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First some topological definitions in terms of nets and sequences:

A topological space $(X, \tau)$ is

  • compact iff every net has a convergent subnet
  • sequentially compact iff every sequence has a convergent subsequence

A uniform space $(X, \mathcal{U})$ is

  • totally bounded iff every net has a Cauchy subnet
  • complete iff every Cauchy net converges
  • sequentially complete iff every Cauchy sequence converges

If $(X, \mathcal{U})$ is a uniform space and $\tau$ its induced topology then it is known that $(X, \tau)$ is compact iff $(X, \mathcal{U})$ is totally bounded and complete.

I found in "Topological Uniform Structures" by Warren Page on p. 44, Example 6.2 the statement: "A sequentially compact uniform space is sequentially complete and totally bounded".

Questions:

  1. Is this statement true?
  2. Is this statement true if $(X, \mathcal{U})$ is the uniform space corresponding to a topological vector space $(X, \tau)$?

I don't think that total boundedness is implied by sequential compactness (how should one argue about nets if one has only information about sequences, e.g. in the open ordinal $[0, \omega_1)$?). The statement should be rather of the form "A sequentially compact uniform space is sequentially complete and sequentially totally bounded" where $(X, \mathcal{U})$ is sequentially totally bounded should be defined as "every sequence has a Cauchy subsequence".

Remarks:

  1. If $(X, \tau)$ is metrizable then the statement is true since compactness is equivalent to sequential compactness. (It also holds if $X$ is a Banach space equipped with its non-metrizable weak topology by Eberlein-Smulian.)

  2. It seems to me that the notion of "sequential total boundedness" is not common in the literature. In ncatlab this notion is referred to as "sequential precompactness".

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  • $\begingroup$ Total boundedness is in general not defined in terms of nets, but in terms of entourages: for every entourage $N \in \mathcal{U}$, there are finitely many $x_1,..,x_n$ such that the $B(x_i, N)$ cover $X$. It's a form of uniform compactness. One can show it is equivalent to the net formulation (every net has a Cauchy subnet). $\endgroup$ – Henno Brandsma Jan 29 '16 at 13:26
  • $\begingroup$ Yes, I'm assuming AC, so that precompactness is equivalent to total boundedness. (A uniform space is precompact iff its completion is compact iff every net has a Cauchy subnet.) $\endgroup$ – yadaddy Jan 29 '16 at 13:28
  • $\begingroup$ Just found an answer to my question (Kelley, General Topology, p. 209, Problem M): In complete uniform spaces, one can strengthen countable compactness to compactness: if $(X, \mathcal{U})$ is complete then $A \subseteq X$ is relatively countably compact if and only if $A$ is relatively compact, i.e. its closure $\overline{A}$ is compact. Since sequential compactness implies countable compactness question 1 (and thus 2) has a positive answer. $\endgroup$ – yadaddy Jan 29 '16 at 13:33
  • $\begingroup$ How does this imply Warren's statement exactly? Please write the details in an answer. Embed everything in the completion? $\endgroup$ – Henno Brandsma Jan 29 '16 at 13:35
  • $\begingroup$ Ah, I see the problem: Kelley's statement relies on the completeness of $(X, \mathcal{U})$ but it is not clear at this stage whether its sequential completeness is enough. However, I also found this answer, which seems not to rely on completeness: math.stackexchange.com/a/104814/167838 $\endgroup$ – yadaddy Jan 29 '16 at 13:47
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Suppose that $(X,\mathcal{U})$ is sequentially compact. So there exists some $N \in \mathcal{U}$ (some entourage) such that $X$ cannot be covered by finitely many $B(x,N)$. So pick $x_0 \in X$, note that $B(x_0, N)$ does not cover $X$ so there is some $x_1 \notin B(x_0,N)$. By recursion, if we have constructed points $x_0,\ldots, x_n$ then we find some $x_{n+1}$ that is not in $\cup_{i=0}^n B(x_i, N)$, as the latter cannot cover $X$.

This sequence $(x_n)$ we constructed this way has a convergent subsequence, by sequential compactness, so $x_{n_k} \rightarrow p$, for some $p \in X$. Then pick some entourage $N'$ such that $N' \circ N' \subseteq N$, then $B(p,N')$ contains a tail of the subsequence, so some $x_{n_{k_1}}, x_{n_{k_2}}, k_1 < k_2$ such that $x_{n_{k_1}}, x_{n_{k_2}} \in B(p,N')$ which implies that $x_{n_{k_2}} \in B(x_{n_{k_1}}, N)$, contradicting how we chose the sequence.

So a sequentially compact (or even limit point compact) uniform space is totally bounded. It's certainly sequentially complete (as a convergent subsequence is certainly Cauchy..).

Added I thought we needed an equivalence, so I wrote

For the converse (a try!) suppose we have a totally bounded and sequentially complete space $(X,\mathcal{U})$. If we have some sequence $(x_n)$, then suppose it has no convergent subsequence (striving for a contradiction), or we can even assume it has no Cauchy subsequence, using sequential completeness. This means that for every subsequence of our sequence there is some entourage $N$ such that infinitely many members of that subsequence are not $N$-close. I think that starting with the complete sequence and the above $N$, find $N'$, an entourage in $\mathcal{U}$, with $N' \circ N' \subseteq N$, and total boundedness gives us a finite cover by finitely many $N'$-balls, we have a subsequence that lives in one of the balls, and which are (by how $N'$ is chosen) $N$-close. This subsequence has its own entourage $N^(1)$ as above, and we can continue recursively again. But I don't see how to conclude this line of reasoning with a final contradiction, though. We get thinner and thinner and closer and closer subsequences, but we have no first countability or some such property to exploit.

So left to right I see but right to left not yet. Maybe someone has ideas?

Added Is there an easy counterexample for the reverse ($\omega_1$ isn't one, maybe some other countably compact and sequentially compact but non-compact space, like a $\Sigma$-product?

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  • $\begingroup$ You mentioned that if a uniform space $X$ is limit point compact then $X$ is totally bounded. I see that this holds if $X$ is countably compact. If $X$ is only limit point compact then we need in addition $T_0$ to get $T_1$ by regularity of $X$ which then gives us countable compactness. In $X = \mathbb{R}^2$ with the pseudometric $d(x,y) := |x_1 - y_1|$ every singleton $\{ x \}$ has an accumulation point ($\{ x \}' = (\{ x_1 \} \times \mathbb{R}) \setminus \{ x \}$). $\endgroup$ – yadaddy Feb 11 '16 at 10:07

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