2
$\begingroup$

enter image description here

One should note that the family here may not countable. If it is countable, the it is a consequence of the following results;

Lemma 1. the product of countable totally bounded metric spaces is totally bounded.

Lemma 2. Every uniform space is uniformly isomorphic to a product of metrizable uniform spaces.

Lemma 3. The product of uniform spaces is uniform induced.

How about the case when uncountable?

$\endgroup$
  • 1
    $\begingroup$ Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ? $\endgroup$ – Max Jul 18 '18 at 8:45
  • $\begingroup$ The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set. $\endgroup$ – Henno Brandsma Jul 18 '18 at 21:12
  • $\begingroup$ The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise. $\endgroup$ – Henno Brandsma Jul 18 '18 at 21:28
1
$\begingroup$

Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.

To check the right to left one, let all $(X_s, \mathcal{U}_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=\prod_{s \in S} X_s$ which is of the form $U=\{(x_s,y_s)_{s \in S}: |x_{s_i} - y_{s_i} | < V_{s_i}, i=1,\ldots,N\}$, where $s_1,\ldots, s_N$ are finitely many indices from $S$ and $V_{s_i} \in \mathcal{U}_{s_i}$. For each $i\in \{1,\ldots,N\}$ we can find a finite set $F_i \subseteq X_{s_i}$ that forms is $V_{s_i}$-dense. Fix any point $p \in X$ for "defaultness" and define $$F = \{(x_s) \in S: \forall i \in \{1,\ldots,N\}: x_{s_i} \in F_i \text{ and } \forall s \in S\setminus\{s_1,\ldots,s_N\}: x_s =p_s\}$$

and note that $|F| = \prod_{i=1}^N |F_i|$ and hence is finite and is the required finite $U$-dense set.

$\endgroup$
0
$\begingroup$

The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:

Theorems used:

(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).

(2) If $X$ is totally bounded, then $X'$ is totally bounded.

(3) A uniform space is compact if and only if it is complete and totally bounded.

Proof:

If $P = \Pi_{s \in S} X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.

If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = \Pi_{s \in S} X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.

$\endgroup$

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.