Question concerning the compactness of the Hilbert Cube Given the Hilbert cube, $H= \displaystyle\prod_{n\in \mathbb{N}} \left[0,\frac{1}{n}\right]\subset \ell^2$. I want to show that it is compact using a method different to the methods already seen on this site, namely:


*

*compactness of Hilbert cube

*Show that the Hilbert cube is compact under the sup metric
I am using the fact that the Hilbert cube with the subspace topology is coarser than the product topology. After that apply Tychonoff to imply compactness. I wish to know if my logic is correct. 
$\textbf{My attempt:}$  Let $u\in U$ be an open set such that 
$$u\in U \subset H =\displaystyle\prod_{n\in \mathbb{N}} \left[0,\frac{1}{n}\right]\subset \ell^2. $$
Without loss of generality, 
$$U = \{x\in H\;|\; ||x-u||<\epsilon \}\qquad \text{ where } ||x-u||=\Bigg(\sum |x_i-u_i^2|\Bigg)^{\frac{1}{2}}.$$
let $N\in \mathbb{N}$ s.t $\displaystyle \sum_{i=N+1} \frac{1}{i^2}< \frac{\epsilon^2}{2}$. We have $\displaystyle \sum_{i=N+1}^\infty |x_i-u_i|^2 < \sum_{i=N+1}\frac{1}{i^2}<\frac{\epsilon^2}{2}.$
Now, let $\alpha_i= \max\left(0, u_i-\frac{\epsilon}{2N}\right)$ and   $\beta_i = \min\left(1, u_i+\frac{\epsilon}{2N}\right).$ The open set 
$$V=\prod_{i=1}^N [\alpha_1,\beta_i].$$ We have
$$u\in V$$
$$V\subset U$$
because $\forall N\in V$,
$$||N-u|| = \left(\sum|v_i-u_i|^2\right)^{\frac{1}{2}}< \left(N\frac{\epsilon^2}{2N}+\frac{\epsilon^2}{2}\right)^{\frac{1}{2}}<\epsilon.$$
Thus we have shown that the Hilbert cube equipped with the subspace topology is is coarser than the product topology, and from Tychonoff's theorem it should be compact. 
 A: I looked it up and as a subspace of $\ell^2$, the topology induced on $H$ from the metric can be shown to be the same as the product topology...
Since it's the product topology,  you can use the Tychonoff theorem directly... 
A: For general interest, here is a proof that the Hilbert-space topology on $H$ is finer than the (Tychonoff) product toology.
Let $p=(p_n)_{n\in \Bbb N}\in S$ where $S$ is a member of the "canonical" base (basis) for the product topology on $H.$ That is, $S=\prod_{n\in N}S_n$ where each $S_n$ is open in $[0,1/n]$ and the set $R=\{n: S_n\ne [0,1/n]\}$ is finite. 
Let $M\in \Bbb N$ such that $\forall n\in R\;(n<M).$ Let $r>0$ where $r$ is small enough that $\forall n\in R\;(\;(-r+p_n,r+p_n)\cap [0,1/n]\subset S_n).$ 
Now for every  $q=(q_n)_{n\in \Bbb N}\in H$  we have $$\|q-p\|<r/M\implies $$ $$\implies\forall n\in R (|q_n-p_n)/n|<r/M)\implies$$ $$\implies \forall n\in R (|q_n-p_n|<n r/M< r)\implies q\in S.$$
So the Hilbert-space topology is finer than the product topology. The proposer has shown the reverse inclusion, so the two topologies are equal.  
