Timeline for For any entourage $U,V$ there's an entourage $W$ such that $U\circ W\subseteq V\circ U$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 4, 2013 at 11:13 | vote | accept | user31968 | ||
| Apr 3, 2013 at 16:46 | comment | added | HenrikRüping | @CC Thank you comments. I agree with your first comment. However your second comment suggests that your definition of $U\circ V$ is my definition of $V \circ U$. (To avoid this I included my definition of $U\circ V$). I still think this is a counterexample. | |
| Apr 3, 2013 at 16:41 | history | edited | HenrikRüping | CC BY-SA 3.0 |
deleted wrong things
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| Apr 3, 2013 at 14:14 | comment | added | user31968 | And for second part $(100,0.5)\in VoU$ because $(100,0)\in U$ and $(0,0.5)\in V$ but $(100,0.5)\notin \{(x,y)| |y-x|<2 \}\cup (\mathbb{R}\times \{0\})$. So $$V\circ U \ne \{(x,y)| |y-x|<2 \}\cup (\mathbb{R}\times \{0\})$$ | |
| Apr 3, 2013 at 14:02 | comment | added | user31968 | My tries teached me: $U o W = (W^{-1}oU^{-1})^{-1}=(WoU)^{-1}$ and not $WoU$. too common mistake I made too! | |
| Apr 3, 2013 at 13:32 | history | answered | HenrikRüping | CC BY-SA 3.0 |