When does every definable nonempty set have a definable element?

  • F. G. Dorais and J. D. Hamkins, “When does every definable nonempty set have a definable element?,” , 2017. (manuscript under review)  
    @ARTICLE{DoraisHamkins:When-does-every-definable-nonempty-set-have-a-definable-element,
    author = {Fran\c{c}ois G. Dorais and Joel David Hamkins},
    title = {When does every definable nonempty set have a definable element?},
    journal = {},
    year = {2017},
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    pages = {},
    month = {},
    note = {manuscript under review},
    abstract = {},
    keywords = {under-review},
    source = {},
    doi = {},
    eprint = {1706.07285},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    url = {http://jdh.hamkins.org/definable-sets-with-definable-elements},
    }

Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.

It is not difficult to see that the models of ZF set theory in which every definable nonempty set has a definable element are precisely the models of $V=\HOD$. Namely, if $V=\HOD$, then there is a definable well-ordering of the universe, and so the $\HOD$-least element of any definable nonempty set is definable; and conversely, if $V\neq\HOD$, then the set of minimal-rank non-OD sets is definable, but can have no definable element.

In this brief article, we shall identify the limit of this elementary observation in terms of the complexity of the definitions. Specifically, we shall prove that $V=\HOD$ is equivalent to the assertion that every $\Pi_2$-definable nonempty set contains an ordinal-definable element, but that one may not replace $\Pi_2$-definability here by $\Sigma_2$-definability.

Theorem. The following are equivalent in any model $M$ of ZF:

  1. $M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.
  2. $M$ thinks there is a definable well-ordering of the universe.
  3. Every definable nonempty set in $M$ has a definable element.
  4. Every definable nonempty set in $M$ has an ordinal-definable element.
  5. Every ordinal-definable nonempty set in $M$ has an ordinal-definable element.
  6. Every $\Pi_2$-definable nonempty set in $M$ has an ordinal-definable element.

Theorem. Every model of ZFC has a forcing extension satisfying $V\neq\HOD$, in which every $\Sigma_2$-definable set has a definable element.

The proof of this latter theorem is reminiscent of several proofs of the maximality principle (see A simple maximality principle), where one undertakes a forcing iteration attempting at each stage to force and then preserve a given $\Sigma_2$ assertion.

This inquiry grew out of a series of questions and answers posted on MathOverflow and the exchange of the authors there.

The definable cut of a model of set theory can be changed by small forcing

Cupid carving his bow -- ParmigianinoIf $M$ is a model of ZFC set theory, let $I$ be the definable cut of its ordinals, the collection of ordinals that are below an ordinal $\delta$ of $M$ that is definable in $M$ without parameters. This would include all the ordinals of $M$, if the definable ordinals happen to be unbounded in $M$, but one can also construct examples where the definable cut is bounded in $M$.  Let $M_I$ be the corresponding definable cut of $M$ itself, the rank-initial segment of $M$ determined by $I$, or in other words, the collection of all sets $x$ in $M$ of rank below a definable ordinal of $M$. Equivalently, $$M_I=\bigcup_{\delta\in I} V_\delta^M.$$ It is not difficult to see that this is an elementary substructure $M_I\prec M$, because we can verify the Tarski-Vaught criterion as follows. If $M\models\exists y\ \varphi(x,y)$, where $x\in M_I$, then let $\delta$ be a definable ordinal above the rank of $x$. In this case, the ordinal $\theta$, which is the supremum over all $a\in V_\delta$ of the minimal rank of a set $y$ for which $\varphi(a,y)$, if there is such a $y$. This supremum $\theta$ is definable, and so since $x\in V_\delta$, the minimal rank of a $y$ such that $\varphi(x,y)$ is at most $\theta$. Consequently, since $\theta\in I$, such a $y$ can be found in $M_I$. So we have found the desired witness inside the substructure, and so it is elementary $M_I\prec M$. Note that in the general case, one does not necessarily know that $I$ has a least upper bound in $M$. Under suitable assumptions, it can happen that $I$ is unbounded in $M$, that $I$ is an ordinal of $M$, or that $I$ is bounded in $M$, but has no least upper bound.

What I am interested in for this post is how the definable cut might be affected by forcing. Of course, it is easy to see that if $M$ is definable in $M[G]$, then the definable cut of $M[G]$ is at least as high as the definable cut of $M$, simply because the definable ordinals of $M$ remain definable in $M[G]$.

A second easy observation is that if the definable cut of $M$ is bounded in $M$, then we could perform large collapse forcing, collapsing a cardinal above $I$ to $\omega$, which would of course make every cardinal of $I$ countable in the extension $M[G]$. In this case, since $\omega_1^{M[G]}$ is definable, it would change the definable cut. So this kind of very large forcing can change the definable cut, making it larger.

But what about small forcing? Suppose that the forcing notion $\newcommand\P{\mathbb{P}}\P$ we intend to forcing with is small in the sense that it is in the definable cut $M_I$. This would be true if $\P$ itself were definable, for example, but really we only require that $\P$ has rank less than some definable ordinal of $M$. Can this forcing change the definable cut?

Let me show at least that the definable cut can never go up after small forcing.

Theorem. If $G\subset\P$ is $M$-generic for forcing $\P$ in the definable cut of $M$, then the definable cut of $M[G]$ is below or the same in the ordinals as it was in $M$.

Proof. Suppose that $G\subset\P$ is $M$-generic, and we consider the forcing extension $M[G]$. We have already proved that $M_I\prec M$ is an elementary submodel. I claim that this relation lifts to the forcing extension $M_I[G]\prec M[G]$. Note first that since $\P\in M_I$ and $M_I$ is a rank initial segment of $M$, it follows that $M_I$ has all the subsets of $\P$ in $M$, and so $G$ is $M_I$-generic. So the extension $M_I[G]$ makes sense. Next, suppose that $M[G]\models\varphi(a)$ for some $a\in M_I[G]$. If $\dot a$ is a name for $a$ in $M_I$, then there is some condition $p\in G$ forcing $\varphi(\dot a)$ over $M$. Since $M_I\prec M$, this is also forced by $p$ over $M_I$, and thus $M_I[G]\models\varphi(a)$ as well, as desired. So $M_I[G]\prec M[G]$, and from this it follows that every definable ordinal of $M[G]$ is in the cut $I$. So the definable cut did not get higher. QED

But can it go down? Not if the model $M$ is definable in $M[G]$, by our earlier easy observation. Consequently,

Theorem. If $M$ is definable in $M[G]$, where $G\subset\P$ is $M$-generic for forcing $\P$ below the definable cut of $M$, then the definable cut of $M[G]$ is the same as the definable cut of $M$.

Proof. It didn’t go down, since $M$ is definable in $M[G]$; and it didn’t go up, since $\P$ was small. QED

What if $M$ is not definable in $M[G]$? Can we make the definable cut go down after small forcing? The answer is yes.

Theorem. If ZFC is consistent, then there is a model $M\models\text{ZFC}$ with a definable notion of forcing $\P$ (hence in the definable cut of $M$), such that if $G\subset\P$ is $M$-generic, then the definable cut of the forcing extension $M[G]$ is strictly shorter than the definable cut of $M[G]$.

Proof. Start with a model of $\text{ZFC}+V=L$, whose definable ordinals are bounded by a cardinal $\delta$. Let’s call it $L$, and let $I$ be the definable cut of $L$, which we assume is bounded by $\delta$. Let $M=L[G]$ be the forcing extension of $L$ obtained by performing an Easton product, adding a Cohen subset to every regular cardinal above $\delta$ in $L$. Since this forcing adds no sets below $\delta$, but adds a Cohen set at $\delta^+$, it follows that $\delta$ becomes definable in $L[G]$. In fact, since the forcing is homogeneous and definable from $\delta$, it follows that the definable ordinals of $L[G]$ are precisely the ordinals that are definable in $L$ with parameter $\delta$. These may be bounded or unbounded in $L[G]$. Now, let $\newcommand\Q{\mathbb{Q}}\Q$ be the Easton product forcing at the stages below $\delta$, and suppose that $G\subset\Q$ is $L[G]$-generic. Consider the model $L[G][H]$. Note that the forcing $\Q$ is definable in $L[G]$, since $\delta$ is definable there. This two-step forcing can be combined into one giant Easton product in $L$, the product that simply forces to add a Cohen subset to every regular cardinal. Since this version of the forcing is homogeneous and definable in $L$, it follows that the definable ordinals of $L[G][H]$ are precisely the definable ordinals of $L$, which are bounded by $I$. In summary, the definable cut of $L[G]$ is strictly above $\delta$, since $\delta$ is definable in $L[G]$, and the forcing $\Q$ has size and rank $\delta$; but the forcing extension $L[G][H]$ has definable cut $I$, which is strictly bounded by $\delta$. So the definable cut was made smaller by small forcing, as claimed. QED

This post is an account of some ideas that Alexander Block and I had noted today during the course of our mathematical investigation of another matter.

Being HOD-of-a-set is invariant throughout the generic multiverse

Iowa State Capitol - Law Library _ Flickr - Photo Sharing!$\newcommand\HOD{\text{HOD}}$The axiom $V=\HOD$, introduced by Gödel, asserts that every set is ordinal definable. This axiom has a subtler foundational aspect than might at first be expected. The reason is that the general concept of “object $x$ is definable using parameter $p$” is not in general first-order expressible in set theory; it is of course a second-order property, which makes sense only relative to a truth predicate, and by Tarski’s theorem, we can have no first-order definable truth predicate. Thus, the phrase “definable using ordinal parameters” is not directly meaningful in the first-order language of set theory without further qualification or explanation. Fortunately, however, it is a remarkable fact that when we allow definitions to use arbitrary ordinal parameters, as we do with $\HOD$, then we can in fact make such qualifications in such a way that the axiom becomes first-order expressible in set theory. Specifically, we say officially that $V=\HOD$ holds, if for every set $x$, there is an ordinal $\theta$ with $x\in V_\theta$, for which which $x$ is definable by some formula $\psi(x)$ in the structure $\langle V_\theta,{\in}\rangle$ using ordinal parameters. Since $V_\theta$ is a set, we may freely make reference to first-order truth in $V_\theta$ without requiring any truth predicate in $V$. Certainly any such $x$ as this is also ordinal-definable in $V$, since we may use $\theta$ and the Gödel-code of $\psi$ also as parameters, and note that $x$ is the unique object such that it is in $V_\theta$ and satisfies $\psi$ in $V_\theta$. (Note that inside an $\omega$-nonstandard model of set theory, we may really need to use $\psi$ as a parameter, since it may be nonstandard, and $x$ may not be definable in $V_\theta$ using a meta-theoretically standard natural number; but fortunately, the Gödel code of a formula is an integer, which is still an ordinal, and this issue is the key to the issue.) Conversely, if $x$ is definable in $V$ using formula $\varphi(x,\vec\alpha)$ with ordinal parameters $\vec\alpha$, then it follows by the reflection theorem that $x$ is defined by $\varphi(x,\vec\alpha)$ inside some $V_\theta$. So this formulation of $V=HOD$ is expressible and exactly captures the desired second-order property that every set is ordinal-definable.

Consider next the axiom $V=\HOD(b)$, asserting that every set is definable from ordinal parameters and parameter $b$. Officially, as before, $V=\HOD(b)$ asserts that for every $x$, there is an ordinal $\theta$, formula $\psi$ and ordinals $\vec \alpha<\theta$, such that $x$ is the unique object in $V_\theta$ for which $\langle V_\theta,{\in}\rangle\models\psi(x,\vec\alpha,b)$, and the reflection argument shows again that this way of defining the axiom exactly captures the intended idea.

The axiom I actually want to focus on is $\exists b\,\left( V=\HOD(b)\right)$, asserting that the universe is $\HOD$ of a set. (I assume ZFC in the background theory.) It turns out that this axiom is constant throughout the generic multiverse.

Theorem. The assertion $\exists b\, (V=\HOD(b))$ is forcing invariant.

  • If it holds in $V$, then it continues to hold in every set forcing extension of $V$.
  • If it holds in $V$, then it holds in every ground of $V$.

Thus, the truth of this axiom is invariant throughout the generic multiverse.

Proof. Suppose that $\text{ZFC}+V=\HOD(b)$, and $V[G]$ is a forcing extension of $V$ by generic filter $G\subset\mathbb{P}\in V$. By the ground-model definability theorem, it follows that $V$ is definable in $V[G]$ from parameter $P(\mathbb{P})^V$. Thus, using this parameter, as well as $b$ and additional ordinal parameters, we can define in $V[G]$ any particular object in $V$. Since this includes all the $\mathbb{P}$-names used to form $V[G]$, it follows that $V[G]=\HOD(b,P(\mathbb{P})^V,G)$, and so $V[G]$ is $\HOD$ of a set, as desired.

Conversely, suppose that $W$ is a ground of $V$, so that $V=W[G]$ for some $W$-generic filter $G\subset\mathbb{P}\in W$, and $V=\HOD(b)$ for some set $b$. Let $\dot b$ be a name for which $\dot b_G=b$. Every object $x\in W$ is definable in $W[G]$ from $b$ and ordinal parameters $\vec\alpha$, so there is some formula $\psi$ for which $x$ is unique such that $\psi(x,b,\vec\alpha)$. Thus, there is some condition $p\in\mathbb{P}$ such that $x$ is unique such that $p\Vdash\psi(\check x,\dot b,\check{\vec\alpha})$. If $\langle p_\beta\mid\beta<|\mathbb{P}|\rangle$ is a fixed enumeration of $\mathbb{P}$ in $W$, then $p=p_\beta$ for some ordinal $\beta$, and we may therefore define $x$ in $W$ using ordinal parameters, along with $\dot b$ and the fixed enumeration of $\mathbb{P}$. So $W$ thinks the universe is $\HOD$ of a set, as desired.

Since the generic multiverse is obtained by iteratively moving to forcing extensions to grounds, and each such movement preserves the axiom, it follows that $\exists b\, (V=\HOD(b))$ is constant throughout the generic multiverse. QED

Theorem. If $V=\HOD(b)$, then there is a forcing extension $V[G]$ in which $V=\HOD$ holds.

Proof. We are working in ZFC. Suppose that $V=\HOD(b)$. We may assume $b$ is a set of ordinals, since such sets can code any given set. Consider the following forcing iteration: first add a Cohen real $c$, and then perform forcing $G$ that codes $c$, $P(\omega)^V$ and $b$ into the GCH pattern at uncountable cardinals, and then perform self-encoding forcing $H$ above that coding, coding also $G$ (see my paper on Set-theoretic geology for further details on self-encoding forcing). In the final model $V[c][G][H]$, therefore, the objects $c$, $b$, $P(\omega)^V$, $G$ and $H$ are all definable without parameters. Since $V\subset V[c][G][H]$ has a closure point at $\omega$, it satisfies the $\omega_1$-approximation and cover properties, and therefore the class $V$ is definable in $V[c][G][H]$ using $P(\omega)^V$ as a parameter. Since this parameter is itself definable without parameters, it follows that $V$ is parameter-free definable in $V[c][G][H]$. Since $b$ is also definable there, it follows that every element of $\HOD(b)^V=V$ is ordinal-definable in $V[c][G][H]$. And since $c$, $G$ and $H$ are also definable without parameters, we have $V[c][G][H]\models V=\HOD$, as desired. QED

Corollary. The following are equivalent.

  1. The universe is $\HOD$ of a set: $\exists b\, (V=\HOD(b))$.
  2. Somewhere in the generic multiverse, the universe is $\HOD$ of a set.
  3. Somewhere in the generic multiverse, the axiom $V=\HOD$ holds.
  4. The axiom $V=\HOD$ is forceable.

Proof. This is an immediate consequence of the previous theorems. $1\to 4\to 3\to 2\to 1$. QED

Corollary. The axiom $V=\HOD$, if true, even if true anywhere in the generic multiverse, is a switch.

Proof. A switch is a statement such that both it and its negation are necessarily possible by forcing; that is, in every set forcing extension, one can force the statement to be true and also force it to be false. We can always force $V=\HOD$ to fail, simply by adding a Cohen real. If $V=\HOD$ is true, then by the first theorem, every forcing extension has $V=\HOD(b)$ for some $b$, in which case $V=\HOD$ remains forceable, by the second theorem. QED

Different models of set theory with the same subset relation

OkonomiyakiRecently Makoto Kikuchi (Kobe University) asked me the following interesting question, which arises very naturally if one should adopt a mereological perspective in the foundations of mathematics, placing a focus on the parthood relation rather than the element-of relation. In set theory, this perspective would lead one to view the subset or inclusion relation $\subseteq$ as the primary fundamental relation, rather than the membership $\in$ relation.

Question. Can there be two different models of set theory, with the same inclusion relation?

We spent an evening discussing it, over delicious (Rokko-michi-style) okonomiyaki and bi-ru, just like old times, except that we are in Tokyo at the CTFM 2015, and I’d like to explain the answer, which is yes, this always happens in every model of set theory.

Theorem. In any universe of set theory $\langle V,\in\rangle$, there is a definable relation $\in^*$, different from $\in$, such that $\langle V,\in^*\rangle$ is a model of set theory, in fact isomorphic to the original universe $\langle V,\in\rangle$, for which the corresponding inclusion relation $$u\subseteq^* v\iff \forall a\, (a\in^* u\to a\in^* v)$$ is identical to the usual inclusion relation $u\subseteq v$.

Proof. Let $\theta:V\to V$ be any definable non-identity permutation of the universe, and let $\tau:u\mapsto \theta[u]=\{\ \theta(a)\mid a\in u\ \}$ be the function determined by pointwise image under $\theta$. Since $\theta$ is bijective, it follows that $\tau$ is also a bijection of $V$ to $V$, since every set is the $\theta$-image of a unique set. Furthermore, $\tau$ is an automorphism of $\langle V,\subseteq\rangle$, since $$u\subseteq v\iff\theta[u]\subseteq\theta[v]\iff\tau(u) \subseteq\tau(v).$$ I had used this idea a few years ago in my answer to the MathOverflow question, Is the inclusion version of Kunen inconsistency theorem true?, which shows that there are nontrivial $\subseteq$ automorphisms of the universe. Note that since $\tau(\{a\})=\{\theta(a)\}$, it follows that any instance of nontriviality $\theta(a)\neq a$ in $\theta$ leads immediately to an instance of nontriviality in $\tau$.

Using the map $\tau$, define $a\in^* b\iff\tau(a)\in\tau(b)$. By definition, therefore, $\tau$ is an isomorphism of $\langle V,\in^*\rangle\cong\langle V,\in\rangle$. Let us show that $\in^*\neq \in$. Since $\theta$ is nontrivial, there is an $\in$-minimal set $a$ with $\theta(a)\neq a$. By minimality, $\theta[a]=a$ and so $\tau(a)=a$. But as mentioned, $\tau(\{a\})=\{\theta(a)\}\neq\{a\}$. So we have $a\in\{a\}$, but $\tau(a)=a\notin\{\theta(a)\}=\tau(\{a\})$ and hence $a\notin^*\{a\}$. So the two relations are different.

Meanwhile, consider the corresponding subset relation. Specifically, $u\subseteq^* v$ is defined to mean $\forall a\,(a\in^* u\to a\in^* v)$, which holds if and only if $\forall a\, (\tau(a)\in\tau(u)\to \tau(a)\in\tau(v))$; but since $\tau$ is surjective, this holds if and only if $\tau(u)\subseteq \tau(v)$, which as we observed at the beginning of the proof, holds if and only if $u\subseteq v$. So the corresponding subset relations $\subseteq^*$ and $\subseteq$ are identical, as desired.

Another way to express what is going on is that $\tau$ is an isomorphism of the structure $\langle V,{\in^*},{\subseteq}\rangle$ with $\langle V,{\in},{\subseteq}\rangle$, and so $\subseteq$ is in fact that same as the corresponding inclusion relation $\subseteq^*$ that one would define from $\in^*$. QED

Corollary. One cannot define $\in$ from $\subseteq$ in a model of set theory.

Proof. The map $\tau$ is a $\subseteq$-automorphism, and so it preserves every relation definable from $\subseteq$, but it does not preserve $\in$. QED

Nevertheless, I claim that the isomorphism type of $\langle V,\in\rangle$ is implicit in the inclusion relation $\subseteq$, in the sense that any other class relation $\in^*$ having that same inclusion relation is isomorphic to the $\in$ relation.

Theorem. Assume ZFC in the universe $\langle V,\in\rangle$. Suppose that $\in^*$ is a class relation for which $\langle V,\in^*\rangle$ is a model of set theory (a weak set theory suffices), such that the corresponding inclusion relation $$u\subseteq^* v\iff\forall a\,(a\in^* u\to a\in^* v)$$is the same as the usual inclusion relation $u\subseteq v$. Then the two membership relations are isomorphic $$\langle V,\in\rangle\cong\langle V,\in^*\rangle.$$

Proof. Since the singleton set $\{a\}$ has exactly two subsets with respect to the usual $\subseteq$ relation — the empty set and itself — this must also be true with respect to the inclusion relation $\subseteq^*$ defined via $\in^*$, since we have assumed $\subseteq^*=\subseteq$. Thus, the object $\{a\}$ is also a singleton with respect to $\in^*$, and so there is a unique object $\eta(a)$ such that $x\in^* a\iff x=\eta(a)$. By extensionality and since every object has its singleton, it follows that $\eta:V\to V$ is both one-to-one and onto. Let $\theta=\eta^{-1}$ be the inverse permutation.

Observe that $a\in u\iff \{a\}\subseteq u\iff \{a\}\subseteq^* u\iff\eta(a)\in^* u$. Thus, $$b\in^* u\iff \theta(b)\in u.$$

Using $\in$-recursion, define $b^*=\{\ \theta(a^*)\mid a\in b\ \}$. The map $b\mapsto b^*$ is one-to-one by $\in$-recursion, since if there is no violation of this for the elements of $b$, then we may recover $b$ from $b^*$ by applying $\theta^{-1}$ to the elements of $b^*$ and then using the induction assumption to find the unique $a$ from $a^*$ for each $\theta(a^*)\in b^*$, thereby recovering $b$. So $b\mapsto b^*$ is injective.

I claim that this map is also surjective. If $y_0\neq b^*$ for any $b$, then there must be an element of $y_0$ that is not of the form $\theta(b^*)$ for any $b$. Since $\theta$ is surjective, this means there is $\theta(y_1)\in y_0$ with $y_1\neq b^*$ for any $b$. Continuing, there is $y_{n+1}$ with $\theta(y_{n+1})\in y_n$ and $y_{n+1}\neq b^*$ for any $b$. Let $z=\{\ \theta(y_n)\mid n\in\omega\ \}$. Since $x\in^* u\iff \theta(x)\in u$, it follows that the $\in^*$-elements of $z$ are precisely the $y_n$’s. But $\theta(y_{n+1})\in y_n$, and so $y_{n+1}\in^* y_n$. So $z$ has no $\in^*$-minimal element, violating the axiom of foundation for $\in^*$, a contradiction. So the map $b\mapsto b^*$ is a bijection of $V$ with $V$.

Finally, we observe that because $$a\in b\iff\theta(a^*)\in b^*\iff a^*\in^* b^*,$$ it follows that the map $b\mapsto b^*$ is an isomorphism of $\langle V,\in\rangle$ with $\langle V,\in^*\rangle$, as desired. QED

The conclusion is that although $\in$ is not definable from $\subseteq$, nevertheless, the isomorphism type of $\in$ is implicit in $\subseteq$, in the sense that any other class relation $\in^*$ giving rise to the same inclusion relation $\subseteq^*=\subseteq$ is isomorphic to $\in$.

Meanwhile, I do not yet know what the situation is when one drops the assumption that $\in^*$ is a class with respect to the $\langle V,\in\rangle$ universe.

Question. Can there be two models of set theory $\langle M,\in\rangle$ and $\langle M,\in^*\rangle$, not necessarily classes with respect to each other, which have the same inclusion relation $\subseteq=\subseteq^*$, but which are not isomorphic?

(This question is now answered! See my joint paper with Kikuchi at Set-theoretic mereology.)

Ehrenfeucht's lemma in set theory

  • G. Fuchs, V. Gitman, and J. D. Hamkins, “Ehrenfeucht’s lemma in set theory,” to appear in Notre Dame Journal of Formal Logic, 2015.  
    @ARTICLE{FuchsGitmanHamkins:EhrenfeuchtsLemmaInSetTheory,
    author = {Gunter Fuchs and Victoria Gitman and Joel David Hamkins},
    title = {Ehrenfeucht's lemma in set theory},
    journal = {to appear in Notre Dame Journal of Formal Logic},
    year = {2015},
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    eprint = {1501.01918},
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This is joint work with Gunter Fuchs and Victoria Gitman. $\newcommand\HOD{\text{HOD}}\newcommand\Ehrenfeucht{\text{EL}}$

Abstract. Ehrenfeucht’s lemma asserts that whenever one element of a model of Peano arithmetic is definable from another, then they satisfy different types. We consider here the analogue of Ehrenfeucht’s lemma for models of set theory. The original argument applies directly to the ordinal-definable elements of any model of set theory, and in particular, Ehrenfeucht’s lemma holds fully for models of set theory satisfying $V=\HOD$. We show that the lemma can fail, however, in models of set theory with $V\neq\HOD$, and it necessarily fails in the forcing extension to add a generic Cohen real. We go on to formulate a scheme of natural parametric generalizations of Ehrenfeucht’s lemma, namely, the principles of the form $\Ehrenfeucht(A,P,Q)$, which asserts that whenever an object $b$ is definable in $M$ from some $a\in A$ using parameters in $P$, with $b\neq a$, then the types of $a$ and $b$ over $Q$ in $M$ are different. We also consider various analogues of Ehrenfeucht’s lemma obtained by using algebraicity in place of definability, where a set $b$ is \emph{algebraic} in $a$ if it is a member of a finite set definable from $a$ (as in J. D. Hamkins and C. Leahy, Algebraicity and implicit definability in set theory). Ehrenfeucht’s lemma holds for the ordinal-algebraic sets, we prove, if and only if the ordinal-algebraic and ordinal-definable sets coincide. Using similar analysis, we answer two open questions posed in my paper with Leahy, by showing that (i) algebraicity and definability need not coincide in models of set theory and (ii) the internal and external notions of being ordinal algebraic need not coincide.

When does every definable set have a definable member? CUNY Set Theory Seminar, October 2014

This will be a talk for the CUNY set theory seminar, October 10, 2014, 12pm  GC 6417.

Abstract. Although the concept of `being definable’ is not generally expressible in the language of set theory, it turns out that the models of ZF in which every definable nonempty set has a definable element are precisely the models of V=HOD.  Indeed, V=HOD is equivalent to the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, this is not true in the case of $\Sigma_2$-definability, because every model of ZFC has a forcing extension satisfying $V\neq\text{HOD}$ in which every $\Sigma_2$-definable set has an ordinal-definable element.

This is joint work with François G. Dorais and Emil Jeřábek, growing out of some questions and answers on MathOverflow, namely,

Definable collections without definable members
A question asked by Ashutosh five years ago, in which François and I gradually came upon the answer together.
Is it consistent that every definable set has a definable member?
A similar question asked last week by (anonymous) user38200
Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable member?
A question I had regarding the limits of an issue in my answer to the previous question.

In this talk, I shall present the answers to all these questions and place the results in the context of classical results on definability, including a review of basic concepts for graduate students.

Large cardinals need not be large in HOD

  • Y. Cheng, S. Friedman, and J. D. Hamkins, “Large cardinals need not be large in HOD,” Annals of Pure and Applied Logic, vol. 166, iss. 11, pp. 1186-1198, 2015.  
    @ARTICLE{ChengFriedmanHamkins2015:LargeCardinalsNeedNotBeLargeInHOD,
    title = "Large cardinals need not be large in {HOD} ",
    journal = "Annals of Pure and Applied Logic ",
    volume = "166",
    number = "11",
    pages = "1186 - 1198",
    year = "2015",
    note = "",
    issn = "0168-0072",
    doi = "10.1016/j.apal.2015.07.004",
    eprint = {1407.6335},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    url = {http://jdh.hamkins.org/large-cardinals-need-not-be-large-in-hod},
    author = "Yong Cheng and Sy-David Friedman and Joel David Hamkins",
    keywords = "Large cardinals",
    keywords = "HOD",
    keywords = "Forcing",
    keywords = "Absoluteness ",
    abstract = "Abstract We prove that large cardinals need not generally exhibit their large cardinal nature in HOD. For example, a supercompact cardinal κ need not be weakly compact in HOD, and there can be a proper class of supercompact cardinals in V, none of them weakly compact in HOD, with no supercompact cardinals in HOD. Similar results hold for many other types of large cardinals, such as measurable and strong cardinals.",
    }

Abstract. We prove that large cardinals need not generally exhibit their large cardinal nature in HOD. For example, a supercompact cardinal $\kappa$ need not be weakly compact in HOD, and there can be a proper class of supercompact cardinals in $V$, none of them weakly compact in HOD, with no supercompact cardinals in HOD. Similar results hold for many other types of large cardinals, such as measurable and strong cardinals.

In this article, we prove that large cardinals need not generally exhibit their large cardinal nature in HOD, the inner model of hereditarily ordinal-definable sets, and there can be a divergence in strength between the large cardinals of the ambient set-theoretic universe $V$ and those of HOD. Our general theme concerns the questions:

Questions.

1. To what extent must a large cardinal in $V$ exhibit its large cardinal properties in HOD?

2. To what extent does the existence of large cardinals in $V$ imply the existence of large cardinals in HOD?

For large cardinal concepts beyond the weakest notions, we prove, the answers are generally negative. In Theorem 4, for example, we construct a model with a supercompact cardinal that is not weakly compact in HOD, and Theorem 9 extends this to a proper class of supercompact cardinals, none of which is weakly compact in HOD, thereby providing some strongly negative instances of (1). The same model has a proper class of supercompact cardinals, but no supercompact cardinals in HOD, providing a negative instance of (2). The natural common strengthening of these situations would be a model with a proper class of supercompact cardinals, but no weakly compact cardinals in HOD. We were not able to arrange that situation, however, and furthermore it would be ruled out by Conjecture 13, an intriguing positive instance of (2) recently proposed by W. Hugh Woodin, namely, that if there is a supercompact cardinal, then there is a measurable cardinal in HOD. Many other natural possibilities, such as a proper class of measurable cardinals with no weakly compact cardinals in HOD, remain as open questions.

CUNY talkRutgers talk | Luminy talk

Local properties in set theory

V_thetaSet-theoretic arguments often make use of the fact that a particular property $\varphi$ is local, in the sense that instances of the property can be verified by checking certain facts in only a bounded part of the set-theoretic universe, such as inside some rank-initial segment $V_\theta$ or inside the collection $H_\kappa$ of all sets of hereditary size less than $\kappa$. It turns out that this concept is exactly equivalent to the property being $\Sigma_2$ expressible in the language of set theory.

Theorem. For any assertion $\varphi$ in the language of set theory, the following are equivalent:

  1. $\varphi$ is ZFC-provably equivalent to a $\Sigma_2$ assertion.
  2. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \theta\, V_\theta\models\psi$,” where $\psi$ is a statement of any complexity.
  3. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \kappa\, H_\kappa\models\psi$,” where $\psi$ is a statement of any complexity.

Just to clarify, the $\Sigma_2$ assertions in set theory are those of the form $\exists x\,\forall y\,\varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. The set $V_\theta$ refers to the rank-initial segment of the set-theoretic universe, consisting of all sets of von Neumann rank less than $\theta$. The set $H_\kappa$ consists of all sets of hereditary size less than $\kappa$, that is, whose transitive closure has size less than $\kappa$.

Proof. ($3\to 2$) Since $H_\kappa$ is correctly computed inside $V_\theta$ for any $\theta>\kappa$, it follows that to assert that some $H_\kappa$ satisfies $\psi$ is the same as to assert that some $V_\theta$ thinks that there is some cardinal $\kappa$ such that $H_\kappa$ satisfies $\psi$.

($2\to 1$) The statement $\exists \theta\, V_\theta\models\psi$ is equivalent to the assertion $\exists\theta\,\exists x\,(x=V_\theta\wedge x\models\psi)$. The claim that $x\models\psi$ involves only bounded quantifiers, since the quantifiers of $\psi$ become bounded by $x$. The claim that $x=V_\theta$ is $\Pi_1$ in $x$ and $\theta$, since it is equivalent to saying that $x$ is transitive and the ordinals of $x$ are precisely $\theta$ and $x$ thinks every $V_\alpha$ exists, plus a certain minimal set theory (so far this is just $\Delta_0$, since all quantifiers are bounded), plus, finally, the assertion that $x$ contains every subset of each of its elements. So altogether, the assertion that some $V_\theta$ satisfies $\psi$ has complexity $\Sigma_2$ in the language of set theory.

($1\to 3$) This implication is a consequence of the following absoluteness lemma.

Lemma. (Levy) If $\kappa$ is any uncountable cardinal, then $H_\kappa\prec_{\Sigma_1} V$.

Proof. Suppose that $x\in H_\kappa$ and $V\models\exists y\,\psi(x,y)$, where $\psi$ has only bounded quantifiers. Fix some such witness $y$, which exists inside some $H_\gamma$ for perhaps much larger $\gamma$. By the Löwenheim-Skolem theorem, there is $X\prec H_\gamma$ with $\text{TC}(\{x\})\subset X$, $y\in X$ and $X$ of size less than $\kappa$. Let $\pi:X\cong M$ be the Mostowski collapse of $X$, so that $M$ is transitive, and since it has size less than $\kappa$, it follows that $M\subset H_\kappa$. Since the transitive closure of $\{x\}$ was contained in $X$, it follows that $\pi(x)=x$. Thus, since $X\models\psi(x,y)$ we conclude that $M\models \psi(x,\pi(y))$ and so hence $\pi(y)$ is a witness to $\psi(x,\cdot)$ inside $H_\kappa$, as desired. QED

Using the lemma, we now prove the remaining part of the theorem. Consider any $\Sigma_2$ assertion $\exists x\,\forall y\, \varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. This assertion is equivalent to $\exists\kappa\, H_\kappa\models\exists x\,\forall y\,\varphi_0(x,y)$, simply because if there is such a $\kappa$ with $H_\kappa$ having such an $x$, then by the lemma this $x$ works for all $y\in V$ since $H_\kappa\prec_{\Sigma_1}V$; and conversely, if there is an $x$ such that $\forall y\, \varphi_0(x,y)$, then this will remain true inside any $H_\kappa$ with $x\in H_\kappa$. QED

In light of the theorem, it makes sense to refer to the $\Sigma_2$ properties as the locally verifiable properties, or perhaps as semi-local properties, since positive instances of $\Sigma_2$ assertions can be verified in some sufficiently large $V_\theta$, without need for unbounded search. A truly local property, therefore, would be one such that positive and negative instances can be verified this way, and these would be precisely the $\Delta_2$ properties, whose positive and negative instances are locally verifiable.

Tighter concepts of locality are obtained by insisting that the property is not merely verified in some $V_\theta$, perhaps very large, but rather is verified in a $V_\theta$ where $\theta$ has a certain closeness to the parameters or instance of the property. For example, a cardinal $\kappa$ is measurable just in case there is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, and this is verified inside $V_{\kappa+2}$. Thus, the assertion “$\kappa$ is measurable,” has complexity $\Sigma^2_1$ over $V_\kappa$. One may similarly speak of $\Sigma^n_m$ or $\Sigma^\alpha_m$ properties, to refer to properties that can be verified with $\Sigma_m$ assertions in $V_{\kappa+\alpha}$. Alternatively, for any class function $f$ on the ordinals, one may speak of $f$-local properties, meaning a property that can be checked of $x\in V_\theta$ by checking a property inside $V_{f(\theta)}$.

This post was made in response to a question on MathOverflow.

Satisfaction is not absolute

  • J. D. Hamkins and R. Yang, “Satisfaction is not absolute,” to appear in the Review of Symbolic Logic, pp. 1-34, 2014.  
    @ARTICLE{HamkinsYang:SatisfactionIsNotAbsolute,
    author = {Joel David Hamkins and Ruizhi Yang},
    title = {Satisfaction is not absolute},
    journal = {to appear in the Review of Symbolic Logic},
    year = {2014},
    volume = {},
    number = {},
    pages = {1--34},
    month = {},
    note = {},
    abstract = {},
    keywords = {},
    source = {},
    eprint = {1312.0670},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    url = {http://jdh.hamkins.org/satisfaction-is-not-absolute},
    doi = {},
    }

$\newcommand\N{\mathbb{N}}\newcommand\satisfies{\models}$

Abstract. We prove that the satisfaction relation $\mathcal{N}\satisfies\varphi[\vec a]$ of first-order logic is not absolute between models of set theory having the structure $\mathcal{N}$ and the formulas $\varphi$ all in common. Two models of set theory can have the same natural numbers, for example, and the same standard model of arithmetic $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$, yet disagree on their theories of arithmetic truth; two models of set theory can have the same natural numbers and the same arithmetic truths, yet disagree on their truths-about-truth, at any desired level of the iterated truth-predicate hierarchy; two models of set theory can have the same natural numbers and the same reals, yet disagree on projective truth; two models of set theory can have the same $\langle H_{\omega_2},{\in}\rangle$ or the same rank-initial segment $\langle V_\delta,{\in}\rangle$, yet disagree on which assertions are true in these structures.

On the basis of these mathematical results, we argue that a philosophical commitment to the determinateness of the theory of truth for a structure cannot be seen as a consequence solely of the determinateness of the structure in which that truth resides. The determinate nature of arithmetic truth, for example, is not a consequence of the determinate nature of the arithmetic structure $\N=\{ 0,1,2,\ldots\}$ itself, but rather, we argue, is an additional higher-order commitment requiring its own analysis and justification.

Many mathematicians and philosophers regard the natural numbers $0,1,2,\ldots\,$, along with their usual arithmetic structure, as having a privileged mathematical existence, a Platonic realm in which assertions have definite, absolute truth values, independently of our ability to prove or discover them. Although there are some arithmetic assertions that we can neither prove nor refute—such as the consistency of the background theory in which we undertake our proofs—the view is that nevertheless there is a fact of the matter about whether any such arithmetic statement is true or false in the intended interpretation. The definite nature of arithmetic truth is often seen as a consequence of the definiteness of the structure of arithmetic $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$ itself, for if the natural numbers exist in a clear and distinct totality in a way that is unambiguous and absolute, then (on this view) the first-order theory of truth residing in that structure—arithmetic truth—is similarly clear and distinct.

Feferman provides an instance of this perspective when he writes (Feferman 2013, Comments for EFI Workshop, p. 6-7) :

In my view, the conception [of the bare structure of the natural numbers] is completely clear, and thence all arithmetical statements are definite.

It is Feferman’s `thence’ to which we call attention.  Martin makes a similar point (Martin, 2012, Completeness or incompleteness of basic mathematical concepts):

What I am suggesting is that the real reason for confidence in first-order completeness is our confidence in the full determinateness of the concept of the natural numbers.

Many mathematicians and philosophers seem to share this perspective. The truth of an arithmetic statement, to be sure, does seem to depend entirely on the structure $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$, with all quantifiers restricted to $\N$ and using only those arithmetic operations and relations, and so if that structure has a definite nature, then it would seem that the truth of the statement should be similarly definite.

Nevertheless, in this article we should like to tease apart these two ontological commitments, arguing that the definiteness of truth for a given mathematical structure, such as the natural numbers, the reals or higher-order structures such as $H_{\omega_2}$ or $V_\delta$, does not follow from the definite nature of the underlying structure in which that truth resides. Rather, we argue that the commitment to a theory of truth for a structure is a higher-order ontological commitment, going strictly beyond the commitment to a definite nature for the underlying structure itself.

We make our argument in part by proving that different models of set theory can have a structure identically in common, even the natural numbers, yet disagree on the theory of truth for that structure.

Theorem.

  • Two models of set theory can have the same structure of arithmetic $$\langle\N,{+},{\cdot},0,1,{\lt}\rangle^{M_1}=\langle\N,{+},{\cdot},0,1,{\lt}\rangle^{M_2},$$yet disagree on the theory of arithmetic truth.
  • Two models of set theory can have the same natural numbers and a computable linear order in common, yet disagree about whether it is a well-order.
  • Two models of set theory that have the same natural numbers and the same reals, yet disagree on projective truth.
  • Two models of set theory can have a transitive rank initial segment in common $$\langle V_\delta,{\in}\rangle^{M_1}=\langle V_\delta,{\in}\rangle^{M_2},$$yet disagree about whether it is a model of ZFC.

The proofs use only elementary classical methods, and might be considered to be a part of the folklore of the subject of models of arithmetic. The paper includes many further examples of the phenomenon, and concludes with a philosophical discussion of the issue of definiteness, concerning the question of whether one may deduce definiteness-of-truth from definiteness-of-objects and definiteness-of-structure.

 

Exploring the Frontiers of Incompleteness, Harvard, August 2013

I will be participating in the culminating workshop of the Exploring the Frontiers of Incompleteness conference series at Harvard University, to take place August 31-September 1, 2013.  Rather than conference talks, the program will consist of extended discussion sessions by the participants of the year-long series, with the discussion framed by very brief summary presentations.  Peter Koellner asked me to prepare such a presentation on the multiverse conception, and you can see the slides in The multiverse perspective in set theory (Slides).

My previous EFI talk was The multiverse perspective on determinateness in set theory, based in part on my paper The set-theoretical multiverse.

Algebraicity and implicit definability in set theory

  • J. D. Hamkins and C. Leahy, “Algebraicity and Implicit Definability in Set Theory,” Notre Dame J. Formal Logic, vol. 57, iss. 3, pp. 431-439, 2016.  
    @article{HamkinsLeahy2016:AlgebraicityAndImplicitDefinabilityInSetTheory,
    author = "Hamkins, Joel David and Leahy, Cole",
    doi = "10.1215/00294527-3542326",
    fjournal = "Notre Dame Journal of Formal Logic",
    journal = "Notre Dame J. Formal Logic",
    number = "3",
    pages = "431--439",
    publisher = "Duke University Press",
    title = "Algebraicity and Implicit Definability in Set Theory",
    volume = "57",
    year = "2016",
    url = {http://jdh.hamkins.org/algebraicity-and-implicit-definability},
    eprint = {1305.5953},
    archivePrefix = {arXiv},
    primaryClass = {math.LO},
    ISSN = {0029-4527},
    MRCLASS = {03E47 (03C55)},
    MRNUMBER = {3521491},
    }

We aim in this article to analyze the effect of replacing several natural uses of definability in set theory by the weaker model-theoretic notion of algebraicity and its companion concept of implicit definability. In place of the class HOD of hereditarily ordinal definable sets, for example, we consider the class HOA of hereditarily ordinal-algebraic sets. In place of the pointwise definable models of set theory, we examine its (pointwise) algebraic models. And in place of G&ouml;del’s constructible universe L, obtained by iterating the definable power set operation, we introduce the implicitly constructible universe Imp, obtained by iterating the algebraic or implicitly definable power set operation. In each case we investigate how the change from definability to algebraicity affects the nature of the resulting concept. We are especially intrigued by Imp, for it is a new canonical inner model of ZF whose subtler properties are just now coming to light. Open questions about Imp abound.

Before proceeding further, let us review the basic definability definitions. In the model theory of first-order logic, an element $a$ is definable in a structure $M$ if it is the unique object in $M$ satisfying some first-order property $\varphi$ there, that is, if $M\models\varphi[b]$ just in case $b=a$. More generally, an element $a$ is algebraic in $M$ if it has a property $\varphi$ exhibited by only finitely many objects in $M$, so that $\{b\in M \mid M\models\varphi[b]\}$ is a finite set containing $a$. For each class $P\subset M$ we can similarly define what it means for an element to be $P$-definable or $P$-algebraic by allowing the formula $\varphi$ to have parameters from $P$.

In the second-order context, a subset or class $A\subset M^n$ is said to be definable in $M$, if $A=\{\vec a\in M\mid M\models\varphi[\vec a]\}$ for some first-order formula $\varphi$. In particular, $A$ is the unique class in $M^n$ with $\langle M,A\rangle\models\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, in the language where we have added a predicate symbol for $A$. Generalizing this condition, we say that a class $A\subset M^n$ is implicitly definable in $M$ if there is a first-order formula $\psi(A)$ in the expanded language, not necessarily of the form $\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, such that $A$ is unique such that $\langle M,A\rangle\models\psi(A)$. Thus, every (explicitly) definable class is also implicitly definable, but the converse can fail. Even more generally, we say that a class $A\subset M^n$ is algebraic in $M$ if there is a first-order formula $\psi(A)$ in the expanded language such that $\langle M,A\rangle\models\psi(A)$ and there are only finitely many $B\subset M^n$ for which $\langle M,B\rangle\models\psi(B)$. Allowing parameters from a fixed class $P\subset M$ to appear in $\psi$ yields the notions of $P$-definability, implicit $P$-definability, and $P$-algebraicity in $M$. Simplifying the terminology, we say that $A$ is definable, implicitly definable, or algebraic over (rather than in) $M$ if it is $M$-definable, implicitly $M$-definable, or $M$-algebraic in $M$, respectively. A natural generalization of these concepts arises by allowing second-order quantifiers to appear in $\psi$. Thus we may speak of a class $A$ as second-order definable, implicitly second-order definable, or second-order algebraic. Further generalizations are of course possible by allowing $\psi$ to use resources from other strong logics.

The main theorems of the paper are:

Theorem. The class of hereditarily ordinal algebraic sets is the same as the class of hereditarily ordinal definable sets: $$\text{HOA}=\text{HOD}.$$

Theorem. Every pointwise algebraic model of ZF is a pointwise definable model of ZFC+V=HOD.

In the latter part of the paper, we introduce what we view as the natural algebraic analogue of the constructible universe, namely, the implicitly constructible universe, denoted Imp, and built as follows:

$$\text{Imp}_0 = \emptyset$$

$$\text{Imp}_{\alpha + 1} = P_{imp}(\text{Imp}_\alpha)$$

$$\text{Imp}_\lambda = \bigcup_{\alpha < \lambda} \text{Imp}_\alpha, \text{ for limit }\lambda$$

$$\text{Imp} = \bigcup_\alpha \text{Imp}_\alpha.$$

Theorem.  Imp is an inner model of ZF with $L\subset\text{Imp}\subset\text{HOD}$.

Theorem.  It is relatively consistent with ZFC that $\text{Imp}\neq L$.

Theorem. In any set-forcing extension $L[G]$ of $L$, there is a further extension $L[G][H]$ with $\text{gImp}^{L[G][H]}=\text{Imp}^{L[G][H]}=L$.

Open questions about Imp abound. Can $\text{Imp}^{\text{Imp}}$ differ from $\text{Imp}$? Does $\text{Imp}$ satisfy the axiom of choice? Can $\text{Imp}$ have measurable cardinals? Must $0^\sharp$ be in $\text{Imp}$ when it exists? (An affirmative answer arose in conversation with Menachem Magidor and Gunter Fuchs, and we hope that $\text{Imp}$ will subsume further large cardinal features. We anticipate a future article on the implicitly constructible universe.)  Which large cardinals are absolute to $\text{Imp}$? Does $\text{Imp}$ have fine structure? Should we hope for any condensation-like principle? Can CH or GCH fail in $\text{Imp}$? Can reals be added at uncountable construction stages of $\text{Imp}$? Can we separate $\text{Imp}$ from HOD? How much can we control $\text{Imp}$ by forcing? Can we put arbitrary sets into the $\text{Imp}$ of a suitable forcing extension? What can be said about the universe $\text{Imp}(\mathbb{R})$ of sets implicitly constructible relative to $\mathbb{R}$ and, more generally, about $\text{Imp}(X)$ for other sets $X$? Here we hope at least to have aroused interest in these questions.

This article arose from a question posed on MathOverflow by my co-author Cole Leahy and our subsequent engagement with it.

Algebraicity and implicit definability in set theory, CUNY, May 2013

This is a talk May 10, 2013 for the CUNY Set Theory Seminar.

Abstract.  An element a is definable in a model M if it is the unique object in M satisfying some first-order property. It is algebraic, in contrast, if it is amongst at most finitely many objects satisfying some first-order property φ, that is, if { b | M satisfies φ[b] } is a finite set containing a. In this talk, I aim to consider the situation that arises when one replaces the use of definability in several parts of set theory with the weaker concept of algebraicity. For example, in place of the class HOD of all hereditarily ordinal-definable sets, I should like to consider the class HOA of all hereditarily ordinal algebraic sets. How do these two classes relate? In place of the study of pointwise definable models of set theory, I should like to consider the pointwise algebraic models of set theory. Are these the same? In place of the constructible universe L, I should like to consider the inner model arising from iterating the algebraic (or implicit) power set operation rather than the definable power set operation. The result is a highly interesting new inner model of ZFC, denoted Imp, whose properties are only now coming to light. Is Imp the same as L? Is it absolute? I shall answer all these questions at the talk, but many others remain open.

This is joint work with Cole Leahy (MIT).

NYlogic abstract | MathOverflow post

Jonas Reitz

Jonas Reitz earned his Ph.D under my supervision in June, 2006 at the CUNY Graduate Center.  He was truly a pleasure to supervise. From the earliest days of his dissertation research, he had his own plan for the topic of the work: he wanted to “undo” forcing, to somehow force backwards, from the extension to the ground model. At first I was skeptical, but in time, ideas crystalized around the ground axiom (now with its own Wikipedia entry), formulated using a recent-at-the-time result of Richard Laver.  Along with Laver’s theorem, Jonas’s dissertation was the beginning of the body of work now known as set-theoretic geology.  Jonas holds a tenured position at the New York City College of Technology of CUNY.

Jonas Reitz


web page | math genealogy | MathSciNet | ar$\chi$iv | google scholar | related posts

Jonas Reitz, “The ground axiom,” Ph.D. dissertation, CUNY Graduate Center, June, 2006.  ar$\chi$iv

Abstract.  A new axiom is proposed, the Ground Axiom, asserting that the universe is not a nontrivial set-forcing extension of any inner model. The Ground Axiom is first-order expressible, and any model of ZFC has a class-forcing extension which satisfies it. The Ground Axiom is independent of many well-known set-theoretic assertions including the Generalized Continuum Hypothesis, the assertion V=HOD that every set is ordinal definable, and the existence of measurable and supercompact cardinals. The related Bedrock Axiom, asserting that the universe is a set-forcing extension of a model satisfying the Ground Axiom, is also first-order expressible, and its negation is consistent. As many of these results rely on forcing with proper classes, an appendix is provided giving an exposition of the underlying theory of proper class forcing.

The mate-in-n problem of infinite chess is decidable, Cambridge, June 2012

This will be a contributed talk at the Turing Centenary Conference CiE 2012 held June 18-23, 2012 in Cambridge, UK.

Abstract.  The mate-in-$n$ problem of infinite chess—chess played on an infinite edgeless board—is the problem of determining whether a designated player can force a win from a given finite position in at most $n$ moves. Although a straightforward formulation of this problem leads to assertions of high arithmetic complexity, with $2n$ alternating quantifiers,  the main theorem of this article nevertheless confirms a conjecture of the second author and C. D. A. Evans by establishing that it is computably decidable, uniformly in the position and in $n$. Furthermore, there is a computable strategy for optimal play from such mate-in-$n$ positions. The proof proceeds by showing that the mate-in-$n$ problem is expressible in what we call the first-order structure of chess $\frak{Ch}$, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable. The structure is also definable in Presburger arithmetic. Unfortunately, this resolution of the mate-in-$n$ problem does not appear to settle the decidability of the more general winning-position problem, the problem of determining whether a designated player has a winning strategy from a given position, since a position may admit a winning strategy without any bound on the number of moves required. This issue is connected with transfinite game values in infinite chess, and the exact value of the omega one of chess $\omega_1^{\rm chess}$ is not known.

Article | Slides | CiE 2012 | Contributed talk schedule

The omega one of infinite chess, New York, 2012

This will be a talk on May 18, 2012 for the CUNY Set Theory Seminar.

Infinite chess is chess played on an infinite edgeless chessboard. The familiar chess pieces move about according to their usual chess rules, and each player strives to place the opposing king into checkmate.  The mate-in-$n$ problem of infinite chess is the problem of determining whether a designated player can force a win from a given finite position in at most $n$ moves.  A naive formulation of this problem leads to assertions of high arithmetic complexity with $2n$ alternating quantifiers—there is a move for white, such that for every black reply, there is a countermove for white, and so on. In such a formulation, the problem does not appear to be decidable; and one cannot expect to search an infinitely branching game tree even to finite depth. Nevertheless, in joint work with Dan Brumleve and Philipp Schlicht, confirming a conjecture of myself and C. D. A. Evans, we establish that the mate-in-$n$ problem of infinite chess is computably decidable, uniformly in the position and in $n$. Furthermore, there is a computable strategy for optimal play from such mate-in-$n$ positions. The proof proceeds by showing that the mate-in-$n$ problem is expressible in what we call the first-order structure of chess, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable.  An equivalent account of the result arises from the realization that the structure of chess is interpretable in the standard model of Presburger arithmetic $\langle\mathbb{N},+\rangle$.  Unfortunately, this resolution of the mate-in-$n$ problem does not appear to settle the decidability of the more general winning-position problem, the problem of determining whether a designated player has a winning strategy from a given position, since a position may admit a winning strategy without any bound on the number of moves required. This issue is connected with transfinite game values in infinite chess, and the exact value of the omega one of chess $\omega_1^{\rm chess}$ is not known. I will also discuss recent joint work with C. D. A. Evans and W. Hugh Woodin showing that the omega one of infinite positions in three-dimensional infinite chess is true $\omega_1$: every countable ordinal is realized as the game value of such a position.

article | slides