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Oct 9 at 17:39 comment added Tomás Pacheco Thank you! However, I don't think point 2 is correct since $\cap_k \Gamma_k =\Gamma_1$ (to be completely rigorous the intersection of these quotients is not well defined, it would make more sense to take the union of the $N_k$'s). And so if we have $\Gamma_2 \neq \{e\}$ and we take $G'$ to be the subgroupoid where remove $[1,e]$ then $G'$ is still Hausdorff and an HLS groupoid for the sequence $N_k$ with k starting at $k=2$.
Oct 9 at 17:13 history edited Omar Mohsen CC BY-SA 4.0
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Oct 9 at 17:10 history edited Omar Mohsen CC BY-SA 4.0
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Oct 9 at 17:09 history edited Omar Mohsen CC BY-SA 4.0
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S Oct 9 at 17:09 review First answers
Oct 9 at 17:34
S Oct 9 at 17:09 history answered Omar Mohsen CC BY-SA 4.0