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Suppose $(G,\mathcal T)$ is a paratopological group and $a,b\in G$ and every neighborhood of $a$ contains $b$. Can we say every neighborhood of $b$ contains $a$?

clearly every closed neighborhood of $b$ contains $a$. So I wonder if any open neighborhood of $1$ contains a closed neighborhood of $1$.


If the answer to the above question is negative I add another assumption:

How if we have a neighborhood $U$ of $1$, with $\overline{U^{-1}}$ compact?

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1 Answer 1

Th answer is negative. Let $\tau=\{[a, +\infty): a\in \Bbb R\}$. Then $(\Bbb R, \tau)$ is a paratopoloical group topology. However, $0$ and $1$ witiness the fact.

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Does $0$ have a neighborhood $U$ with $\overline{-U}$ compact? –  user31967 Sep 13 '13 at 10:02

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