Timeline for Is being close to a Halting set computable?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2018 at 21:20 | vote | accept | Stella Biderman | ||
| Oct 6, 2018 at 18:00 | comment | added | Andrej Bauer | I meant that for every $k$ there exists an encoding. I supplemented my answer with a construction that works for all $k$'s at once. | |
| Oct 6, 2018 at 18:00 | history | edited | Andrej Bauer | CC BY-SA 4.0 |
added 355 characters in body
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| Oct 6, 2018 at 15:44 | comment | added | Stella Biderman | @tomasz Cool, that’s what I had figured but the wording seemed to imply the latter to me for some reason. | |
| Oct 6, 2018 at 14:48 | comment | added | tomasz | @StellaBiderman: For every $k$ there is an encoding. | |
| Oct 6, 2018 at 12:58 | comment | added | Stella Biderman | Thank you for the further details! I’m a little confused by your third paragraph; the implicit order of quantifiers is throwing me. Do you mean that, for any $k$, there exists an encoding such that $d(x, H)=k$ is undecidable? Or that there exists an encoding such that, for any $k$, $d(x, H)$ is undecidable? | |
| Oct 6, 2018 at 8:41 | history | answered | Andrej Bauer | CC BY-SA 4.0 |