Timeline for Can we represent computable functions by r.e. sets ?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2011 at 4:52 | vote | accept | Cheng Peng | ||
| Dec 27, 2011 at 16:30 | comment | added | Joel David Hamkins | One doesn't need forcing to construct the set $A$, if you follow my instructions. You build a set $A$ by ensuring for each c.e. set $W_e$ either that $A\intersect W_e$ is finite or that $A-W_e$ is finite. Given such a set $A$, any bijection $f:\mathbb{N}\to A$ has the desired property, but $f$ may not be computable. | |
| Dec 27, 2011 at 16:25 | vote | accept | Cheng Peng | ||
| Dec 27, 2011 at 16:25 | |||||
| Dec 27, 2011 at 16:24 | comment | added | Cheng Peng | thank you very much for your helpful answer. I guess the answer is no but I cannot find a counter example. I am a neophyte who learn computability theory and I am unfamiliar with forcing. Thanks again. | |
| Dec 26, 2011 at 19:57 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Added negative answer in general case; deleted 8 characters in body
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| Dec 26, 2011 at 12:53 | history | undeleted | Joel David Hamkins | ||
| Dec 26, 2011 at 12:52 | history | deleted | Joel David Hamkins | ||
| Dec 26, 2011 at 12:48 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |