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$(G,\mathcal T)$ is a semitopological group with $$(\forall x\in G)(\forall U\in \mathcal T)(\exists V \text{ neighborhood of }1)(x\notin U^cV)$$ Then $G$ is quasitopological (ie inverse is continuous).

Any idea to prove it or to find missing minimal necessary assumptions.

1 Answers1

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It seems the following.

To satisfy your condition, $x$ should be not in $U^c$, that is $x\in U$. Now if we take into account this correction, we obtain a corollary for $x=1$: for each neighborhood $U$ of the unit there exists a each neighborhood $U$ of the unit such that $1\not\in U^cV$, that is $V^{-1}\subset U$, which is equivalent to the continuity of the inversion.

Alex Ravsky
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  • thanks. Let me ask a different question. Let $(R,\mathcal T)$ be topological ring. Can you give a reference which verifies convergence and continuity of $f(x)=\sum_{n=1}^\infty a_nx^n$? thanks. –  Feb 22 '14 at 09:05
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    @CutieKrait I don’t know an exact reference. I think you can find relevant info when you google something like “convergence in normed rings”. – Alex Ravsky Feb 23 '14 at 11:49