Timeline for Is every convex subset of a Borel-linearly ordered space measurable?
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 22, 2015 at 0:00 | comment | added | Julian Newman | @NateEldredge: Yes, I have had precisely this thought - in particular, the "well-ordered cofinal subsets" argument probably requires at least $\omega_1$-dependent choice, whereas perhaps the embedding of $X$ into $2^\xi$ doesn't need anything like this. | |
| Sep 21, 2015 at 20:14 | comment | added | Nate Eldredge | @JulianNewman: In fact, maybe we can give a more direct proof by embedding $X$ into $2^\xi$, which I think would let us construct the desired countable cofinal set more explicitly. | |
| Sep 21, 2015 at 20:06 | comment | added | Julian Newman | @NateEldredge: Yes, thank you very much for this. I was actually about to post a comment myself that I have read somewhere that every Borel-totally ordered space $X$ admits no copies of $\omega_1$, since there exists a countable ordinal $\xi$ such that $X$ can be embedded into $2^\xi$. [I don't know whether this embedding is just an order embedding or has additional measurable-structure-respecting properties.] So then, by precisely the argument in your comment and in your answer, we have the desired result. | |
| Sep 21, 2015 at 20:00 | vote | accept | Julian Newman | ||
| Sep 21, 2015 at 19:47 | history | edited | Nate Eldredge | CC BY-SA 3.0 |
mention convexity
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| Sep 21, 2015 at 19:42 | comment | added | Joel David Hamkins | Bravo! ${{{{}}}}$ | |
| Sep 21, 2015 at 19:24 | history | answered | Nate Eldredge | CC BY-SA 3.0 |