Timeline for Does limitation of size imply axiom of powerset in Morse-Kelly if the generalized continuum hypothesis is included in Morse-Kelley set theory?
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17 events
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| Nov 25, 2023 at 14:25 | comment | added | M. Solomon | Note that the set case of the axiom (gch) is more controversial than the proper class case (limitation of size). | |
| Nov 25, 2023 at 14:05 | comment | added | M. Solomon | But the common spirit connecting the two cases is that both say a power class has cardinality as close as possible to the cardinality of the class it’s the power class of. If C is a set by cantor’s theorem the cardinality of its power class P is greater so the cardinality of P is the next higher cardinality after the cardinality of C. If C is a proper class the cardinality of P is the cardinality of C. Also the power class property is instrumental in proving limitation of size from global axiom of choice, so we claim the power class property appears naturally in the axiom. | |
| Nov 25, 2023 at 13:47 | comment | added | M. Solomon | And in fact your definition works as well (in a way better) than mine since V is the only proper class case we care about and V satisfies your definition. I only pointed out the difference in case it had to do with your apparent objection to the axiom that combines gch and limitation of size based on power class concept. I think your objection is you view the set and the proper class classes of this axiom as being very different in spirit, due to difficulty of defining a reasonable power class concept for proper classes. Differences are given in my above comments. | |
| Nov 25, 2023 at 0:24 | comment | added | Hanul Jeon | @M.Solomon I put my interpretation of "power class" in my earlier comment, and I noticed latter that you did not use this definition. My example still works under your notion, however. | |
| Nov 24, 2023 at 16:12 | comment | added | M. Solomon | And that the case of the axiom where where C is a set and P is the power set of C is the usual gch, which says that the cardinality of P is the next highest cardinality after the cardinality of C, whereas the case where C is a proper class yields that the cardinality of V, and hence of any proper class, is the next highest cardinality after the cardinalities of the sets. Of course by limitation of size there is no cardinality higher than the cardinality of C if C is a proper class. | |
| Nov 24, 2023 at 15:31 | comment | added | M. Solomon | It is true that the only proper class case of the axiom that matters is the V case, which gives limitation of choice. | |
| Nov 23, 2023 at 13:38 | comment | added | M. Solomon | I just noticed that your above argument that V is the only proper class that is a power class uses a different concept of power class than I used. You say that C is a power class if it contains each subset of C. I say that P is a power class if it equals the class of all subsets of some class C. P and C don’t have to be the same. For example let C = the class of all single element sets. C is a proper class that does not satisfy the condition in the axiom but the power class P of C, which is not V, does satisfy that condition by limitation of size and the power class property. | |
| Nov 22, 2023 at 0:21 | comment | added | Hanul Jeon | @M.Solomon I did not say your GCH is not contradictory with the usual one. But it is not the right formulation of GCH for "class-sized cardinals," and I guess there is no such. | |
| Nov 21, 2023 at 22:44 | vote | accept | M. Solomon | ||
| Nov 21, 2023 at 22:43 | comment | added | M. Solomon | I think you have shown that you answered my question negatively for my formulation of gch also. Thanks! But I still don’t understand how my formulation of gch contradicts the usual one. | |
| Nov 21, 2023 at 22:35 | comment | added | Hanul Jeon | As I stated before, the power class of a given class does not represent the "true" collection of all possible "subclasses" of a given class (which will be a third-order object rather than a second-order object, or in other words, a class.) Thus the powerclass of a proper class does not have the "right" size, which could be out of your intention. | |
| Nov 21, 2023 at 22:31 | comment | added | Hanul Jeon | Even your current formulation of GCH is valid over the model $L_{\omega_1}$ I suggested: Take any subset $X\subseteq L_{\omega_1}$, and consider the powerclass $P(X)$ of $X$ computed over $L_{\omega_1}$, which is equal to the set of all countable subsets of $X$, whose cardinality is $\omega_1$. If $|S|<|P(X)|$ holds over $(L_{\omega_1}, \mathcal{P}(L_{\omega_1}))$, then $|S|<|P(X)| = \omega_1$ holds over $L$. This implies $S$ is countable, so it has the same cardinality with some member of $P(X)$. | |
| Nov 21, 2023 at 22:26 | comment | added | Hanul Jeon | The issue is that unlike that the powerset of a set has a larger cardinality than the given set, the powerclass of a given class is not necessarily "larger" than the given set. A powerclass of a class is more similar to, something like, a collection of "small" subsets of a given set (e.g., For an inaccessible cardinal $\kappa$, the collection of all subsets of $\kappa$ of cardinality $<\kappa$, which is completely different from the genuine powerset of $\kappa$.) | |
| Nov 21, 2023 at 22:20 | comment | added | M. Solomon | I do consider the power class of a set to be the class of all subsets of that class. My formulation of gch is Let P be the power class of a class C. If |S| < |P| then S has the cardinality of some element of P (in other words the cardinalities of the subclasses of P are already the cardinalities of the elements of P or the cardinality of P). The case where C is a set is the usual gch, right? The case where C=V gives the axiom of limitation of size. Am I missing something? | |
| Nov 21, 2023 at 21:59 | comment | added | Hanul Jeon | @M.Solomon I do not see why your GCH is really the GCH we commonly say, since it follows from the Global Choice that is compatible with the failure of the usual GCH, if my understanding on your power class is correct: I understand your power class as a class $C$ containing all subsets of $C$. Then you can see by induction that $V_\alpha\subseteq C$ for every ordinal $\alpha$, so $C=V$. If a class $S$ satisfies $|S|<|V|$, and if we have the Global Choice, then $S$ must be a set so it has the same cardinality with a member of $V$. | |
| Nov 21, 2023 at 21:50 | comment | added | M. Solomon | As far as formulation of gch, I wanted to use a formulation of gch and limitation of size together that does not explicitly refer to sets: given any power class P( a class equal to the power class of some class) and any class S with |S| <|P|, S has the same cardinality as some member of P. I’m pretty sure the answer to my question is no also for this formulation. Does your proof work for this formulation of gch also? Actually I like that gch and limitation of size can be bundled together that natural way as limiting cardinalities of subclasses of power classes. | |
| Nov 21, 2023 at 18:19 | history | answered | Hanul Jeon | CC BY-SA 4.0 |