# An elementary embedding which fixes the ordinals, but is not the identity.

In Woodin's "In the Search of Ultimate-$$L$$", in section $$3.5$$ on extenders, he uses an example which I have a hard time understanding. He essentially constructs two transitive models of $$\mathsf{ZFC}^-$$ with an elementary embedding between them, which is cofinal and is the identity on the ordinals but is not the identity!

I would appreciate any clarification on this. First let me quote the relevant bit:

Let $$L[G]$$ be a generic extension of $$L$$ for adding $$\omega_2^L$$ many Cohen reals and let $$L[G][g]$$ be a generic extension of $$L[G]$$ for adding $$\omega_2^L$$-many Cohen reals.
Now define $$M=L_{\omega_2^L}(\mathbb{R}^{L[G]})[G]$$ and define $$N=L_{\omega_2^L}(\mathbb{R}^{L[G][g]})[G][g],$$ where each is viewed as a transitive set. Thus in each case we are constructing over the reals from an additional predicate. Note that $$P(\omega)$$ exists in both $$M$$ and $$N$$ but for example, $$P(\omega_1)$$ does not exist in either $$M$$ or $$N$$.

It follows by the homogeneity of Cohen forcing that both $$M$$ and $$N$$ are models of ZFC\Powerset with the usual formulation of the Axiom of Choice and that the natural map $$\pi:M\rightarrow N,$$ where $$\pi(\mathbb{R}^M)=\mathbb{R}^N$$ is an elementary embedding.

Now I think I understand the construction and I think that the natural map is given by exactly setting $$\pi(\mathbb{R}^M)=\mathbb{R}^N$$ and $$\pi|\mathbb{R}^M=\operatorname{id}_{\mathbb{R}^M}$$ and for other sets in $$M$$, since we build it level by level with definable subsets and we know where to send each parameter, we know where to send each $$x\in M$$ level by level.

Now what I have trouble with, would be to check that elementarity and that these are models of $$\mathsf{ZFC}^-$$. As hinted, I tried to use homogeneity of the Cohen forcing to show elementarity, but for formulas that have parameters in $$M$$, I don't see how we can get rid of the parameters. And also between the axioms of $$\mathsf{ZFC}^-$$, choice is the one which is the most elusive for me, as although I don't have a formal argument for the others, but they seem plausible and usually in these circumstances we don't have choice. So my questions are: How can we show that $$1) \mbox{ }\pi$$ is elementary and $$2) \mbox{ } M$$ and $$N$$ are models of $$\mathsf{ZFC}^-$$, especially choice?

EDIT: I have doubts on why $$\pi$$ is even well-defined, as I have defined it above. Going level by level we may encounter several formulas with other parameters which define the same $$x\in M$$. And checking this is crucial in another sense, since later on when we prove elementarity, these different formulas should indeed correspond in a well-defined manner.

EDIT II: I was informed about this post from MO, which actually solves half of my question! By the argument in that proof $$\pi$$ is readily seen to be well-defined and elementary. But I am still not sure how we can show that $$M$$ and $$N$$ satisfy $$\mathsf{ZFC}^-$$, especially choice.

• Here is my idea for (b): Since $\omega_2^L=\omega_2^{L[G]}$, $L_{\omega_2^L}(\mathbb{R}^{L[G]})$ is a model of $\mathsf{ZFC^-}$ and the forcing for adding $\omega_2^L$ is a class forcing over that model. Thus it suffices to show that the forcing is pretame, and I think it follows from that the forcing has ccc. Nov 16, 2020 at 0:44
• @HanulJeon, thanks! Although I have a few questions. First is that I think the $[G]$ above is meant as a relative constructibility predicate and not class forcing, but are they the same?(I think they may be.) Also how do you get $\mathsf{ZFC}^-$ in the first part, especially choice? (The other stuff are kinda ok but AC is different, I should have mentioned this in the post.) Also I don't know what pretameness is, do you have a reference? Nov 16, 2020 at 8:08
• For the last question, you may refer to Sy Friedman's Fine Structure and Class Forcing (other options are Kameryn Williams' doctoral dissertation or just googling pretame forcing.) Nov 16, 2020 at 10:19
• In general, adding uncountably many Cohen reals to $L$ and considering $L(\Bbb R)$, then we get $\sf AC_{WO}$, and in general the failure of choice comes into play only when we try to well-order the reals, and not before it. So only when we get to the power set of the real numbers. But that part is beyond the reach of $L_{\omega_2}(\Bbb R)$, intuitively speaking. What is unclear to me is why are we adding $\aleph_2$ Cohen reals? Any uncountable number of them should suffice, in principle. Nov 16, 2020 at 14:51
• The answer by Andrés shows that I was right, and any uncountably many Cohen reals would do the trick. For the AC part, I suspect that this is a more delicate analysis of the failure of choice, and showing that it won't happen until the $\omega_2$th step of the relative constructible hierarchy. Nov 19, 2020 at 19:21

Edit: I have added an argument that shows choice in $$M$$ in terms of existence of choice functions.

Replacement in M: Here is an argument that shows $$ZF^-$$ in $$M$$ (the same works for $$N$$ of course):

Let $$\mathbb R^G$$ denote $$\mathbb R^{L[G]}$$. In a first step, we will see that in $$M$$ any definable map $$f:\mathbb R^G\rightarrow \omega_2$$ is bounded. In a second step I will explain how replacement follows. In $$M$$, all sets are definable from ordinal parameters, reals and proper initial segments of $$G$$. Since any real and any proper inital segment of $$G$$ is in an intermediate extension $$L[G\upharpoonright \alpha]$$ and since $$L[G]$$ is a further $$Add(\omega, \omega_2)$$-extension from there, we might as well assume that all parameters in the definition of $$f$$ are from the ground model and will suppress them in what follows. If $$\dot x$$ is a $$Add(\omega, \omega)$$-name, then by ccc, there is $$\alpha_{\dot x}<\omega_2$$ so that $$1\Vdash f^{L_{\check\omega_2}(\mathbb R^\dot G)[G]}(\dot x)<\check\alpha_{\dot x}$$. If $$\dot y$$ is any $$Add(\omega, \omega_2)$$-nice name for a real then there is an automorphism $$\pi$$ of $$Add(\omega, \omega_2)$$ that turns $$\dot y$$ into an $$Add(\omega, \omega)$$-name $$\hat\pi(\dot y)$$. But then $$1\Vdash f^{L_{\check\omega_2}(\mathbb R^\dot G)[\dot G]}(\hat\pi(\dot y))<\check\alpha_{\hat\pi(\dot y)}$$ implies $$1\Vdash f^{L_{\check\omega_2}(\mathbb R^{\widehat{\pi^{-1}}(\dot G)})[\widehat{\pi^{-1}}(\dot G)]}(\dot y)<\check\alpha_{\hat\pi(\dot y)}$$ Note that $$f$$ does not change as all the parameters in its definiton are in the ground model. Furthermore, it is easy to see that $$1\Vdash L_{\omega_2}(\mathbb R^{\widehat{\pi^{-1}}(\dot G)})[\widehat{\pi^{-1}}(\dot G)] = L_{\omega_2}(\mathbb R^{\dot G})[\dot G]$$ using that $$\pi$$ is definable in $$M$$. Thus $$f^M(y)<\alpha_{\hat\pi(\dot y)}$$ so that $$f^M$$ is bounded by $$\operatorname{sup}\{\alpha_{\dot x}\mid \dot x \text{ is a }Add(\omega, \omega)\text{-name}\}<\omega_2$$ We can use this to show that reflection is true in $$M$$.

Claim: For any $$\in$$-formula $$\varphi$$ there is a $$M$$-definable club of $$\alpha<\omega_2$$ with $$\varphi$$ absulute between $$L_\alpha(\mathbb R^G)[G]$$ and $$M$$.

Proof: We will do it by induction on the complexity of $$\varphi$$. The only difficult case is if $$\varphi = \exists x \psi(x)$$. Note that in $$M$$, any set is the surjective image of $$\mathbb R^G\times\omega_1$$ in $$M$$ (as all $$L_\alpha(\mathbb R^G)[G]$$, $$\alpha<\omega_2$$ are such an image). Even better, $$\omega_1$$ is the surjective image of $$\mathbb R^G$$ in $$M$$ so that all sets are the surjective image of a function with domain $$\mathbb R^G$$. Let $$D$$ be an $$M$$-definable club that witnesses our induction hypothesis for $$\psi$$. We can close some $$L_\beta(\mathbb R^G)[G]$$ with $$\beta\in D$$ under witnesses for $$\varphi$$ as follows: Find a surjection $$f:\mathbb R^G\rightarrow [L_\beta(\mathbb R^G)[G]]^{<\omega}$$ in $$M$$ and map a real $$x$$ to the least level of the hierachy in which a witness to $$\varphi(\vec p)$$ (if there is one) exists where $$f(x)=\vec p$$. This function must be bounded so that there is a least $$\gamma\in D$$ so that $$L_\gamma(\mathbb R^G)[G]$$ contains witnesses for $$\varphi$$ with parameters from $$L_\beta(\mathbb R^G)[G]$$. This map $$\beta\mapsto \gamma$$ is definable in $$M$$ and the closure points of this map form a club with the desired properties.$$\Box$$

Now replacement (and even better, collection) is an easy consequence of the reflection principle above.

Choice in M: The axiom of choice in $$M$$ (in terms of existence of choice functions) follows by somewhat similar arguments:

First note that for any $$\alpha<\omega_2$$ there is a map $$C_\alpha:\mathbb R^G\rightarrow\mathbb R^G$$ in $$M$$ so that $$C_\alpha(x)$$ is (better: codes canonically a) $$Add(\omega, \omega)$$-generic over $$L[G\upharpoonright\alpha, x]$$ for any $$x$$. For any $$x\in\mathbb R^G$$ there is some $$\alpha\leq\beta<\alpha+\omega_1$$ so that the section of $$G$$ starting at $$\beta$$ with length $$\omega$$ is generic over $$L[G\upharpoonright\alpha][x]$$. Let $$C_\alpha(x)$$ be this section with $$\beta$$ as small as possible. Then $$C_\alpha$$ is definable over $$(L_{\alpha+\omega_1}(\mathbb R^G)[G];\in, G)$$ and hence is in $$M$$.

[Here the use of the predicate $$G$$ is crucial! No such function $$C$$ exists in $$L_{\omega_2}(\mathbb R^G)$$, so in particular the map $$x\mapsto \{y\mid y$$ is $$Add(\omega, \omega)$$-generic over $$L[x]\}$$ does not have a choice function there.]

Now let $$f\in M$$ be a function we want to find a choice function for. Since replacement holds in $$M$$ and all sets there are surjective images of $$\mathbb R^G$$ we may assume that $$f:\mathbb R\rightarrow\mathcal P(\mathbb R)^M\setminus\{\emptyset\}$$. Let $$\alpha<\omega_2$$ be large enough so that $$f$$ is definable in $$M$$ from paraters in $$L[G\upharpoonright\alpha]$$ (remeber that these parameters can be chosen to be ordinals, reals and initial segments of $$G$$ so such an $$\alpha$$ exists).

Claim: If $$x\in \mathbb R^G$$ and $$g\in M$$ is $$Add(\omega,\omega)$$-generic over $$L[G\upharpoonright\alpha, x]$$ then $$f(x)\cap\mathbb R^{L[G\upharpoonright\alpha, x, g]}\neq\emptyset$$.

Proof: For notational reasons we assume $$\alpha=0$$, so that $$L[G\upharpoonright\alpha] = L$$. Work in $$L[x]$$. Note that $$L[x]$$ is either $$L$$ or a Cohen extension thereof. Also $$g$$ is generic over $$L[x]$$. Thus $$G$$ factors into $$h\times H^\prime$$ so that $$L[x]=L[h]$$ and $$H^\prime$$ is $$Add(\omega, \omega_2)$$-generic over $$L[x]$$ with $$L[G]=L[x][H^\prime]$$. Similarly, we can take an isomorphis $$\mu:Add(\omega, \omega_2)\rightarrow Add(\omega, \omega)\times Add(\omega, \omega_2)$$ so that $$H^\prime$$ factors again (via $$\pi^{-1}$$) into $$g\times H$$. In $$L[x]$$ we can find a nice name $$\dot y$$ so that $$1\Vdash^{L[x]}_{Add(\omega, \omega_2)} \dot y \in f(\check x)^{L_{\check\omega_2}(\mathbb R^{\dot G})[\dot G]}$$ Using the ccc, we can further find an automorphism $$\pi$$ of $$Add(\omega, \omega_2)$$ so that $$\mu\circ\pi$$ shifts $$\dot y$$ completely into the first factor $$Add(\omega, \omega)$$. With the same strategy as in the boundedness arguments, using that all parameters defining $$f$$ are in the ground model, we can see that in fact $$z=\widehat{\mu\circ\pi}(\dot y)^{g\times H^\prime}\in f(x)$$ But the evaluation of $$\widehat{\mu\circ\pi}(\dot y)$$ depends only on $$g$$ so that $$z\in L[x][g]$$ witnesses the claim to hold.

If $$\alpha\neq 0$$, replace $$L$$ by $$L[G\upharpoonright\alpha]$$ in the argument above, work from there and use that $$L[G]$$ is still a $$Add(\omega,\omega_2)$$-extension from there. $$\Box$$

It is now straight forward to define a choice function for $$f$$ in $$M$$: Just map a real $$x$$ to the $$<_{L[x, C_\alpha(x)]}$$-least element of $$f(x)$$ (where $$<_{L[x, C_\alpha(X)]}$$ denotes the canonical wellorder on the reals of $$L[x, C_\alpha(x)]$$).

• Thanks Andreas! I have one small question/remark: this proof works perfectly for $L_{\omega_2}(\mathbb{R}^G)$, but here $M=L_{\omega_2}(\mathbb{R}^G)[G]$, and I guess adding the definability predicate doesn't change your proof, right? Maybe the $G$ predicate in $M$ helps with choice? Nov 20, 2020 at 16:09
• Oh yes thanks! I actually misread the definition of $M$, the same proof works fortunately. I will edit my answer. And you are right: Choice fails in $L_{\omega_2}(\mathbb R^G)$, so the predicate is necessary. Nov 20, 2020 at 16:32
• The line that you crossed out isn't wrong per se, is it? Since we will automatically have $\alpha\lt\omega_1$ and then we can define the section in any $L_\beta(\mathbb{R})[G]$ where, $\alpha<\beta$. Also as a small question, we use $\omega_2$ for the height of $M$ exactly because the number of $Add(\omega, \omega)$ names is $\omega_1$ in the boundedness argument, right? Also thanks for this detailed and insightful answer. Nov 20, 2020 at 19:17
• No, the way I phrased it $\alpha$ will not be automatically $<\omega_1$, in fact the map sending $x$ to the least $\alpha$ with $x\in L[G\upharpoonright\alpha]$ is a surjection onto $\omega_2$ and thus not in $M$ . Yes that is exactly the reason for that height. In $L_{\omega_1}(\mathbb R^G)[G]$ there is a definable surjection from the reals onto $\omega_1$ so that is not a model of $ZF^-$. That was a really fun question! Nov 20, 2020 at 20:08
• I have edited my answer, I hope its more clear now. (This also changed some deatails in the proof of replacement). Note that the whole $G$ is not allowed in the definition of $f$, you only consider $M$ as an $\in$-structure. With $G$ as a predicate, $M$ can define a surjection from $\mathbb R^G$ onto $\omega_2$ by mapping the section of $G$ at $\alpha$ to $\alpha$. Nov 21, 2020 at 13:26