Timeline for Is there inconsistency with having countable models of Z with these internalizing properties?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21 at 12:16 | comment | added | Joel David Hamkins | Yes, that is exactly what I had in mind. | |
| Jul 21 at 12:09 | vote | accept | Zuhair Al-Johar | ||
| Jul 21 at 12:03 | comment | added | Zuhair Al-Johar | So, we take any missing subset $S$ of $\omega$, then we externally define a functin $f^1$ from $e$ to $S$ that is bijective, then we define another bijective function $f^2$ from the odds to the complementary set of $S$ relative to $\omega$. Then the union of these two functions would be a function from $\omega \to \omega$ that will contradict the stipulated condition. I think, that's what you are saying. Nice! | |
| Jul 21 at 11:58 | comment | added | Joel David Hamkins | Externally, the even numbers are bijective with any infinite subset of $\omega$, so we can define such an $f$ by mapping $2n$ to the $n$th element of the missing set. So $f[e]$ will be that missing set. And then we define $f(2n+1)$ so as to make it a bijection $f:\omega\to\omega$. (Note: the missing subsets will be infinite/coinfinite.) | |
| Jul 21 at 11:44 | comment | added | Zuhair Al-Johar | What's the proof that such a function exists in every countable transitive model of $\sf Z$? Because your argument begins with an assumption that clearly violates the needed condition. | |
| Jul 21 at 11:19 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |