I am undergraduate student, this is my first class on "real analysis" I don't know about compactness yet. This comes from chapter 7 of NL Carothers "Principles of Analysis."
The Hilbert cube $H^{\infty}$is the collection of all real sequences $x=(x_{n})$ with $|x_{n}|\leq1$ for all $n=1,2,...$. Consider the metric space $(H^{\infty},d)$ where $d(x,y)=\sum_{n=1}^{\infty}\frac{|x_{n}-y_{n}|}{2^{-n}}$.
Here is my attempt:
To prove that the Hilbert cube is complete with the metric $d$ above it is sufficient to show that every infinite, totally bounded subset of $H^{\infty}$ has a limit point in $H^{\infty}.$ Let $S$ be an infinite, totally bounded subset of $H^{\infty}$. Such a set exists since the collection of all real sequences $x=x_{n}$ with $|x_{n}|\leq1$ and is isomorphic to $\{0,1\}^{\mathbb{N}}$. $A$ contains a Cauchy sequence $(x_{n})$ since $A$ is totally bounded (recall: every sequence has a Cauchy subsequence). This sequence is comprised of distinct points, that is $x_{n}\neq x_{m}$ when $n\neq m.$ Set $A_{n}=\{x_{k}:k\geq n\}.$ Then we have that $A\supset A_{1}\supset A_{2}\supset...$. Each $A_{n}$ is nonempty and $diam(A_{n})\rightarrow0.$ But clearly $cl(A_{n})\supset cl(A_{n+1})\neq\emptyset$ for each $n$ and $diam(cl(A_{n}))=diam(A_{n})\rightarrow0$ as $n\rightarrow\infty$. Therefore there is an $x\in\cap_{n=1}^{\infty}cl(A_{n})\neq\emptyset.$ Now since $x_{n}\in A_{n}\implies d(x_{n},x)\leq diam(cl(A_{n}))\rightarrow0.$ That is $x_{n}\rightarrow x$ and so $x$ is a limit point of $A.$
I am really so lost though. I wrote this by following along with the proof in NL Carothers chapter 7 on statements equivalent to saying a metric space $(M,d)$ is complete. I just don't know how I know that $A$ exists. I am so stuck there.
I just "feel" like its probably true that the collection of all real sequences $x=x_{n}$ with $|x_{n}|\leq1$ is isomorphic to $\{0,1\}^{\mathbb{N}}$. I actually don't know this, but intuitively in my head it makes since. And if that's true I also feel like it's probably also true that $A$ exists. But this isn't really "rigorous" it's just a hunch on top of another hunch...