I'm trying to show that the Hilbert cube is compact, preferably by showing that it is sequentially compact.
My text defines the Hilbert cube as:
$H=\{(x_1,x_2,…) \in [0,1]^{\infty} : for \ each \ n \in \mathbb{N}, |x_n|\leq \dfrac{1}{2^n}\}$
I'm trying to show that it is compact with respect to the metric:
$d(x,y)=\underset{n}{\textrm {sup}}|x_n – y_n|$
I tried showing that it was compact vis sequential compactness by this solution here: Prove that the hilbert cube is compact
However, my solution was wrong and I understand why. I am not sure how to approach showing that it is complete and totally bounded either, but I see that it is also a possible way to solve this.
Best Answer
Let's work on completeness.
Let $(x_j)$ be a Cauchy sequence with respect to the given metric in the Hilbert cube.
Let's think of $x_j$ as the $j^{\text{th}}$ row of a matrix, and $x^k$ as the $k^{\text{th}}$ column, consisting of all the $k^{\text{th}}$ entries of the elements of the sequence (which are themselves sequences). The $k^{\text{th}}$ entry of the $j^{\text{th}}$ element is $x_{jk}$.
A sketch of the first part of the proof is:
Now we just have to show that this limit $x^*$ is legitimate. This means verifying two things:
Let's attack (2). Fix $\epsilon > 0$. We can take, for each $i$, $N_i$ such that $n > N_i$ implies $|x^*_i - x_{ni}| < \epsilon$. This is the definition of convergence in $\mathbb{R}$ applied to each column. Now the important thing is to finally use the definition of the Hilbert cube, which implies that there exists $N^*$ such that $i > N^*$ implies $|x^*_i - x_{ni}| < \epsilon$, independent of $n$. This is because the entries of a point in the Hilbert cube get really small, indeed by definition we have $|x_{ni} \leq \frac{1}{2^i}|$ independent of $n$, therefore $|x^*_i - x_{ni}| \leq \frac{1}{2^i}$, independent of $n$. Having established this, set $N = \max_{i \leq N^*}N_i$. You can confirm that $n > N$ implies $|x^*_i - x_{ni}| < \epsilon$ for all $i$. Hence we have $d(x^*, x_n) = \sup_i|x_i^* - x_{ni}| < \epsilon$ for $n > N$. That's convergence!
The important thing is that we can construct an infinite class of something (here the $N_i$) but only worry about finitely many of them due to the Hilbert cube's property that entries of its elements get really small.
See if you can get total boundedness using this as inspiration, it's a nice problem and the flavor is similar!
Hint (if you want):