2
$\begingroup$

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:

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.

$\endgroup$
1
  • $\begingroup$ I have not checked the details of your work but you are correct that a coarser topology than a compact topology is also compact $\endgroup$ Apr 16 '18 at 22:17
1
$\begingroup$

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...

$\endgroup$
1
  • $\begingroup$ I have given as an answer (because it won/t fit in a comment), a proof that the Hilbert-space topology is finer than the product topology, which is the other half of proving the two topologies are equal. But a coarser topology than a compact topology is also compact, so for the purposes of the Q, the proposer's approach is sufficient. $\endgroup$ Apr 16 '18 at 22:33
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.