Prove that the hilbert cube is compact

Question

I'm trying to show that the Hilbert cube is compact and want to know if I'm taking the right approach. 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 need to show that it is compact with respect to the metric:

$d(x,y)=\underset{n}{\textrm {sup}}|x_n - y_n|$

In order for the sequence in $H$ to converge, we need each of it's components to converge.

Since, each component $x_n$ is bounded (by $0$ and $\dfrac{1}{2^n}$) for any point in $H$, a sequence (in $H$) of such points would form a sequence $(x_n)$ for each component $n$ and each of those sequences would be bounded.

Therefore, by Bolzano-Weierstrass, they have convergent subsequences.

We can show that for each sequence of points in $H$, we have a subsequence of the sequence of $x_1$s that converges, say ${x_1}_k$, we then have a sequence $({x_1}_k, {x_2}_k,...)$.

For the sequence of ${x_2}_k$s, there exists a convergent subsequence ${{{x_2}_k}_j}$. Since all subsequences of a convergent subsequence also converge, we have that both $({{{x_1}_k}_j})$ and $({{{x_2}_k}_j})$ converges in $({{{x_1}_k}_j}, {{{x_2}_k}_j},...)$.

Repeating this process ad infinitum, we get convergent subsequences for all $x_n$ so that the sequence in $H$ has a subsequence that converges as well and, therefore, $H$ is compact.

Is this correct? If it is, is that enough to show that $H$ is compact under the given supremum metric?

Answer 1

No. This is not correct. As an aside, your paragraph

Since, each $x_n$ is bounded (by $0$ and $1\over 2^n$) for any point in $H$, a sequence of such points would form a sequence $(x_n)$ for each $n$ and each of those sequences would be bounded.

is so poorly constructed, I could not figure out what even the intent was until after I had read through the rest of the the post and had enough evidence to reconstruct it.

But the actual error in the argument is this: There is not necessarily a relation between the convergent subsequences in the different indices.

For example, the convergent subsequence chosen for $(x_{1n})$ may be $(x_{1(2k)})$, while the convergent subsequence chosen for $(x_{2n})$ may be $(x_{2(2k+1)})$. Thus the convergent subsequences for your first two indices would not ever come from the same points. There is no subsequence $(x_{n_k})$ of $(x_n)$ itself such that $\pi_1(x_{n_k})$ is the $k^{th}$ in your $(x_{1n})$ subsequence and $\pi_2(x_{n_k})$ is the $k^{th}$ in your $(x_{2n})$ subsequence.

You could try to work around this as follows:

Let $n^1_k$ define a subsequence of $(x_n)$ such that $\pi_1(x_{n^1_k})$ converges. Let $n^2_k$ define a subsequence of $(x_{n^1_k})$ such that $\pi_2(x_{n^2_k})$ converges. Etc.

But this still doesn't work. While $\pi_j(x_{n^m_k})$ is guaranteed to converge for each $j \le m$, there is no guarantee of a sequence that works for all indices. For example, it could be that $n^m_1 = m$ for all $m$. Thus any subsequence you tried for the points themselves would start out less than an infinite number of your index-wise convergent subsequences.

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Since convergence of sequences depends on the tails of the sequences, you might by careful argument find a way past those obstacles, but it would be tricky to do so. I suggest you try to prove this by other means than sequential compactness.

Instead, I suggest trying to prove that $H$ is complete and totally bounded.