Timeline for Order-Perserving Bijection $f:A\to A^*$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Feb 6, 2016 at 22:47 | vote | accept | Julian Rachman | ||
| Feb 5, 2016 at 22:00 | comment | added | Julian Rachman | @J.-E.Pin I assume that you have your own answer now then? | |
| Feb 5, 2016 at 9:39 | comment | added | Julian Rachman | @J.-E.Pin Actually, yes. | |
| Feb 5, 2016 at 9:04 | comment | added | J.-E. Pin | I suspect that the order you want to use on $A^*$ is not the subword ordering but the following order: $a_1 \cdots a_n \leqslant b_1 \cdots b_m$ if there exists $i_1 < \cdots < i_n$ such that $a_k \leqslant b_{i_k}$ for all $k$. | |
| Feb 4, 2016 at 16:31 | answer | added | Joel David Hamkins | timeline score: 1 | |
| Feb 4, 2016 at 7:38 | comment | added | Andrej Bauer | Ok, I edited the question. | |
| Feb 4, 2016 at 7:38 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
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| Feb 4, 2016 at 7:37 | comment | added | Julian Rachman | @AndrejBauer Yes. I know that if $A$ is finite, then this is not possible. | |
| Feb 4, 2016 at 7:36 | comment | added | Andrej Bauer | Did you mean to write that $A$ is infinite? | |
| Feb 4, 2016 at 7:33 | history | asked | Julian Rachman | CC BY-SA 3.0 |