Timeline for Infinite sequences/ordered tuples of proper classes in NBG
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 27 at 10:54 | comment | added | Shthephathord23 | Let us continue this discussion in chat. | |
| Feb 26 at 14:12 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 26 at 14:08 | comment | added | Zuhair Al-Johar | @Shthephathord23, Yes! That is correct. Because of global choice a proper class is a class of cardinality $On$. I'm speaking within MK and not NBG. But for although all proper classes are of the same cardinality, namely $On$, yet working in MK, there are more proper classes than sets. For example you cannot have a tuple of all proper classes, because the largest tuple you can have is of cardinality $On$ while you have more proper classes than elements of $On$. | |
| Feb 26 at 13:42 | comment | added | Shthephathord23 | But wouldn't the fact that all proper classes have the "same length" be provable from global choice(which is an axiom of NBG as it says on wikipedia), so then all proper classes have the same "cardinality"? | |
| Feb 26 at 13:36 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 26 at 13:35 | vote | accept | Shthephathord23 | ||
| Feb 26 at 13:35 | comment | added | Shthephathord23 | Yes, that is what the predicate $P$ would mean, sorry for not mentioning. | |
| Feb 26 at 13:31 | comment | added | Zuhair Al-Johar | @Shthephathord23, what is $P$ in $ \{ x \mid (\exists \alpha)(P(\alpha, x))\}$, does it mean "pair" standing for an ordered pair whose first projection is $\alpha$ and second projection is $x$. If this is it, then of course you can get unions this way, because each $\alpha$ index a class (whether set or not), and you get the union of all those indexed by ordinals that satisfy an arbitrary formula $\varphi$ for example. | |
| Feb 26 at 13:26 | comment | added | Zuhair Al-Johar | @Shthephathord23. Yes you can have a tuple of order $On$ itself. Of course $On$ is the class of all set ordinals, and not the class of all ordinals, since $On$ itself is an ordinal and it is a proper class. Remember you can speak of tuples of proper classes as long as its order is $\leq On$, but there are more proper classes than sets! This is known in MK. You can indeed define arbitrary unions along tuples but those tuples cannot exceed $On$ in length. Of course there are some extensions of MK that states existence of well ordering over all proper classes, but this is another issue. | |
| Feb 26 at 13:16 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 26 at 12:58 | comment | added | Shthephathord23 | However, what I have called "union class" could be defined easily as $\{ x \mid (\exists \alpha)(P(\alpha, x))\}$, so it is not an issue. Your answer is very helpful, but please confirm to me whether or not can we index tuples by proper classes, such as I have done with $On$ and point out why not if that's the case. Either way the answer was helpful. | |
| Feb 26 at 12:55 | comment | added | Shthephathord23 | Your answer does really answer my question, but I would like to give you some background as to what I am actually trying to do. I want tuples of arbitrary length(i guess with the observation above, even proper-class length) of proper classes, because I actually want to define arbitrary unions and arbitrary generalized Cartesian products of families of proper classes and I thought that defining a tuple was the way to begin. But, talking outside the formal language, the definition above, assumes we have "the union" already figured out, and then we partition it accordingly in tuples. | |
| Feb 26 at 12:49 | comment | added | Shthephathord23 | So, what I am understanding is that if we have some class of values, call it $C$(i conceptually think of it as the "union" of all my outputs, I know this is wrong since we can't do sum classes of proper classes) we can take any ordinal $\alpha$ and if we want a $\beta$-indexed tuple then it would be $\alpha \times X \cap \beta \times X$? And I guess this method does generalize for tuples indexed by a proper class, since we can take $On \times X$ and we can talk about any tuple indexed by a proper subclass of $On$. Here $On = \{ x \mid Ord(x)\}$(i.e. $On$ is the class of all ordinals). | |
| Feb 26 at 12:38 | comment | added | Zuhair Al-Johar | @Shthephathord23, Yes! Since you want an infinite tuple of proper classes, then each entry is a proper class and thus nonempty, so you need that domain to be $\omega$ itself, where $\omega$ is the set of all naturals. | |
| Feb 26 at 12:19 | comment | added | Shthephathord23 | ok, but I still don't understand the construction. From what I can follow, is the class of all first projection the domain(i.e. $\{ x \vert (\exists y)( (x, y) \in A \times B) \}$)? | |
| Feb 26 at 12:11 | comment | added | Zuhair Al-Johar | @Shthephathord23, the Kuratowski-Cartesian product between classes $A$ and $B$ (denoted as $A \times B$) is the class of all Kuratowski-ordered pairs $\langle a,b \rangle$ where $a \in A$ and $b \in B$. The Kuratowski ordered pair of $a$ and $b$ denoted as $\langle a,b \rangle$ is the set $\{\{a\},\{a,b\}\}$. So, to write it formally: $A \times B = \{ \langle a, b\rangle \mid a \in A \land b \in B \}$ | |
| Feb 26 at 12:09 | comment | added | Shthephathord23 | Can you elaborate on the "Kuratowski-Cartesian product"? I am not an expert in the study of set theory. | |
| Feb 26 at 11:44 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Feb 26 at 11:37 | history | answered | Zuhair Al-Johar | CC BY-SA 4.0 |