# The universal definition — it can define any mathematical object you like, in the right set-theoretic universe

In set theory, we have the phenomenon of the universal definition. This is a property $\phi(x)$, first-order expressible in the language of set theory, that necessarily holds of exactly one set, but which can in principle define any particular desired set that you like, if one should simply interpret the definition in the right set-theoretic universe. So $\phi(x)$ could be defining the set of real numbes $x=\mathbb{R}$ or the integers $x=\mathbb{Z}$ or the number $x=e^\pi$ or a certain group or a certain topological space or whatever set you would want it to be. For any mathematical object $a$, there is a set-theoretic universe in which $a$ is the unique object $x$ for which $\phi(x)$.

The universal definition can be viewed as a set-theoretic analogue of the universal algorithm, a topic on which I have written several recent posts:

Let’s warm up with the following easy instance.

Theorem. Any particular real number $r$ can become definable in a forcing extension of the universe.

Proof. By Easton’s theorem, we can control the generalized continuum hypothesis precisely on the regular cardinals, and if we start (by forcing if necessary) in a model of GCH, then there is a forcing extension where $2^{\aleph_n}=\aleph_{n+1}$ just in case the $n^{th}$ binary digit of $r$ is $1$. In the resulting forcing extension $V[G]$, therefore, the real $r$ is definable as: the real whose binary digits conform with the GCH pattern on the cardinals $\aleph_n$. QED

Since this definition can be settled in a rank-initial segment of the universe, namely, $V_{\omega+\omega}$, the complexity of the definition is $\Delta_2$. See my post on Local properties in set theory to see how I think about locally verifiable and locally decidable properties in set theory.

If we push the argument just a little, we can go beyond the reals.

Theorem. There is a formula $\psi(x)$, of complexity $\Sigma_2$, such that for any particular object $a$, there is a forcing extension of the universe in which $\psi$ defines $a$.

Proof. Fix any set $a$. By the axiom of choice, we may code $a$ with a set of ordinals $A\subset\kappa$ for some cardinal $\kappa$. (One well-orders the transitive closure of $\{a\}$ and thereby finds a bijection $\langle\mathop{tc}(\{a\}),\in\rangle\cong\langle\kappa,E\rangle$ for some $E\subset\kappa\times\kappa$, and then codes $E$ to a set $A$ by an ordinal pairing function. The set $A$ tells you $E$, which tells you $\mathop{tc}(\{a\})$ by the Mostowski collapse, and from this you find $a$.) By Easton’s theorem, there is a forcing extension $V[G]$ in which the GCH holds at all $\aleph_{\lambda+1}$ for a limit ordinal $\lambda<\kappa$, but fails at $\aleph_{\kappa+1}$, and such that $\alpha\in A$ just in case $2^{\aleph_{\alpha+2}}=\aleph_{\alpha+3}$ for $\alpha<\kappa$. That is, we manipulate the GCH pattern to exactly code both $\kappa$ and the elements of $A\subset\kappa$. Let $\phi(x)$ assert that $x$ is the set that is decoded by this process: look for the first stage where the GCH fails at $\aleph_{\lambda+2}$, and then extract the set $A$ of ordinals, and then check if $x$ is the set coded by $A$. The assertion $\phi(x)$ did not depend on $a$, and since it can be verified in any sufficiently large $V_\theta$, the assertion $\phi(x)$ has complexity $\Sigma_2$. QED

Let’s try to make a better universal definition. As I mentioned at the outset, I have been motivated to find a set-theoretic analogue of the universal algorithm, and in that computable context, we had a universal algorithm that could not only produce any desired finite set, when run in the right universe, but which furthermore had a robust interaction between models of arithmetic and their top-extensions: any set could be extended to any other set for which the algorithm enumerated it in a taller universe. Here, I’d like to achieve the same robustness of interaction with the universal definition, as one moves from one model of set theory to a taller model. We say that one model of set theory $N$ is a top-extension of another $M$, if all the new sets of $N$ have rank totally above the ranks occuring in $M$. Thus, $M$ is a rank-initial segment of $N$. If there is a least new ordinal $\beta$ in $N\setminus M$, then this is equivalent to saying that $M=V_\beta^N$.

Theorem. There is a formula $\phi(x)$, such that

1. In any model of ZFC, there is a unique set $a$ satisfying $\phi(a)$.
2. For any countable model $M\models\text{ZFC}$ and any $a\in M$, there is a top-extension $N$ of $M$ such that $N\models \phi(a)$.

Thus, $\phi(x)$ is the universal definition: it always defines some set, and that set can be any desired set, even when moving from a model $M$ to a top-extension $N$.

Proof. The previous manner of coding will not achieve property 2, since the GCH pattern coding started immediately, and so it would be preserved to any top extension. What we need to do is to place the coding much higher in the universe, so that in the top extension $N$, it will occur in the part of $N$ that is totally above $M$.

But consider the following process. In any model of set theory, let $\phi(x)$ assert that $x$ is the empty set unless the GCH holds at all sufficiently large cardinals, and indeed $\phi(x)$ is false unless there is a cardinal $\delta$ and ordinal $\gamma<\delta^+$ such that the GCH holds at all cardinals above $\aleph_{\delta+\gamma}$. In this case, let $\delta$ be the smallest such cardinal for which that is true, and let $\gamma$ be the smallest ordinal working with this $\delta$. So both $\delta$ and $\gamma$ are definable. Now, let $A\subset\gamma$ be the set of ordinals $\alpha$ for which the GCH holds at $\aleph_{\delta+\alpha+1}$, and let $\phi(x)$ assert that $x$ is the set coded by the set $A$.

It is clear that $\phi(x)$ defines a unique set, in any model of ZFC, and so (1) holds. For (2), suppose that $M$ is a countable model of ZFC and $a\in M$. It is a fact that every countable model of ZFC has a top-extension, by the definable ultrapower method. Let $N_0$ be a top extension of $M$. Let $N=N_0[G]$ be a forcing extension of $N_0$ in which the set $a$ is coded into the GCH pattern very high up, at cardinals totally above $M$, and such that the GCH holds above this coding, in such a way that the process described in the previous paragraph would define exactly the set $a$. So $\phi(a)$ holds in $N$, which is a top-extension of $M$ as no new sets of small rank are added by the forcing. So statement (2) also holds. QED

The complexity of the definition is $\Pi_3$, mainly because in order to know where to look for the coding, one needs to know the ordinals $\delta$ and $\gamma$, and so one needs to know that the GCH always holds above that level. This is a $\Pi_3$ property, since it cannot be verified locally only inside some $V_\theta$.

A stronger analogue with the universal algorithm — and this is a question that motivated my thinking about this topic — would be something like the following:

Question. Is there is a $\Sigma_2$ formula $\varphi(x)$, that is, a locally verifiable property, with the following properties?

1. In any model of ZFC, the class $\{x\mid\varphi(x)\}$ is a set.
2. It is consistent with ZFC that $\{x\mid\varphi(x)\}$ is empty.
3. In any countable model $M\models\text{ZFC}$ in which $\{x\mid\varphi(x)\}=a$ and any set $b\in M$ with $a\subset b$, then there is a top-extension $N$ of $M$ in which $\{x\mid\varphi(x)\}=b$.

An affirmative answer would be a very strong analogue with the universal algorithm and Woodin’s theorem about which I wrote previously. The idea is that the $\Sigma_2$ properties $\varphi(x)$ in set theory are analogous to the computably enumerable properties in computability theory. Namely, to verify that an object has a certain computably enumerable property, we run a particular computable process and then sit back, waiting for the process to halt, until a stage of computation arrives at which the property is verified. Similarly, in set theory, to verify that a set has a particular $\Sigma_2$ property, we sit back watching the construction of the cumulative set-theoretic universe, until a stage $V_\beta$ arrives that provides verification of the property. This is why in statement (3) we insist that $a\subset b$, since the $\Sigma_2$ properties are always upward absolute to top-extensions; once an object is placed into $\{x\mid\varphi(x)\}$, then it will never be removed as one makes the universe taller.

So the hope was that we would be able to find such a universal $\Sigma_2$ definition, which would serve as a set-theoretic analogue of the universal algorithm used in Woodin’s theorem.

If one drops the first requirement, and allows $\{x\mid \varphi(x)\}$ to sometimes be a proper class, then one can achieve a positive answer as follows.

Theorem. There is a $\Sigma_2$ formula $\varphi(x)$ with the following properties.

1. If the GCH holds, then $\{x\mid\varphi(x)\}$ is empty.
2. For any countable model $M\models\text{ZFC}$ where $a=\{x\mid \varphi(x)\}$ and any $b\in M$ with $a\subset b$, there is a top extension $N$ of $M$ in which $N\models\{x\mid\varphi(x)\}=b$.

Proof. Let $\varphi(x)$ assert that the set $x$ is coded into the GCH pattern. We may assume that the coding mechanism of a set is marked off by certain kinds of failures of the GCH at odd-indexed alephs, with the pattern at intervening even-indexed regular cardinals forming the coding pattern.  This is $\Sigma_2$, since any large enough $V_\theta$ will reveal whether a given set $x$ is coded in this way. And because of the manner of coding, if the GCH holds, then no set is coded. Also, if the GCH holds eventually, then only a set-sized collection is coded. Finally, any countable model $M$ where only a set is coded can be top-extended to another model $N$ in which any desired superset of that set is coded. QED

Update.  Originally, I had proposed an argument for a negative answer to the question, and I was actually a bit disappointed by that, since I had hoped for a positive answer. However, it now seems to me that the argument I had written is wrong, and I am grateful to Ali Enayat for his remarks on this in the comments. I have now deleted the incorrect argument.

Meanwhile, here is a positive answer to the question in the case of models of $V\neq\newcommand\HOD{\text{HOD}}\HOD$.

Theorem. There is a $\Sigma_2$ formula $\varphi(x)$ with the following properties:

1. In any model of $\newcommand\ZFC{\text{ZFC}}\ZFC+V\neq\HOD$, the class $\{x\mid\varphi(x)\}$ is a set.
2. It is relatively consistent with $\ZFC$ that $\{x\mid\varphi(x)\}$ is empty; indeed, in any model of $\ZFC+\newcommand\GCH{\text{GCH}}\GCH$, the class $\{x\mid\varphi(x)\}$ is empty.
3. If $M\models\ZFC$ thinks that $a=\{x\mid\varphi(x)\}$ is a set and $b\in M$ is a larger set $a\subset b$, then there is a top-extension $N$ of $M$ in which $\{x\mid \varphi(x)\}=b$.

Proof. Let $\varphi(x)$ hold, if there is some ordinal $\alpha$ such that every element of $V_\alpha$ is coded into the GCH pattern below some cardinal $\delta_\alpha$, with $\delta_\alpha$ as small as possible with that property, and $x$ is the next set coded into the GCH pattern above $\delta_\alpha$. This is a $\Sigma_2$ property, since it can be verified in any sufficiently large $V_\theta$.

In any model of $\ZFC+V\neq\HOD$, there must be some sets that are no coded into the $\GCH$ pattern, for if every set is coded that way then there would be a definable well-ordering of the universe and we would have $V=\HOD$. So in any model of $V\neq\HOD$, there is a bound on the ordinals $\alpha$ for which $\delta_\alpha$ exists, and therefore $\{x\mid\varphi(x)\}$ is a set. So statement (1) holds.

Statement (2) holds, because we may arrange it so that the GCH itself implies that no set is coded at all, and so $\varphi(x)$ would always fail.

For statement (3), suppose that $M\models\ZFC+\{x\mid\varphi(x)\}=a\subseteq b$ and $M$ is countable. In $M$, there must be some minimal rank $\alpha$ for which there is a set of rank $\alpha$ that is not coded into the GCH pattern. Let $N$ be an elementary top-extension of $M$, so $N$ agrees that $\alpha$ is that minimal rank. Now, by forcing over $N$, we can arrange to code all the sets of rank $\alpha$ into the GCH pattern above the height of the original model $M$, and we can furthermore arrange so as to code any given element of $b$ just above that coding. And so on, we can iterate it so as to arrange the coding above the height of $M$ so that exactly the elements of $b$ now satisfy $\varphi(x)$, but no more. In this way, we will ensure that $N\models\{x\mid\varphi(x)\}=b$, as desired. QED

I find the situation unusual, in that often results from the models-of-arithmetic context generalize to set theory with models of $V=\HOD$, because the global well-order means that models of $V=\HOD$ have definable Skolem functions, which is true in every model of arithmetic and which sometimes figures implicitly in constructions. But here, we have the result of Woodin’s theorem generalizing from models of arithmetic to models of $V\neq\HOD$.  Perhaps this suggests that we should expect a fully positive solution for models of set theory.

Further update. Woodin and I have now established the fully general result of the universal finite set, which subsumes much of the preliminary early analysis that I had earlier made in this post. Please see my post, The universal finite set.

# An equivalent formulation of the GCH

The continuum hypothesis CH is the assertion that the size of the power set of a countably infinite set $\aleph_0$ is the next larger cardinal $\aleph_1$, or in other words, that $2^{\aleph_0}=\aleph_1$. The generalized continuum hypothesis GCH makes this same assertion about all infinite cardinals, namely, that the power set of any infinite cardinal $\kappa$ is the successor cardinal $\kappa^+$, or in other words, $2^\kappa=\kappa^+$.

Yesterday I received an email from Geoffrey Caveney, who proposed to me the following axiom, which I have given a name.   First, for any set $F$ of cardinals, define the $F$-restricted power set operation $P_F(Y)=\{X\subseteq Y\mid |X|\in F\}$ to consist of the subsets of $Y$ having a cardinality allowed by $F$.  The only cardinals of $F$ that matter are those that are at most the cardinality of $Y$.

The Alternative GCH is the assertion that for every cardinal number $\kappa$, there is a set $F$ of cardinals such that the $F$-restricted power set $P_F(\kappa)$ has size $\kappa^+$.

Caveney was excited about his axiom for three reasons. First, a big part of his motivation for considering the axiom was the observation that the equation $2^\kappa=\kappa^+$ is simply not correct for finite cardinals $\kappa$ (other than $0$ and $1$) — and this is why the GCH makes the assertion only for infinite cardinals $\kappa$ — whereas the alternative GCH axiom makes a uniform statement for all cardinals, including the finite cardinals, and it gets the right answer for the finite cardinals. Specifically, for any natural number $n$, we can let $F=\{0,1\}$, and then note that $n$ has exactly $n+1$ many subsets of size in $F$. Second, Caveney had also observed that the GCH implies his axiom, since as we just mentioned, it is true for the finite cardinals and for infinite $\kappa$ we can take $F=\{\kappa\}$, using the fact that every infinite cardinal $\kappa$ has $2^\kappa$ many subsets of size $\kappa$ (we are working in ZFC). Third, Caveney had noticed that his axiom implies the continuum hypothesis, since in the case that $\kappa=\aleph_0$, there would be a family $F$ for which $P_F(\aleph_0)$ has size $\aleph_1$. But since there are only countably many finite subsets of $\aleph_0$, it follows that $F$ must include $\aleph_0$ itself, and so this would mean that $\aleph_0$ has only $\aleph_1$ many infinite subsets, and this implies CH.

To my way of thinking, the natural question to consider was whether Caveney’s axiom was actually weaker than GCH or not. At first I noticed that the axiom implies $2^{\aleph_1}=\aleph_2$ and similarly $2^{\aleph_n}=\aleph_{n+1}$, getting us up to $\aleph_\omega$. Then, after a bit I noticed that we can push the argument through all the way.

Theorem. The alternative GCH is equivalent to the GCH.

Proof. We’ve already argued for the converse implication, so it remains only to show that the alternative GCH implies the GCH. Assume that the alternative GCH holds.

We prove the GCH by transfinite induction. For the anchor case, we’ve shown already above that the GCH holds at $\aleph_0$, that is, that CH holds. For the successor case, assume that the GCH holds at some $\delta$, so that $2^\delta=\delta^+$, and consider the case $\kappa=\delta^+$. By the alternative GCH, there is a family $F$ of cardinals such that $|P_F(\kappa)|=\kappa^+$. If every cardinal in $F$ is less than $\kappa$, then $P_F(\kappa)$ has size at most $\kappa^{<\kappa}=(\delta^+)^\delta=2^\delta=\delta^+=\kappa$, which is too small. So $\kappa$ itself must be in $F$, and from this it follows that $\kappa$ has at most $\kappa^+$ many subsets of size $\kappa$, which implies $2^\kappa=\kappa^+$. So the GCH holds at $\kappa$, and we’ve handled the successor case. For the limit case, suppose that $\kappa$ is a limit cardinal and the GCH holds below $\kappa$. So $\kappa$ is a strong limit cardinal. By the alternative GCH, there is a family $F$ of cardinals for which $P_F(\kappa)=\kappa^+$. It cannot be that all cardinals in $F$ are less than the cofinality of $\kappa$, since in this case all the subsets of $\kappa$ in $P_F(\kappa)$ would be bounded in $\kappa$, and so it would have size at most $\kappa$, since $\kappa$ is a strong limit. So there must be a cardinal $\mu$ in $F$ with $\newcommand\cof{\text{cof}}\cof(\kappa)\leq\mu\leq\kappa$. But in this case, it follows that $\kappa^\mu=\kappa^+$, and this implies $\kappa^{\cof(\kappa)}=\kappa^+$, since by König’s theorem it is always at least $\kappa^+$, and it cannot be bigger if $\kappa^\mu=\kappa^+$. Finally, since $\kappa$ is a strong limit cardinal, it follows easily that $2^\kappa=\kappa^{\cof(\kappa)}$, since every subset of $\kappa$ is determined by it’s initial segments, and hence by a $\cof(\kappa)$-sequence of bounded subsets of $\kappa$, of which there are only $\kappa$ many. So we have established that $2^\kappa=\kappa^+$ in the limit case, completing the induction. So we get all instances of the GCH.
QED

# Embeddings of the universe into the constructible universe, current state of knowledge, CUNY Set Theory Seminar, March 2015

This will be a talk for the CUNY Set Theory Seminar, March 6, 2015.

I shall describe the current state of knowledge concerning the question of whether there can be an embedding of the set-theoretic universe into the constructible universe.

Question.(Hamkins) Can there be an embedding $j:V\to L$ of the set-theoretic universe $V$ into the constructible universe $L$, when $V\neq L$?

The notion of embedding here is merely that $$x\in y\iff j(x)\in j(y),$$ and such a map need not be elementary nor even $\Delta_0$-elementary. It is not difficult to see that there can generally be no $\Delta_0$-elementary embedding $j:V\to L$, when $V\neq L$.

Nevertheless, the question arises very naturally in the context of my previous work on the embeddability phenomenon, Every countable model of set theory embeds into its own constructible universe, where the title theorem is the following.

Theorem.(Hamkins) Every countable model of set theory $\langle M,\in^M\rangle$, including every countable transitive model of set theory, has an embedding $j:\langle M,\in^M\rangle\to\langle L^M,\in^M\rangle$ into its own constructible universe.

The methods of proof also established that the countable models of set theory are linearly pre-ordered by embeddability: given any two models, one of them embeds into the other; or equivalently, one of them is isomorphic to a submodel of the other. Indeed, one model $\langle M,\in^M\rangle$ embeds into another $\langle N,\in^N\rangle$ just in case the ordinals of the first $\text{Ord}^M$ order-embed into the ordinals of the second $\text{Ord}^N$. (And this implies the theorem above.)

In the proof of that theorem, the embeddings $j:M\to L^M$ are defined completely externally to $M$, and so it was natural to wonder to what extent such an embedding might be accessible inside $M$. And I realized that I could not generally refute the possibility that such a $j$ might even be a class in $M$.

Currently, the question remains open, but we have some partial progress, and have settled it in a number of cases, including the following, on which I’ll speak:

• If there is an embedding $j:V\to L$, then for a proper class club of cardinals $\lambda$, we have $(2^\lambda)^V=(\lambda^+)^L$.
• If $0^\sharp$ exists, then there is no embedding $j:V\to L$.
• If $0^\sharp$ exists, then there is no embedding $j:V\to L$ and indeed no embedding $j:P(\omega)\to L$.
• If there is an embedding $j:V\to L$, then the GCH holds above $\aleph_0$.
• In the forcing extension $V[G]$ obtained by adding $\omega_1$ many Cohen reals (or more), there is no embedding $j:V[G]\to L$, and indeed, no $j:P(\omega)^{V[G]}\to V$. More generally, after adding $\kappa^+$ many Cohen subsets to $\kappa$, for any regular cardinal $\kappa$, then in $V[G]$ there is no $j:P(\kappa)\to V$.
• If $V$ is a nontrivial set-forcing extension of an inner model $M$, then there is no embedding $j:V\to M$. Indeed, there is no embedding $j:P(\kappa^+)\to M$, if the forcing has size $\kappa$. In particular, if $V$ is a nontrivial forcing extension, then there is no embedding $j:V\to L$.
• Every countable set $A$ has an embedding $j:A\to L$.

This is joint work of myself, W. Hugh Woodin, Menachem Magidor, with contributions also by David Aspero, Ralf Schindler and Yair Hayut.

See my related MathOverflow question: Can there be an embedding $j:V\to L$ from the set-theoretic universe $V$ to the constructible universe $L$, when $V\neq L$?

Talk Abstract

# Brent Cody

Brent Cody earned his Ph.D. under my supervision at the CUNY Graduate Center in June, 2012.  Brent’s dissertation work began with the question of finding the exact consistency strength of the GCH failing at a cardinal $\theta$, when $\kappa$ is $\theta$-supercompact.  The answer turned out to be a $\theta$-supercompact cardinal that was also $\theta^{++}$-tall.  After this, he quickly dispatched more general instances of what he termed the Levinski property for a variety of other large cardinals, advancing his work towards a general investigation of the Easton theorem phenomenon in the large cardinal context, which he is now undertaking.  Brent held a post-doctoral position at the Fields Institute in Toronto, afterwards taking up a position at the University of Prince Edward Island.  He is now at Virginia Commonwealth University.

Brent Cody

Brent Cody, “Some Results on Large Cardinals and the Continuum Function,” Ph.D. dissertation for The Graduate Center of the City University of New York, June, 2012.

Abstract.  Given a Woodin cardinal $\delta$, I show that if $F$ is any Easton function with $F”\delta\subseteq\delta$ and GCH holds, then there is a cofinality preserving forcing extension in which $2^\gamma= F(\gamma)$ for each regular cardinal $\gamma<\delta$, and in which $\delta$ remains Woodin.

I also present a new example in which forcing a certain behavior of the continuum function on the regular cardinals, while preserving a given large cardinal, requires large cardinal strength beyond that of the original large cardinal under consideration. Specifically, I prove that the existence of a $\lambda$-supercompact cardinal $\kappa$ such that GCH fails at $\lambda$ is equiconsistent with the existence of a cardinal $\kappa$ that is $\lambda$-supercompact and $\lambda^{++}$-tall.

I generalize a theorem on measurable cardinals due to Levinski, which says that given a measurable cardinal, there is a forcing extension preserving the measurability of $\kappa$ in which $\kappa$ is the least regular cardinal at which GCH holds. Indeed, I show that Levinski’s result can be extended to many other large cardinal contexts. This work paves the way for many additional results, analogous to the results stated above for Woodin cardinals and partially supercompact cardinals.