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In Zermelo set theory, the axiom of constructibility V=L seems to imply every set has a transitive closure. But does that argument have to assume transitive closure in the first place, to get an adequate theory of constructibility?

Actually I am most intersted in this over Mathias's set theory MAC in https://www.dpmms.cam.ac.uk/~ardm/maclane.pdf. I don't know if that makes a difference.

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  • $\begingroup$ In Mathias's paper, MAC includes transitive closure by definition. Do you mean ZBQC? $\endgroup$ Jul 7 '14 at 10:10
  • $\begingroup$ @EmilJeřábek Sure, or in other words I want to know if adding V=L to MAC makes Tco redundant. $\endgroup$ Jul 7 '14 at 12:47
  • $\begingroup$ TCo is transitive containment, which is not exactly the same as transitive closure. You need something like powersets with bounded comprehension or $Pi_1$-comprehension to get the transitive closure from TCo. $\endgroup$ Jul 8 '14 at 11:46
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    $\begingroup$ Most formulations of $V = L$ make transitive containment redundant. Whether you use the Gödel $L_\alpha$, Jensen $J_\alpha$ or other hierarchies, these are transitive by design so formal statements for V = L, like $\forall x \exists \alpha (Ord(\alpha) \land x \in L_\alpha)$, immediately imply transitive containment. Transitive closure is a bit tricky but it often follows too. $\endgroup$ Jul 8 '14 at 11:51
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(I'm a student so what i'm going to write next can hide any kind of errors...but i hope not. Moreover this is one of my first answers here, so tell me if it is someting wrong in my exposition or anything else. I'll be gratefull with anyone who show me errors in my argument. Thanks.)

I think you don't need $V=L$ to conclude $\forall x\in V\quad tc(x)\in V$ (where $tc(x)$ is the transitive closure of $x$).

As Kunen has proved in his "Set Theory" (III-Th.3.6), in $ZF^-$ (i.e. ZF without Foundation) you can prove that $$x\in WF\equiv tc(x)\in WF,$$ where WF is the class of well founded sets. Then (by III-Th.4.1) you have that in $ZF^-$ you can prove that

$$\text{axiom of Foundationn}\equiv V=WF.$$

so if you are in $ZF$ you have Foundation and so you have $V=WF$ (i.e. $\forall\,x\quad x\in WF$) and so you have $\forall x\quad tc(x)\in V$ that is the same as $\forall\,x\; \exists\,y\quad y=tc(x)$. cvd

Note that, if you work in $ZF^-$, you mantain the result in the case of sets in $WF$ (i.e $\forall x\in WF\quad tc(x)\in WF(\subseteq V)$) but, in general you cannot say that for all $x\in V$. At this step if you assume $V=L$ than you can use VI-Lemma 1.11 (i.e $L(\alpha)\subseteq R(\alpha)$ for all $\alpha$) and so you can conclude that $L\subseteq WF (\subseteq V)$ and by VI-Lemma 1.6 (i.e. $L(\alpha)$ is transitive and $\forall\,\xi\leq\alpha\;\quad L(\xi)\subset L(\alpha)$) and the recursive costruction of $L$, you can conclude both $L$ is transitive and $\forall\,x\in L\quad ct(x)\in L$

Moreover the definition of $L$ is based on $Df$ (and/or $En$; see Def.V-1.1,1.4) and by Lemma V-1.7 thay are absolute for transitive model of "ZF without Power Axiom", and $WF$ is a model of whole ZF.

So, in my point of view, this "prove" that $L$ is closed under transitive closure (and so V is close too, if you assume L=V) with or without Foundaton, and you don't need to assume any kind of transitive closure of it.

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  • $\begingroup$ Thank you. But I am asking when V is closed under transitive closure, not just when there is inner model with transitive closures. $\endgroup$ Jul 6 '14 at 23:09
  • $\begingroup$ ZF^- proves the existence of transitive closures, and you are using this fact throughout the whole argument, starting with the “As Kunen has proved...”. So, I fail to see how this answers the question. $\endgroup$ Jul 6 '14 at 23:13
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    $\begingroup$ Corrado, my understanding is that Mathias has proved that Z and also Mac do not prove that every set has a transitive closure. That was very surprising, since everyone thinks of $V_{\omega+\omega}$ as a "typical" kind of model of Z, but Mathias's models are far weirder. The question here is whether V=L is sufficient to rule out those pathologies. Of course the expectation is that it is, but this probably depends on exactly how $V=L$ is formulated, since in these weak theories, there are issues. $\endgroup$ Jul 6 '14 at 23:26

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