I want to show that the Hilbert cube which is:
$H=\{(x_1,x_2,…) \in [0,1]^{\infty} : for \ each \ n \in \mathbb{N}, |x_n|\leq \dfrac{1}{2^n}\}$
is compact with respect to the metric:
$d(x,y)=sup|x_n – y_n|$
where $x=(x_n)$ and $y=(y_n)$
I found some proofs of compactness of Hilbert cube but non of them were helpful to me because they consist advanced topological contents which I'm not allowed to use them because this question is in our analysis course. for example one of the easiest way to prove is using Tychonoff theorem but I'm not allowed to do it.
My idea is to prove that $[0,1]^{\infty}$ is complete and totally bounded(so it is compact) and then after that I have to prove that the Hilbert cube is a closed subset of $[0,1]^{\infty}$. so it is also compact. but unfortunately I can't develope this idea.
Also the question forced us to use a special metric, Is the kind of metric important at all?
Best Answer
You should be careful with blindly applying Tychonoff's theorem. After all, that theorem also says that the "unit cube" $C = \{(x_1,...) : x \in [0,1] \}$ is compact, even though the sequence $(x_n)_i = \delta_{in}$ lacks a converging subsequence. This kind of nonsense cannot happen with your cube because you've imposed a condition as you go further down. So, what this says is that the topology induced by your choice of metric is certainly not the one induced by Tychonoff!
You can actually work "backwards" and try to see why such sequences won't happen in your cube and use that to argue that it is compact. Lost in a Maze's argument will also work.