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A compact paratopological group is a topological group. How to prove it?

An abelian paratopological group is a topological group. Is this right?

A paratopological group is a topological semigroup that is algebraically a group. In other words, it is a group $G$ with a topology such that the group's product operation is a continuous function from $G × G$ to $G$. This differs from the definition of a topological group in that the group inverse is not required to be continuous.

Thanks a lot.

David Chan
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1 Answers1

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The answer to the first question is yes, even if $G$ is not assumed Hausdorff. For a proof, see this answer. The $T_1$ case is elementary, though the general case seems to require some work.

The answer to the second question is no. Another answer in the same thread gives an example of a topology on $\mathbb{R}$ for which addition is continuous, but $x\mapsto -x$ is not continuous.

If you want an example of an abelian Hausdorff paratopological group that is not a topological group, there appears to be an example in the first link I posted, which uses some kind of $p$-adic positive cones on $\mathbb{Z}$.

Andrew Dudzik
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    So the answer to the first question is "yes and no"? I didn't realize paratopological groups were related to paraconsistent logic. – Mario Carneiro Jun 17 '15 at 08:02
  • @MarioCarneiro In both cases I was obviously referring to the first question that I hadn't yet answered. I've updated to the more standard terminology of static ordering. – Andrew Dudzik Jun 17 '15 at 08:25
  • A standard example of an (abelian) paratopological group failing to be a topological group is the Sorgenfrey line, that is the real line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $[a,b)$, $a<b$). – Alex Ravsky Jun 18 '15 at 04:34