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I have seen that the definition of precompact sets in a topological group $G$ is a bit tricky. Can someone please explain? I saw that it has to do something with totally bounded sets. Is there a more 'natural' definition of precompactness? For example, is it true to say that if I have a subset $K$ in $G$, such that the closure of $K$ is compact in $G$, then $K$ is a precompact set in $G$?

Daniel Fischer
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User666x
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For example, is it true to say that if I have a subset $K$ in $G$, such that the closure of $K$ is compact in $G$, then $K$ is a precompact set in $G$?

Yes, that is true. Such sets are called "relatively compact (in $G$)", which is a slightly more restrictive condition than precompactness.

While precompactness is an intrinsic property of a uniform space - the most often encountered examples of uniform spaces are metric spaces and topological groups [and subsets of these] - relative compactness is an extrinsic property, it refers to the ambient space. But every relatively compact subset of a uniform space is precompact.

A uniform space $(X,\mathscr{U})$ is called precompact (or totally bounded) if for every entourage $V \in \mathscr{U}$, $X$ can be covered by finitely many sets which are "small of order $V$". A set $M$ is small of order $V$ if $M\times M \subset V$. This shows that every subspace of a precompact space is again precompact, and it is easily seen that every quasicompact uniform space is precompact. For metric spaces, we can use the entourages $A_\varepsilon = \{ (x,y) : d(x,y) \leqslant \varepsilon\}$, and then a set is "small of order $A_\varepsilon$" if and only if its diameter is $\leqslant \varepsilon$.

If we restrict our attention to Hausdorff spaces, we have the simple characterisation that a Hausdorff uniform space is precompact if and only if its completion is compact [and that a space is compact if and only if it is complete and precompact].

Is there a more 'natural' definition of precompactness?

For topological groups $G$, we can characterise precompactness as follows:

$S \subset G$ is precompact if and only if for every neighbourhood $U$ of $e$, there is a finite set $F_U$ such that $$S\subset \bigcup_{x\in F_U} x\cdot U.$$

I'm not sure whether that meets your expectations, but it seems a little easier to understand than the general formulation for uniform spaces.

And for subsets complete Hausdorff topological groups, the concepts of precompactness and relative compactness coincide. If a (Hausdorff) topological group is not complete, it has precompact subsets that are not relatively compact.

Daniel Fischer
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  • ♦ one question. Fu is a finite subset of G? – User666x Jul 26 '15 at 10:58
  • Yes, but one can also require it to be a finite subset of $S$, that is equivalent. – Daniel Fischer Jul 26 '15 at 11:01
  • ♦ so in metric space this is just the totally bounded subspaces of X? – User666x Jul 28 '15 at 12:21
  • Yes. Although total boundedness is less commonly used with non-metric spaces, people also speak of totally bounded subsets/subspaces of arbitrary uniform spaces. Total boundedness and precompactness are different names for the same property. – Daniel Fischer Jul 28 '15 at 14:20
  • @ Daniel Fischer♦ can you please give me the precise definition of a precompact set in a uniform space $(X,\mathscr{U})$? Thank you in advance! – User666x Aug 13 '15 at 16:44
  • For an entourage $U\in \mathscr{U}$, one says that a set $M$ is small of order $U$ if $M\times M \subset U$. Then $P\subset X$ is precompact if for every $U\in \mathscr{U}$, $P$ can be covered by finitely many sets that are small of the order $U$. – Daniel Fischer Aug 13 '15 at 17:33
  • I've seen everywhere that left relatively dense implies pre-compact (ie the characterisation offered). I think the proof should be simple, but I can't find one or do it. Can you please show the equivalence? Thanks! – abe.nong Dec 05 '16 at 11:44
  • @abe.nong I'm not sure what equivalence you mean? That a topological group is precompact if and only if every neighbourhood of $e$ is left relatively dense? Something else? – Daniel Fischer Dec 05 '16 at 13:06
  • Yes, that's what I meant, sorry! The iff that you stated in your comment just then. – abe.nong Dec 05 '16 at 13:12
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    @abe.nong The inversion $\iota \colon G \to G;, x \mapsto x^{-1}$ gives an isomorphism (not group isomorphism [unless $G$ is abelian], an isomorphism of uniform spaces) between $(G,\mathscr{U}_l)$ and $(G,\mathscr{U}_r)$, where $\mathscr{U}_l$ is the left uniformity, generated by the $V_l := { (x,y) : x^{-1}y \in V}$, where $V$ runs over the neighbourhoods of $e$, and $\mathscr{U}_r$ is the right uniformity, generated by the $V_r := { (x,y) : x y^{-1} \in V}$. To see that, note $$(\iota\times\iota)(V_l)={(x^{-1},y^{-1}):x^{-1}y\in V} = {(\xi,\eta):\xi\eta^{-1}\in V} = V_r.$$ – Daniel Fischer Dec 05 '16 at 13:42
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    Thus $G$ is precompact in the left uniformity if and only if it is precompact in the right uniformity. Therefore it suffices look at the left uniformity. Now if $(G,\mathscr{U}_l)$ is precompact, by definition of precompactness, for every neighbourhood $V$ of $e$ there are finitely many sets $W_1,\dotsc, W_n$ that are small of order $V_l$ such that $G=\bigcup W_k$. Small of order $V_l$ means $W_k\times W_k \subset V_l$, and this means $W_k^{-1}\cdot W_k \subset V$. For each $k$, pick $x_k\in W_k$, then $x_k^{-1}W_k\subset W_k^{-1}W_k\subset V$, so $W_k\subset x_kV$, and we have our finite set – Daniel Fischer Dec 05 '16 at 13:43
  • Thanks so much! My problem was more the other direction. That is, every neighbourhood of e being left relatively dense implied pre-compactness. Thanks again! – abe.nong Dec 05 '16 at 14:22
  • Or in particular, when we look at a basic entourage $V_\ell$ generated by a neighbourhood $V$, what are the sets $A_i$ such that $A_i \times A_i \subset V_\ell$? Using relative density on $V$ hasn't worked for me, as $V \times V \not\subset V_\ell$ as $x, y \in V \not\implies x^{-1}y \in V$. Cheers! – abe.nong Dec 05 '16 at 14:30
  • Nevermind, of course I don't use $V$, i just use the existence of $U$ such that $U^{-1}U \subset V$ by the neighbourhood filter axioms for a topological group to get $U\times U \subset V_\ell$. Thank you very much for your help! – abe.nong Dec 05 '16 at 14:50
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    $F = {x_1,\dotsc,x_n}$ with $G = F\cdot V$, that is, $V$ is left relatively dense. Conversely, if every neighbourhood of $e$ is left relatively dense, we want to show that $(G,\mathscr{U}_l)$ is precompact. So take a neighbourhood $W$ of $e$, and we want to cover $G$ by finitely many sets small of order $W_l$. Since $x\cdot V$ is small of order $(V^{-1}\cdot V)_l$, it suffices to take a neighbourhood $V$ of $e$ with $V^{-1}\cdot V \subset W$, and by the left relative denseness of $V$, we have a finite $F$ with $F\cdot V = G$. The finitely many sets $xV$, $x\in F$ do what we want. – Daniel Fischer Dec 05 '16 at 15:07
  • @abe.nong Sorry, very flaky internet connection today. It broke down while I was typing, and thus cut off the one direction. It broke down again immediately after I managed to post that. – Daniel Fischer Dec 05 '16 at 16:52