Timeline for Proper classes subnumerous to $V$ in a model of a Morse-Kelley related theory
Current License: CC BY-SA 3.0
42 events
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| Sep 26, 2016 at 19:29 | comment | added | Andreas Blass | @NoahSchweber You're being too generous (in your comment of Sep. 22) about the agreement of the L and V-in-L hierarchies. Consider, for example, the $L$-cardinal $\kappa=(\omega_1)^{(L)}$. $L$ thinks $L_\kappa$ has cardinality $(\omega_1)^{(L)}$, whereas $L$ thinks $(V_\kappa)^{(L)}$ is way bigger, $(\beth_{\omega_1})^{(L)}$ to be precise. The $L$-hieracrchy won't match the $V$-in-$L$ hierarchy until you get to a fixed-point of the $\beth$ sequence in $L$. | |
| Sep 23, 2016 at 5:14 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 23, 2016 at 5:07 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 23, 2016 at 4:52 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 19:57 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 18:26 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 18:20 | comment | added | Zuhair Al-Johar | OK, I've edited the answer to that effect. | |
| Sep 22, 2016 at 18:14 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 14:34 | comment | added | Noah Schweber | @Zuhair Yes, that's right. (But note that it's not "according to how I'm putting things together" - this is just what those symbols actually mean!) Also, note that the $L$-hierarchy does match up with the $V$-hierarchy relative to $L$, eventually. Specifically, if $\kappa$ is an $L$-cardinal then $L_\kappa=(V_\kappa)^L$; in particular, this means that $\bigcup_{\alpha\in ON} L_\alpha=\bigcup_{\alpha\in ON} (V_\alpha^L)=L$. The discrepancy comes when we examine individual (non-$L$-cardinal) levels. | |
| Sep 22, 2016 at 14:31 | comment | added | Zuhair Al-Johar | @Noah, so according to how you are putting things it appears that I should have spoken in terms of the $V$-hierarchy relative to $L$. So when it is said that $L$ satisfies ZFC, then it is the $V$-hierarchy relative to $L$ that is meant. | |
| Sep 22, 2016 at 14:23 | comment | added | Zuhair Al-Johar | I see! then I must correct my answer. I think that $ZF+GCH$ can interpret $MK^-$, since the stage $V_{w+w+1}$ can act as a model of $"MK^- + \text{the first two conditions of my question}"$ no need to go to the constructible universe, however it still fails the third condition. | |
| Sep 22, 2016 at 13:18 | comment | added | Noah Schweber | However, their widths are different. If $V\not=L$, then $V_\alpha\supsetneq (V_\alpha)^L$ in general; but even if $V$ does equal $L$, we will always have $L_\alpha\subsetneq (V_\alpha)^L$ for many $\alpha$! (Equality will hold if $\alpha$ is an $L$-cardinal, but otherwise not.) Even inside $L$, the $L$-hierarchy (not the $V$-hierarchy relative to $L$!) is seen to be incredibly thin. Here's an easy way to see that the $L$-hierarchy has to be thin, even in $L$: show that every real in $L_{\omega+1}$ is arithmetic. But ZFC proves there are non-arithmetic reals . . . | |
| Sep 22, 2016 at 13:15 | comment | added | Noah Schweber | (cont'd) I strongly suggest you consult a standard reference on the subject, e.g. Devlin's book. (And indeed it is true that $L$ knows that $L_\alpha$ is countable for every $L$-countable $\alpha$; see e.g. math.stackexchange.com/questions/48726/…) The $L$ hierarchy looks very different from the $V$ hierarchy, even as computed inside $L$. Here are the basic facts: there are three hierarchies you care about, $\{L_\alpha\}$, $\{V_\alpha\}$, and $\{(V_\alpha)^L\}$. All three have the same height: $M_\alpha\cap ON=\alpha$ for each choice of $M$. | |
| Sep 22, 2016 at 13:12 | comment | added | Noah Schweber | @Zuhair I understand that, but your statements are still incorrect; you do not understand the construction of $L$. Remember that $L_{\alpha+1}$ is the definable powerset of $L_\alpha$; this does not mean that $L$ thinks that $L_{\alpha+1}=\mathcal{P}(L_\alpha)$! Indeed, that statement is extremely false. For example, the constructible reals $L\cap\mathbb{R}$ are not $L_{\omega+1}\cap\mathbb{R}$, but rather $L_{\omega_1}\cap\mathbb{R}$; indeed, for any $L$-countable $\alpha$ there is a real $r\in L$ with $r\not\in L_\alpha$. This is a standard fact about the constructible hierarchy. | |
| Sep 22, 2016 at 4:48 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 4:18 | comment | added | Zuhair Al-Johar | I'm specifically using "within" to mean provability within the model and not how it externally appears. | |
| Sep 22, 2016 at 4:16 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 4:14 | comment | added | Zuhair Al-Johar | I'm speaking about stages of the constructible universe $L$ of Godel as how it is seen within that model and not as how it is seen externally. $L$ is a model of ZFC, so it proves within it that the power set of x is strictly greater in cardinality than x, so each stage would be strictly larger than its predecessor. That's obvious. | |
| Sep 22, 2016 at 4:04 | comment | added | Zuhair Al-Johar | Yes I'm speaking about what $L$ thinks, I'm not speaking external to $L$. I'm saying it is provable "within" $L$. This is different from what we are having externally , since externally $L_\alpha$ is countable, but "internally" I mean as how $L$ is seeing matters within, it is as I said. | |
| Sep 22, 2016 at 3:57 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 3:56 | comment | added | Noah Schweber | @zuhair see my comment, the answer is still incorrect. | |
| Sep 22, 2016 at 3:55 | comment | added | Zuhair Al-Johar | I've corrected my answer, I also clarified what is meant by having higher cardinalities, it is as seen within the model. Of course the number of formulas is countable. I hope that helps. | |
| Sep 22, 2016 at 3:53 | comment | added | Noah Schweber | I believe you are confusing $L_\alpha $ with $(V_\alpha)^L $; the $\alpha $th level of the $L $ hierarchy is vastly smaller in general than what $L $ thinks is the $\alpha $th level of the cumulative hierarchy. In particular, your first-paragraph sentence beginning "clearly it is provable" is completely false as written. | |
| Sep 22, 2016 at 3:52 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 3:41 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 3:33 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 3:05 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 2:47 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 22, 2016 at 2:38 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 15:27 | comment | added | Emil Jeřábek | The number of formulas is relevant in that the definition of the stages of L involves taking definable subsets of the previous stages. Asaf is perfectly right, $|L_\alpha|=|\alpha|$ for infinite ordinals $\alpha$, and standard ordinal multiplication makes $2\cdot\omega=\omega$ (the well order you get by putting $\omega$ many pairs after each other is still $\omega$). | |
| Sep 21, 2016 at 14:21 | comment | added | Zuhair Al-Johar | I think there is a confusion here, I'll edit my answer as to avoid this confusion. There is nothing to do with the number of formulas here, we are just taking different stages of Godel's constructible universe $L$. | |
| Sep 21, 2016 at 14:17 | comment | added | Asaf Karagila♦ | (I see now that someone downvoted your answer. It wasn't me.) | |
| Sep 21, 2016 at 14:17 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 14:16 | comment | added | Asaf Karagila♦ | No. $2\omega=\omega$ and $\omega+\omega=\omega2$, the latter of which is indeed greater than $\omega$. Secondly, there are only countably many formulas in the language of set theory, and it follows that for an infinite $\alpha$, $|\alpha|=|L_\alpha|$. Unless your $2w$ is some uncountable ordinal, you're just wrong. I'm not saying that your idea cannot work, just that the way you present it contains several mistakes. | |
| Sep 21, 2016 at 14:09 | comment | added | Zuhair Al-Johar | $2w > w$ where > is for ordinal greater than (not cardinal greater than), and definitely $L_{2w}$ is much bigger than $L_w$, the later is nothing but the set of all constructible hereditarily finite sets, while the fomrer is the union of all naturally indexed iterative constructible powers of the later. The cardinality of $L_w$ is $\aleph_0$, the cardinality of $L_{w+n}$ is $\aleph_n$, and that of $L_{2w}$ is $\aleph_w$ | |
| Sep 21, 2016 at 14:00 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 13:50 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 13:50 | comment | added | Asaf Karagila♦ | (1) $2\omega=\omega$, you probably mean $\omega2$; (2) $L_{\omega2+1}$ is countable, it knows about no infinite cardinal except $\omega$ itself. | |
| Sep 21, 2016 at 13:48 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 13:42 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 13:03 | history | edited | Zuhair Al-Johar | CC BY-SA 3.0 |
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| Sep 21, 2016 at 12:52 | history | answered | Zuhair Al-Johar | CC BY-SA 3.0 |