I have the following definition:


A set A in a metric space (M, d) is said to be totally bounded if, given any $\epsilon >0$, there exist finitely many points $x_1, ..., x_n \in M$ such that $A \subset \bigcup_{i = 1}^{n} B_\epsilon(x_i)$.


The Hilbert cube $H^\infty$ is the collection of all real sequences $x = (x_n)$ with $|x_n|\leq 1$ for $ n = 1, 2, ...$

Furthermore the question refers to an exercise:

A metric space is called seperable if it contains a countable dense subset.

Question: How do I prove that $H^\infty$ is totally bounded?

I don't have any clue how I should handle this, thanks in advance!

  • 2
    $\begingroup$ What metric on $H^\infty$ do you take? $\endgroup$ – Wojowu Oct 7 '17 at 13:51
  • 1
    $\begingroup$ Can you explain why your definition differs from the usual one of the Hilbert cube? It's sort of important, since the traditional Hilbert cube is compact, of course. $\endgroup$ – user436658 Oct 7 '17 at 14:42


With your definition, $H^{\infty}$ is not totally bounded, as you can find an infinite subset $\{e_n\}$ such that $d(e_m, e_n) \ge 1$ for all $m\ne n$. ( for whatever reasonable norm you are considering).

Are you sure it's not " $|x_n|\le \frac{1}{n}$ " ?

  • $\begingroup$ The definition in my book says $|x_n| \leq 1$ unfortunately.. Could you give an example of such an infinite subset $\{e_n\}$? $\endgroup$ – titusAdam Oct 8 '17 at 8:39

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