Timeline for The existence of the Mostowski collapses for a non-set-like well-founded relation from which there is a homomorphism to $Ord$
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 16, 2017 at 20:21 | vote | accept | Guozhen Shen | ||
| Jul 16, 2017 at 20:21 | |||||
| Jul 16, 2017 at 20:21 | vote | accept | Guozhen Shen | ||
| Jul 16, 2017 at 20:21 | |||||
| Jul 16, 2017 at 20:20 | vote | accept | Guozhen Shen | ||
| Jul 16, 2017 at 20:20 | |||||
| Dec 4, 2016 at 7:16 | comment | added | Guozhen Shen | I solved this problem negatively. | |
| Dec 1, 2016 at 3:43 | comment | added | Noah Schweber | @Sets Hm, good point. I'm about to go to sleep, but I'll come back tomorrow and see if I can fix this. | |
| Dec 1, 2016 at 3:41 | comment | added | Guozhen Shen | If $R$ is transitive, the function in (2) is also a function in (3), and your proof works well. But if $R$ is not transitive, your proof yields a function in (2), which need not to be a function in (3), since in this case, the range of the function in (3) need not to be included in $Ord$. | |
| Dec 1, 2016 at 3:39 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Dec 1, 2016 at 3:29 | comment | added | Guozhen Shen | The $\mu(x)=\mathrm{max}(MS(x))$ is still an ordinal, but as the example above, the desired $G(x)$ could be a set which is not an ordinal. If we consider the class $MSC(x)$ of all possible values of the Mostwoski collapse of $R\cap a^2$ at $x$ ($MSC(x)$ is a subset of $V_{H(x)+1}$), this set need not to have a maximal member, and we can not choose a right member of $MSC(x)$ to define $G(x)$. | |
| Dec 1, 2016 at 2:56 | comment | added | Noah Schweber | @Sets Quite right, I was sloppy. However, it's easily fixed: in fact, the Mostowski spectrum always has a maximum value, and we may take that. | |
| Dec 1, 2016 at 2:56 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Nov 30, 2016 at 7:45 | comment | added | Guozhen Shen | The function $\mu$ you have defined satisfies that $\mu(x)=sup^{+}\{\mu(y)\ |\ y\ R\ x\}$ but not that $\mu(x)=\{\mu(y)\ |\ y\ R\ x\}$. For example, if $x\ R\ y$, $y\ R\ z$ and not $x\ R\ z$, then $\mu(z)$ should be $\{\{0\}\}$, which is not an ordinal. | |
| Nov 30, 2016 at 5:20 | comment | added | Noah Schweber | @Sets Quite right, I misread your desired property; fixed. | |
| Nov 30, 2016 at 5:20 | history | undeleted | Noah Schweber | ||
| Nov 30, 2016 at 5:20 | history | edited | Noah Schweber | CC BY-SA 3.0 |
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| Nov 30, 2016 at 5:17 | history | deleted | Noah Schweber | via Vote | |
| Nov 30, 2016 at 4:59 | comment | added | Guozhen Shen | $F\circ H$ need not to have the desired property, because $H(y)<H(x)$ need not to imply $y\ R\ x$. | |
| Nov 30, 2016 at 4:30 | history | answered | Noah Schweber | CC BY-SA 3.0 |