Tychonoff's theorem vs closed ball If Tychonoff's theorem is true, why closed ball in $\mathbb{R}^n$ is not compact?
The theorem says that if $X_i$ is compact, for every $i\in I$, so $\prod_{i\in I}X_i$ is compact. Then take $n\in\mathbb{N}$ and we have $\prod_{n\in\mathbb{N}}[-1,1]_i$ in $\mathbb{R}^\infty$ is not compact. But, what??
 A: An infinite product of copies of $[-1,1]$, say, is indeed compact in the product topology, but a Banach space topology is not like a product topology; the "closest" we can get to that kind of a topology in an infinite-dimensional Banach space is the so-called weak-* topology (which is not metrisable most of the time, so quite unlike the norm topology, which of course is), and there we can prove the Banach-Alaoglu theorem which shows that the ball in that topology is compact and in which the Tychonoff theorem is a key ingredient of the proof...
A: $\prod_{n \in \mathbb{N}}[-1,1]$ is indeed compact under product topology. I believe you misunderstand the structure of a closed ball, that is a closed ball $B_{d}(x, \epsilon)$ is defined to be the set $\{y: d(x,y) \leqslant \epsilon\}$ for some $\epsilon$, not $\prod_{n \in \mathbb{N}}[-1,1]$ as you claimed. And the set $B_{d}(x, \epsilon)$ is clearly not compact.
Proof Let $\gamma < \epsilon$. Consider the set $B = \{ (\gamma,0,0,...), (0,\gamma,0,...), (0,0,\gamma,0,...) \} \subset B_{d}(x, \epsilon)$ is closed in $B_{d}(x, \epsilon)$ and has discrete topology, therefore it can not has limit point in $B_{d}(x, \epsilon)$. This contradicts with the face that compactness implies limit point compactness.
Hope this helps!
