Timeline for Does any locally compact topological group which is not Hausdorff have a Haar measure?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| May 23 at 13:24 | comment | added | Emil Jeřábek | @ResearchMath I don’t think you need that; the $N$-invariance of open subsets follows quite easily from the definitions. Shifting things to the origin, let $U\ni1$ be open, and assume for contradiction that $x\in N\setminus U$. Then $xU^{-1}$ is an open neighbourhood of $x$ that avoids $1$, contradicting $x\in N=\overline{\{1\}}$. | |
| May 21 at 18:30 | vote | accept | ResearchMath | ||
| May 21 at 18:29 | comment | added | ResearchMath | (for those wondering) In (3) use is made of the following: Let $G$ be a topological group. For every neighborhood $U$ of $1$, there is a neighborhood $V$ of $1$ such that $\overline{V} \subset U$ (see e.g. 'Abstract Harmonic Analysis Volume I' by Hewitt and Ross, Corollary (4.7)). | |
| May 21 at 11:28 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
added 12 characters in body
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| May 21 at 9:22 | history | answered | YCor | CC BY-SA 4.0 |