I know that, in metric spaces, compactness implies completeness. In fact, (i) compactness is equivalent to the fact (ii) every infinite set has an accumulation point and to the fact that (iii) any sequence admits a convergent subsequence, and this third fact applied to Cauchy sequences implies completeness. In my Geometry 1 notes, I have examples of topological spaces where (i) holds and (iii) doesn't, and viceversa. Completeness does not imply compactness if the space is not totally limited, i.e. if there exists $\epsilon>0$ such that the space cannot be covered by a finite number of balls of radius $\epsilon$. I would like to have counterexamples to the equivalence of (i) and (ii) and of (ii) and (iii). But more importantly, which is the title of the question, I would like to know the relations between (i), (ii), (iii) and completeness in a topological group, or in a group in general. So:

  • Are there any equivalences between the three conditions in (topological) groups?
  • Or any implications?
  • Can you provide counterexamples to any non-equivalence?
  • What is the relation between compactness and completeness in groups? Or more specifically, is there any implication, or are there conditions that give an equivalence/implication?
  • I know that there can't be an equivalence as topological groups which are also metric spaces exist and for metric spaces the equivalence is easily disproved with $\mathbb{R}$; an implication, however, holds for metric spaces; if that implication is not valid for topological groups, or for groups, can you provide a counterexample?

Yes, this question is huge, I realize that. Any reference is very welcome. Please avoid Google Books, as I've had problems with that kind of reference.

  • $\begingroup$ I assume completeness was intended, given content. However, completeness is a term that arises in the metric space context, so it doesn't exactly make sense to ask about it outside of metric spaces (well there are uniform structures, but I defer to someone else who actually knows about those) $\endgroup$ – JHance Dec 29 '14 at 20:38
  • 1
    $\begingroup$ What is your definition of compactness? "Every open cover has a finite subcover." Or something else? $\endgroup$ – quid Dec 29 '14 at 20:39
  • $\begingroup$ @quid yes that is my definition of compactness. $\endgroup$ – MickG Dec 29 '14 at 20:46
  • $\begingroup$ @Jhance yes I meant completeness, i.e. that any Cauchy sequence converges. Here is a link on Uniform spaces. Anyway, completeness means that every Cauchy sequence converges. In topological groups, we have a notion of convergence and of Cauchy sequence, hence a notion of completeness. Maybe it's called in another way. In fact, in any group there is a notion of Cauchy sequence, and from that one can get a topology, as illustrated here. So in any group we have a notion of completeness. $\endgroup$ – MickG Dec 29 '14 at 20:55
  • $\begingroup$ Again, maybe it has another name in that context. Maybe it hasn't been investigated. I have no idea. But I am curious :). $\endgroup$ – MickG Dec 29 '14 at 20:56

It seems the following.

At first we consider compactness.

(i) is called compactness,

(ii) is called countably compactness,

(iii) is called sequentially compactness.

It is easy to check that a space is countably compact provided it is sequentially compact or compact. In particular, each compact space which is not sequentially compact (for instance, $\{0;1\}^\frak c$ or $\beta N$ [Eng, Ex.3.10.38] ) is an example of a countably compact space which is not sequentially compact. Nevertheless, each compact space with countable character or of cardinality less than $2^{\aleph_1}$ is sequentially compact. Form the other side, each sequentially compact space which is not compact (for instance, $\omega_1$ endowed with the order topology or a $\Sigma$-product of an uncountable family of compact metrizable spaces with at least two points each) is an example of a countably compact space which is not compact. I recall that a space $X$ is $\Sigma$-product of a family $\{X_\alpha:\alpha\in A\}$ if there exists a point $x\in\prod X_\alpha$ such that $X=\{y\in \prod X_\alpha: |\{\alpha:y_\alpha\ne x_\alpha\}|\le\omega\}.$ From the other side, by Chaber Theorem, each regular countably compact space with $G_\delta$- diagonal is a metrizable compact [Gru, p.422 of the file or p.433 of the book].

For topological groups the situation does not change essentially. We already have an implication that a space is countably compact provided it is sequentially compact or compact. None of them can be reversed for topological groups, because $\Bbb Z_2^\frak c$ is a compact topological group which is not sequentially compact, and a $\Sigma$-product $\{y\in \Bbb Z_2^{\frak c}: |\{\alpha:y_\alpha\ne 0\}|\le\omega\}$ is a sequentially compact topological group which is not compact.

Now we consider completeness. For the sake of simplicity, in the following we assume all topological groups to be Hausdorff. A topological group $G$ is H-closed if $G$ is closed in every topological group containing $G$ as a topological subgroup. A topological group $G$ is H-closed if and only if it is Raikov-complete, that is complete with respect to the upper uniformity which is defined as the least upper bound $\mathcal L\vee\mathcal R$ of the left and the right uniformities on $G$. Recall that the sets $\{(x,y): x^{-1}y\in U\}$, where $U$ runs over a base at unit of $G$, constitute a base of entourages for the left uniformity $\mathcal L$ on $G$. In the case of the right uniformity $\mathcal R$, the condition $x^{-1}y\in U$ is replaced by $yx^{-1}\in U$. The Raikov completion $\hat G$ of a topological group $G$ is the completion of $G$ with respect to the upper uniformity $\mathcal L\vee\mathcal R$. For every topological group $G$ the space $\hat G$ has a natural structure of a topological group. The group $\hat G$ can be defined as a unique (up to an isomorphism) Raikov complete group containing $G$ as a dense subgroup. Each locally compact topological group is Raikov complete [AT, Th. 3.6.24]. From the other side, a countably compact topological group is compact iff it is Raikov complete [AT, Th. 3.7.2, 3.7.15].

According to the classical Birkhoff-Kakutani Theorem [AT,3.3.12], a topological group is metrizable if and only if it is first countable. Because a topological space is metrizable by a complete metric iff it is Čech-complete and metrizable [Eng, Th. 4.3.26] and a first countable topological group is Raikov complete iff it is Čech-complete ([Bro] or [Cho]), we see that a first countable topological group is Raikov complete iff it is metrizable by a complete metric.

Also every metric on a compact space is totally bounded (in your terms, totally limited) and complete [Eng, Th. 4.3.27-28]. So a metrizable space is compact iff on it there exists a metric which is both totally bounded and complete [Eng, Th. 4.3.29]. Also a metrizable space is compact iff each metric on it is complete.


[AT] Alexander V. Arhangel'skii, Mikhail G. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.

[Bro] L.G. Brown, Topologically complete groups, Proc. Amer. Math. Soc. 35 (1972), 593--600.

[Cho] M.M. Choban, On completions of topological groups, Vestnik Moskov. Univ. Ser. Mat. Mekh. 1 (1970), 33--38 (in Russian).

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

[Gru] Gary Gruenhage. Generalized metric spaces in Handbook of set-theoretic topology, ed. K. Kunen, J. Vaughan, North-Holland, 1984.

  • $\begingroup$ Excellent. One thing: what does the exponent C in $\mathbb{Z}_2^{\mathfrak{C}}$ and its brother $\{0;1\}^{\mathfrak{C}}$ mean? $\endgroup$ – MickG Jan 1 '15 at 21:32
  • $\begingroup$ @MickG $\frak c=2^\omega=|\Bbb R|$ $\endgroup$ – Alex Ravsky Jan 2 '15 at 1:39

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.