '12 at 21:56

1 Answer

We show that $B$, the unit ball in $X$, is precompact and complete.

• Precompactness: we fix $\varepsilon>0$ and $N$ such that $\frac 1{2^N}\leq \varepsilon$. Using the fact that $[-1,1]^N$ is precompact, we get $v_1,\dots,v_k\in [-1,1]^N$ such that $[-1,1]^N\subset \bigcup_{j=1}^kB(v_j,\varepsilon)$, where $B$ is the ball in $\Bbb R^N$ endowed with the supremum norm. Then take $w_j:=(v_j,0,\dots,)$.

• Completeness: consider $\{x^{(n)}\}\subset B$ a Cauchy sequence for $\lVert\cdot\rVert$. For each fixed $k$, $\{x^{(n)}_k\}_n$ is a Cauchy sequence in $\Bbb R$, hence converges to some $x_k$. For a fixed $\varepsilon>0$, we take $N$ such that $\frac 1{2^N}\leq \varepsilon$ and since all the terms of the sequence are bounded by $1$ we just work with a finite number of entries.

-