Suppose that $G$ is a regular feebly compact Moore quasitopological group. Must $G$ be a topological group? This was previously posted here on MathSE also.

A semitopological group $G$ is a group $G$ with a topology such that the product map of $G\times G$ into $G$ is separately continuous. If G is a semitopological group and the inverse map of $G$ onto itself associating $x^{-1}$ with arbitrary $x\in G$ is continuous, then $G$ is called a quasitopological group. If $G$ is a quasitopological group and the product map of $G\times G$ into $G$ is jointly continuous, then $G$ is called a topological group.

A space is feebly compact if every family of locally finite non-empty open sets is finite.

Note that if $G$ is Tychonoff, the answer to this question is true,since $G$ will be Cech complete.


Yes. The paper [KKM] contains more general results. In particular, by Corollary 1 each regular semitopological group which is a cover semi-complete Baire space is a topological group. It remains to note that each $p-\sigma$-fragmentable space (see [Bou] for the definition) (in particlar, each $p$-space, so each Moore space) is cover semi-complete [KKM] and a (quasi)regular feebly compact space is Baire.


[Bou] Ahmed Bouziad, Every Cech-analytic Baire semi-topological group is a topological group, Proc. Amer. Math. Soc. 124 (1996), 953-959.

[KKM] P. S. Kenderov, I. S. Kortezov, W. B. Moors Topological games and topological groups, Topology Appl. 109 (2001) P.157-165. (I have a preprint, to which I referred).

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    $\begingroup$ It is very useful for me, thank you!! $\endgroup$ – Paul Jun 15 '18 at 7:33

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