User trevor wilson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:21:38Z http://mathoverflow.net/feeds/user/1682 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125108/stationary-many-subsets-of-kappa-whose-order-type-is-a-cardinal-and-whose-in Stationary many subsets of $\kappa^+$ whose order type is a cardinal and whose intersection with $\kappa$ is an inaccessible cardinal Trevor Wilson 2013-03-21T00:04:13Z 2013-03-21T01:38:16Z <p>Is anything known about the consistency strength of the following statement?</p> <ul> <li>$\kappa$ is a Mahlo cardinal and there is a stationary set of $a \in \mathcal{P}_\kappa(\kappa^+)$ such that $a \cap \kappa$ is an inaccessible cardinal and the order type of $a$ is $(a \cap \kappa)^+$?</li> </ul> <p>This statement follows from $\kappa^+$-supercompactness of $\kappa$ and also from subcompactness of $\kappa$. It is a strengthening of the principle $\text{Pr}(\kappa^+)$ that was shown by Burke in "Generic embeddings and the failure of box" to imply $\neg \square_\kappa$, so a lower bound on the consistency strength is the existence of two Mahlo cardinals.</p> http://mathoverflow.net/questions/112733/relation-between-neg-square-kappa-and-the-tree-property-at-kappa Relation between $\neg \square(\kappa)$ and the tree property at $\kappa$. Trevor Wilson 2012-11-18T00:28:24Z 2013-03-19T19:17:21Z <p>If $\kappa$ is an inaccessible cardinal then the tree property at $\kappa$ is equivalent to weak compactness of $\kappa$, which implies that $\square(\kappa)$ fails---that is, that every coherent sequence of clubs of length $\kappa$ can be threaded.</p> <p>I am wondering about other implications involving square and the tree property, namely:</p> <ul> <li><p>If $\kappa$ is an inaccessible cardinal and $\square(\kappa)$ fails, must $\kappa$ have the tree property (and therefore be weakly compact?)</p></li> <li><p>If $\kappa$ is a regular cardinal, does $\neg \square(\kappa)$ imply that $\kappa$ has the tree property?</p></li> <li><p>If $\kappa$ is a regular cardinal with the tree property, must $\square(\kappa)$ fail?</p></li> </ul> <p>(By the way, I am aware of the relative consistency result that if $\square(\kappa)$ fails for some regular cardinal $\kappa$, then $\kappa$ is weakly compact in $L$ and so in particular it has the tree property in $L$.)</p> http://mathoverflow.net/questions/112733/relation-between-neg-square-kappa-and-the-tree-property-at-kappa/125008#125008 Answer by Trevor Wilson for Relation between $\neg \square(\kappa)$ and the tree property at $\kappa$. Trevor Wilson 2013-03-19T19:17:21Z 2013-03-19T19:17:21Z <p>To complete the accepted answer to my question (which addresses parts 2 and 3) perhaps I should mention here that the answer to part 1 is no: If $\delta$ is a supercompact cardinal and $\kappa$ is the least inaccessible cardinal above $\delta$ then $\square(\kappa)$ fails but $\kappa$ is not weakly compact.</p> http://mathoverflow.net/questions/120463/homogeneous-namba-like-forcing Homogeneous Namba-like forcing Trevor Wilson 2013-01-31T21:11:24Z 2013-02-01T00:47:41Z <p>Let $\kappa \ge \aleph_3$ be a regular cardinal that is countably closed ($\alpha^\omega &lt; \kappa$ for every $\alpha &lt; \kappa$.) I'm mostly interested in the case that $\kappa$ is strongly inaccessible. Can there be a homogeneous notion of forcing that makes $\text{cof}(\kappa^{+V}) &lt; \kappa$ without adding any bounded subsets of $\kappa$?</p> <p>If there is a Woodin cardinal above $\kappa$ then the stationary tower forcing could do this except that it is (probably) not homogeneous.</p> <p>If there is a forcing notion as desired then I believe the results of the paper "Stacking mice" would give a non-domestic mouse, so some large cardinals would be required to show that such a forcing exists. Can we get one from, <em>e.g.</em> a supercompact cardinal?</p> http://mathoverflow.net/questions/110871/can-measures-be-added-by-forcing Can measures be added by forcing? Trevor Wilson 2012-10-28T01:50:12Z 2012-11-03T16:04:46Z <p>The Lévy-Solovay theorem says that small forcings do not create measures. J.D. Hamkins has generalized this to a larger class of forcings called gap forcings. I would assume this cannot be generalized to <em>all</em> forcings, but I cannot think of a counterexample. Is there a forcing notion that creates a $\kappa$-complete (or even countably complete) measure $\mu$ on some uncountable cardinal $\kappa$ such that $\mu \cap V$ is not in $V$?</p> http://mathoverflow.net/questions/109925/proof-of-ad-every-set-of-reals-is-suslin-implies-ad-mathbbr Proof of "AD + every set of reals is Suslin" implies AD$_\mathbb{R}$ Trevor Wilson 2012-10-17T16:31:29Z 2012-10-19T21:50:53Z <p>Could someone point me toward a proof that "ZF + AD + every set of reals is Suslin" (+ $\mathsf{DC}_\mathbb{R}$?) implies $\mathsf{AD}_\mathbb{R}$, either with a reference or a hint?</p> <p>I am interested how local the proof is; that is, which sets of reals need to be Suslin in order to show the determinacy of a particular real game.</p> http://mathoverflow.net/questions/107441/does-every-nonempty-definable-finite-set-have-a-definable-member Does every nonempty definable finite set have a definable member? Trevor Wilson 2012-09-18T06:31:12Z 2012-09-18T15:18:26Z <p>I asked this on MSE yesterday ( <a href="http://math.stackexchange.com/q/197873/39378" rel="nofollow">http://math.stackexchange.com/q/197873/39378</a> ) but no one has answered it yet. I hope it's not too soon to post it here.</p> <p>Here are a few ways to formalize the question, so you can pick your favorite and answer it. Assume whatever large cardinals you like.</p> <p>(1) Is it consistent with ZFC that there is an inaccessible cardinal $\delta$ and a nonempty finite set that is first-order definable without parameters over $(V_\delta,\in)$ but has no elements that are first-order definable without parameters over $(V_\delta,\in)$?</p> <p>(2) Is there any model of ZFC that has a finite nonempty set, first-order definable without parameters over the model, with no element that is first-order definable without parameters over the model?</p> <p>(3) Is it consistent with ZFC that there is an ordinal-definable finite nonempty set with no ordinal-definable member? (I am aware of the question <a href="http://mathoverflow.net/questions/17608/a-question-about-ordinal-definable-real-numbers" rel="nofollow">http://mathoverflow.net/questions/17608/a-question-about-ordinal-definable-real-numbers</a>, but that question asks about sets of real numbers and I already know the answer to my question for sets of real numbers, or indeed for sets of subsets of any ordinal, because they are definably linearly ordered.)</p> <p>(4) Any of the above formulations with ZFC replaced by ZF.</p> http://mathoverflow.net/questions/106681/strong-ideals-that-are-not-pre-saturated Strong ideals that are not pre-saturated Trevor Wilson 2012-09-08T19:14:05Z 2012-09-11T19:11:30Z <p>An ideal on $\omega_1$ is <em>strong</em> if it is precipitous and the associated generic elementary embedding always maps $\omega_1$ to $\omega_2$. This definition is from Baumgartner and Taylor, <em>Saturation Properties of Ideals in Generic Extensions II</em> (available online at <a href="http://www.jstor.org/stable/1998900" rel="nofollow">http://www.jstor.org/stable/1998900</a>.) Every pre-saturated ideal on $\omega_1$ is strong (this was well-known even before the terminology was introduced, I think) and in this paper the authors ask whether the converse is true.</p> <p>Does anyone know the status of this question: "is every strong ideal on $\omega_1$ pre-saturated"?</p> http://mathoverflow.net/questions/106705/2d-problems-which-are-easier-to-solve-in-3d/106822#106822 Answer by Trevor Wilson for 2D Problems Which are Easier to Solve in 3D Trevor Wilson 2012-09-10T15:29:23Z 2012-09-10T15:29:23Z <p>Desargues' Theorem is a statement about triangles in the plane that is easier to prove using solid geometry.</p> http://mathoverflow.net/questions/106681/strong-ideals-that-are-not-pre-saturated/106755#106755 Answer by Trevor Wilson for Strong ideals that are not pre-saturated Trevor Wilson 2012-09-09T21:16:48Z 2012-09-09T21:24:46Z <p>It is consistent that the nonstationary ideal on $\omega_1$ is strong but not pre-saturated. Baumgartner and Taylor proved in the aforementioned paper that strong ideals are preserved by c.c.c. forcing and asked whether the same is true for pre-saturated ideals. The answer to this question is negative, implying a negative answer to the question I posted above. Apparently this was first proved by Veličković in the paper <em>Forcing axioms and stationary sets</em> (which I cannot seem to access online) from ZFC + SPFA. Another example of a c.c.c. forcing that destroys pre-saturation may be found in a more recent paper by Larson and Yorioka, <em>Another c.c.c. forcing that destroys presaturation</em>, assuming the consistency of ZF + AD.</p> <p>I don't know if a negative answer can be forced from only a Woodin cardinal (which is equiconsistent with the existence of a presaturated ideal and also with the existence of a strong ideal.)</p> http://mathoverflow.net/questions/106522/why-cant-an-explicit-well-ordering-of-the-reals-be-ruled-out-in-zf/106529#106529 Answer by Trevor Wilson for Why can't an explicit well-ordering of the reals be ruled out in ZF? Trevor Wilson 2012-09-06T16:41:39Z 2012-09-06T16:41:39Z <p>It sounds like you're taking the word "explicit" (which has no precise mathematical meaning) to mean "can be proved to exist by ZF". This is problematic because a theory (e.g., ZF) proves statements. It does not construct objects, or prove that a particular object exists, although it may prove <em>statements</em> asserting the existence of objects with certain properties.</p> <p>For example, although every model of "ZF + $V=L$" has a definable wellordering of its reals, this does not help us construct wellorderings of the reals in models of "ZF + $V \ne L$". The most serious obstacle in this case is that the two models we are considering could have different sets of reals. So although ZF does prove that $L$ satisfies "there is a definable wellordering of the reals," this only gives us a definable wellordering of the reals of $L$. It is consistent that $L$ does not contain all the reals, and even that it contains only countably many reals.</p> http://mathoverflow.net/questions/106395/infinity-borel-sets-in-zfc Infinity-Borel sets in ZFC Trevor Wilson 2012-09-05T03:03:31Z 2012-09-05T03:03:31Z <p>The notion of an $\infty$-Borel set of reals is useful in the study of AD. Under ZFC it becomes trivial: every set of reals is $\infty$-Borel. However, the notion of an $\infty$-Borel <em>code</em> is still interesting. For example, an $\infty$-Borel code $S$ gives us an absolute way to define a set of reals $(A_S)^{V[g]}$ in a generic extension $V[g]$, albeit without some of the nice properties of uB-codes. So I have two related questions:</p> <p>(1) What are some applications of $\infty$-Borel codes in ZFC?</p> <p>(2) What are some applications where an $\infty$-Borel code is used to define a set of reals in a generic extension?</p> <p>I am discounting the special case of $\omega$-Borel codes, that is, ordinary codes for Borel sets.</p> http://mathoverflow.net/questions/106113/in-search-of-a-set-theory-with-specific-properties/106365#106365 Answer by Trevor Wilson for In search of a set theory with specific properties Trevor Wilson 2012-09-04T16:57:41Z 2012-09-04T16:57:41Z <p>Here is a CW answer incorporating the answers in the comments. Both NF (New Foundations) and NFU (New Foundations with urelements, obtained from NF by weakening the axiom of extensionality) satisfy conditions (1) through (3).</p> http://mathoverflow.net/questions/102854/weakly-homogeneous-trees-under-ad/104632#104632 Answer by Trevor Wilson for Weakly homogeneous trees under AD Trevor Wilson 2012-08-13T18:41:22Z 2012-08-14T01:30:56Z <p>I hope it's okay to post an answer to my own question. I am essentially repeating Woodin's proof that he just showed me. Any errors were probably introduced by me.</p> <p>Recall that under AD all measures on ordinals are countably complete and ordinal-definable. By the coding of measures theorem of Kechris assuming AD and that there is a Suslin cardinal above $\kappa$, there are fewer than $\Theta$ many measures on $\kappa$ (or $\kappa^{&lt;\omega}$ for that matter.) So by Turing determinacy we get a fine, countably complete measure $U$ on the set of measures on $\kappa^{&lt;\omega}$.</p> <p>Fix a tree $T$ on $\omega\times \kappa$. As usual, given a real $x \in \omega^\omega$ we let $T_x = \lbrace s \in \kappa^{&lt;\omega} : (x \restriction |s|, s) \in T\rbrace$.</p> <p>Claim: $U$-almost all $\sigma$ witness the weak homogeneity of $T$.</p> <p>Proof: For each $\sigma$ we define a game $G_{\sigma}$, closed for Player I, for which Player I has a winning strategy iff $\sigma$ does not witness the weak homogeneity of $T$.</p> <ul> <li> I plays: $(x_0, \alpha_0, \beta_0)$, $(x_1, \alpha_1,\beta_1),\ldots$ </li> <li> II plays: $\mu_0$, $\mu_1,\ldots$ </li> </ul> <p>Let $x$, $\vec{\alpha}$, $\vec{\beta}$, and $\vec{\mu}$ denote the resulting sequence of moves.</p> <ul> <li> Rules for I: $\vec{\alpha} \in [T_x]$ and the sequence $\vec{\beta} \in \mathrm{Ord}^\omega$ continuously witnesses that the tower $\vec{\mu}$ is illfounded. </li> <li> Rules for II: $\vec{\mu}$ is a tower of measures in $\sigma$ concentrating on $T_x$. </li> </ul> <p>If both players follow the rules until the end, we say that player I wins.</p> <p>If $\sigma$ does not witness that $T$ is weakly homogeneous, say $x \in p[T]$ but there is no wellfounded tower of measures in $\sigma$ concentrating on $T_x$, then there is a continuous witness to the illfoundedness of towers of measures in $\sigma$ concentrating on $T_x$. (This is proved by an argument similar to what follows but using a fine, countably complete measure on the set of subsets of $\kappa^{&lt;\omega}$.)</p> <p>So if $\sigma$ does not witness that $T$ is weakly homogeneous, then player I has a winning strategy in $G_\sigma$. (The $x$ and $\vec{\alpha}$ are fixed in advance and the $\vec{\beta}$ comes from $\vec{\mu}$ via the continuous witness mentioned above.) Assume toward a contradiction that for $U$-almost every $\sigma$, player I has a winning strategy in $G_\sigma$. The game is closed, the moves are ordinals and measures, and all measures are ordinal-definable, so for such $\sigma$ player I has a winning strategy $F(\sigma)$ of playing the least move leading to a subgame where he or she still has a winning strategy.</p> <ul> <li> Define the integer $x_0$ to be the one played by $F(\sigma)$ on the first turn for $U$-almost every $\sigma$. </li> <li> Define a measure $\mu_0$ on $\kappa$ by $A \in \mu_0 \iff \forall^*_U \sigma\; (\alpha^{\sigma}_0 \in A)$ where $\alpha^{\sigma}_0$ is the ordinal $\alpha_0$ played by $F(\sigma)$ on the first turn. </li> <li> Define the integer $x_1$ to be the one played by $F(\sigma)$ on the second turn against $\mu_0$ for $U$-almost every $\sigma$. </li> <li> Define a measure $\mu_1$ on $\kappa^2$ by $A \in \mu_1 \iff \forall^*_U \sigma\; ((\alpha^{\sigma}_0,\alpha^{\sigma}_1) \in A)$ where $\alpha^{\sigma}_1$ is the ordinal $\alpha_1$ played by $F(\sigma)$ against $\mu_0$ on the second turn. </li> </ul> <p>Continuing in this way, we get a real $x \in \omega^\omega$ and a sequence of measures $\vec{\mu}$. One can easily check that $\vec{\mu}$ is a tower of measures. Each $\mu_i$ concentrates on $T_x$ because $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in T_x$. It is a wellfounded tower because if $A_i \in \mu_i$ for all $i&lt;\omega$ then by countable completeness of $U$ there is a $\sigma$ such that $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in A_i$ for all $i&lt;\omega$. However, by countable completeness of $U$ there is a $\sigma$ such that $\vec{\mu}$ is a legal play by player II against player I's winning strategy $F(\sigma)$, so player I's moves $\beta^\sigma_i$ continuously witness the illfoundedness of $\vec{\mu}$. Contradiction.</p> http://mathoverflow.net/questions/102854/weakly-homogeneous-trees-under-ad Weakly homogeneous trees under AD Trevor Wilson 2012-07-22T03:42:03Z 2012-08-14T01:30:56Z <p>If AD$_\mathbb{R}$ holds and $\kappa &lt; \Theta$ then every tree $T$ on $\kappa$ is weakly homogeneous (Martin–Woodin, "Weakly homogeneous trees.") I recall hearing that the hypothesis can be weakened to "AD holds and there is a Suslin cardinal above $\kappa$." Is this correct? If so, does anyone know where a proof can be found, or have the idea of the proof?</p> <p>It may help to keep in mind that the main consequence of $AD_\mathbb{R} + \kappa &lt; \Theta$ used in Martin–Woodin is the existence of a normal fine measure on $\mathcal{P}_{\omega_1}(\mathcal{P}(\kappa^{&lt;\omega}) \cup \mathrm{meas}(\kappa^{&lt;\omega}))$, where $\mathrm{meas}(\kappa^{&lt;\omega})$ denotes the set of countably complete measures on $\kappa^{&lt;\omega}$ (equivalently, on $\kappa^n$ for some $n&lt;\omega$.)</p> <p>In other words, the proof uses that $\omega_1$ is $\mathcal{P}(\kappa^{&lt;\omega}) \cup \mathrm{meas}(\kappa^{&lt;\omega})$-supercompact. If we only assume "AD holds and there is a Suslin cardinal above $\kappa$" then because the measures are essentially ordinals one can show that $\omega_1$ is $\mathrm{meas}(\kappa^{&lt;\omega})$-supercompact, but I see no way to show that $\omega_1$ is $\mathcal{P}(\kappa^{&lt;\omega})$-supercompact. So perhaps a more substantial modification to the Martin–Woodin proof is required.</p> http://mathoverflow.net/questions/77074/two-cardinal-diamond-principles-and-saturation-of-the-nonstationary-ideal Two-cardinal diamond principles and saturation of the nonstationary ideal Trevor Wilson 2011-10-03T20:16:29Z 2012-08-12T00:22:02Z <p>In the paper "Stationary reflection and the club filter", the author Masahiro Shioya says that the club filter on $P_{\omega_1}(\lambda)$ cannot be $2^\lambda$-saturated for $\lambda > \omega_1$, citing Shelah's book "Nonstructure Theory" (in preparation). I have three questions:</p> <p>1) Is there a published reference for this result?</p> <p>2) Does the theorem apply to $P_{\omega_1}(\lambda) | S$ for an arbitrary stationary set $S$?</p> <p>3) Does the proof go through a two-cardinal diamond principle? I.e., did Shelah prove (in ZFC) that $\lozenge_{\omega_1,\lambda}$ holds for $\lambda > \omega_1$? What about $\lozenge_{\omega_1,\lambda}(S)$ for arbitrary stationary $S$?</p> <p>I am particularly interested in the case $\lambda = 2^{\omega} = \omega_2$. In this case $\lozenge_{\omega_1,\lambda}(S)$ was proved by Donder and Matet in the paper "Two cardinal versions of diamond" for stationary sets $S$ of the form $\lbrace a \in P_{\omega_1}(\lambda) : \sup a \in B\rbrace$ where $B \subset \lambda$ is a stationary set consisting of points of cofinality $\omega$. Does this hold for arbitrary stationary $S$?</p> http://mathoverflow.net/questions/104464/tarski-grothendieck-set-theory-the-axiom-of-pairing-and-the-axiom-of-specificati/104497#104497 Answer by Trevor Wilson for Tarski-Grothendieck set theory, the axiom of pairing and the axiom of specification Trevor Wilson 2012-08-11T15:30:08Z 2012-08-11T15:30:08Z <p>Assume all the axioms of TG except for pairing. We will show that pairing and specification (a.k.a. separation) both follow, with the caveat in the following paragraph.</p> <p>We will also assume that the empty set exists. This does not seem to follow from the TG axioms as defined in the linked question <a href="http://mathoverflow.net/questions/102846/" rel="nofollow">http://mathoverflow.net/questions/102846/</a> or on Wikipedia <a href="https://en.wikipedia.org/wiki/Tarski-Grothendieck_set_theory" rel="nofollow">https://en.wikipedia.org/wiki/Tarski-Grothendieck_set_theory</a>, which do not seem to imply either "there is a set" or "if there is a set then there is an empty set." The natural way to proceed seems to be to postulate the existence of the empty set.</p> <p>(1) For pairing, suppose we are given sets $x$ and $y$ and we want to form the pair $\lbrace x,y \rbrace$. Let $A$ be a Tarski set containing $x$. Then $A$ also contains some element other than $x$ (this can be seen by considering separately the two cases $x = \emptyset$ and $x \ne \emptyset$.) Define a class function $F$ with domain $A$ by $$F(z) = \begin{cases} x &amp; \text{if z = x}\newline y &amp;\text{if z \in A and z \ne x}. \end{cases}$$ The range of $F$ is $\lbrace x,y \rbrace$, which is a set by replacement.</p> <p>(2) For specification, let $A$ be a set and $\varphi$ be a formula. We want to show that the class $B = \lbrace x \in A : \varphi(x)\rbrace$ is a set. If $B$ is empty then we are done because the empty set exists. So we may assume that there is $x_0 \in A$ such that $\varphi(x_0)$ holds. Then $B$ is the range of the class function $F$ with domain $A$ defined by $$F(x) = \begin{cases} x &amp; \text{if x \in A and \varphi(x)}\newline x_0 &amp;\text{if x \in A and \neg \varphi(x)}, \end{cases}$$ so it is a set by replacement. </p> http://mathoverflow.net/questions/104138/companion-of-the-pointclass-of-inductive-sets Companion of the pointclass of inductive sets Trevor Wilson 2012-08-06T19:14:51Z 2012-08-06T20:05:53Z <p>This question is about the notion of a companion for a Spector class, as defined in Moschovakis's book <i>Elementary Induction on Abstract Structures.</i> I am interested in Spector classes on $\mathbb{R}$, which are just a type of boldface pointclass. The smallest one is IND, the class of (boldface) inductive sets, which I will consider as a typical example.</p> <p>The companion of a Spector class $\bf \Gamma$ on $\mathbb{R}$ is a structure $(M,\in,R)$ with certain properties (listed in the book) such that $\bf \Gamma$ is the class of pointsets that are $\Sigma_1$-definable in $M$ with real parameters from the relation $R$. The companion of $\bf \Gamma$ is not unique, but its underlying set $M$ is unique and also the class of relations on $M$ that are $\Sigma_1$-definable from $R$ in $M$ with real parameters is unique.</p> <p>The pointclass IND can also be described as the class of pointsets that are $\Sigma_1$-definable over $M$ from parameters in $\mathbb{R} \cup \lbrace\mathbb{R} \rbrace$ where $M = L_\kappa(\mathbb{R})$ is the least admissible level of $L(\mathbb{R})$. We must allow $\lbrace\mathbb{R} \rbrace$ itself as a parameter here or we would just get the ${\bf \Sigma}^1_2$ sets. For any companion $(M,\in,R)$ of IND we must have $M = L_\kappa(\mathbb{R})$.</p> <p>Question: Is there a companion $(M,\in,R)$ of IND where the relation $R$ has a simple definition over $M = L_\kappa(\mathbb{R})$ (simpler than in Moschovakis's general construction of a companion?) Maybe something that is already studied in the fine structure of $L(\mathbb{R})$?</p> http://mathoverflow.net/questions/103917/set-existence-question/103952#103952 Answer by Trevor Wilson for set existence question Trevor Wilson 2012-08-04T15:14:05Z 2012-08-04T17:16:43Z <p>Let $F$ denote the function $\lbrace (i,S_i) : i &lt; \omega \rbrace$ in whatever sense it may exist---I hope this abuse of notation will not be confusing.</p> <p>The undefinability of truth does "get in the way" in the sense that it shows that there cannot be such a function $F$ that is definable. If we let $C$ be the set of $i$ such that $\sigma_i$ is a sentence ($x$ does not appear) then from $F$ we could define $\lbrace i \in C : S_i = \mathbb{R}\rbrace$, which is essentially a truth set and therefore cannot be definable by Tarski's theorem.</p> <p>EDIT: it looks like I am using "definable" to mean "definable without parameters" and Joel is using it to mean "definable with parameters." As Joel points out in the comments and in his answer, it is <i>possible</i> for a truth set, and indeed the desired function $F$, to be definable from ordinal parameters.</p> <p>To answer the precise question you stated, ZFC does not <i>prove</i> the existence of such a function $F$. As Andreas mentions in the comments to Bjørn's answer, there are models of ZFC in which all sets are definable. (This is not a first-order property of the model.) Any such model $M$ cannot satisfy "the desired function exists" regardless of how we try to formalize this. One way to see this is that if there were such a function $F \in M$, then externally we could use the fact that every set in $M$ is definable in $M$ to show that $ran(F) = \mathcal{P}(\mathbb{R})^M$. This statement about the range is absolute to $M$, contradicting Cantor's theorem in $M$.</p> http://mathoverflow.net/questions/103895/can-omega-1-be-supercompact Can $\omega_1$ be supercompact? Trevor Wilson 2012-08-03T20:27:15Z 2012-08-03T20:35:00Z <p>Is "ZF + $\omega_1$ is supercompact" consistent relative to "ZFC + there is a supercompact cardinal"?</p> <p>In particular, if $\delta$ is supercompact, does it remain so in $V(\mathbb{R} \cap V[G])$ where $G \subset Col(\omega,&lt;\delta)$ is $V$-generic? This seems to be the case for measurability but I am having trouble proving it for supercompactness. It seems likely that someone else has tried this, so I though I'd ask here.</p> <p>The appropriate definition of supercompactness in ZF is the one in terms of normal fine measures, where normality is defined using diagonal intersections.</p> <p>I am aware that $\omega_1$ has some amount of supercompactness under AD. I am interested in a more direct proof using forcing, which I hope will give (full) supercompactness.</p> http://mathoverflow.net/questions/103807/countable-structures-with-uncountable-many-automorphisms/103809#103809 Answer by Trevor Wilson for Countable structures with uncountable many automorphisms Trevor Wilson 2012-08-02T18:43:38Z 2012-08-02T18:50:35Z <p>Let $\vec{a} \in A^{&lt;\omega}$. There are uncountably many automorphisms $f$ and only countably many possible values for $f(\vec{a})$, so there must be two different automorphisms $f_1$ and $f_2$ with $f_1(\vec{a}) = f_2(\vec{a})$. Then $f_2^{-1} \circ f_1$ is a nontrivial automorphism fixing $\vec{a}$.</p> <p>In fact, there are uncountably many automorphisms moving $\vec{a}$ the same way, so fixing one and composing the others with its inverse gives uncountably many nontrivial automorphisms fixing $\vec{a}$.</p> http://mathoverflow.net/questions/103223/sigma-algebra-generated-by-od-sets sigma-algebra generated by OD sets Trevor Wilson 2012-07-26T17:44:59Z 2012-07-26T19:49:42Z <p>Assume $V=L(\mathbb{R})$ and the Axiom of Determinacy. Is every set of reals generated by ordinal-definable sets of reals under the operations of countable union and intersection?</p> <p>The class of sets generated in this way is Wadge-cofinal and not wellorderable (it contains $\lbrace x\rbrace$ for every $x \in \mathbb{R}$) so there don't seem to be obvious limitations on its extent.</p> <p>This question came up when I was trying to answer Asaf Karagila's "bonus question" here: <a href="http://mathoverflow.net/questions/102000/generating-family-for-the-lebesgue-sigma-algebra" rel="nofollow">http://mathoverflow.net/questions/102000/generating-family-for-the-lebesgue-sigma-algebra</a></p> http://mathoverflow.net/questions/103116/theta-and-the-hartogs-of-2-mathbb-r/103118#103118 Answer by Trevor Wilson for $\Theta$ and the Hartogs of $2^\mathbb R$ Trevor Wilson 2012-07-25T19:51:41Z 2012-07-25T21:14:38Z <p>Yes, it is consistent with AD. In $L(\mathbb{R})$, if AD holds then $\Theta$ injects into $\mathcal{P}(\mathbb{R})$, or equivalently, into $2^{\mathbb{R}}$. The OD sets of reals are Wadge-cofinal and $\Theta$ is regular, so we can define an injection $F : \Theta \to \mathcal{P}(\mathbb{R})$ by induction. Let $F(\alpha)$ be the $&lt;_{OD}$-least OD set of reals with Wadge rank greater than the Wadge ranks of all the sets of reals $F(\gamma)$, $\gamma &lt; \alpha$.</p> <p>Going further, we can determine the Hartogs number $\aleph(2^{\mathbb{R}})$ in all models of $AD + V=L(\mathcal{P}(\mathbb{R}))$. The proof splits into two cases corresponding to the length of the Solovay sequence being a successor ordinal or a limit ordinal.</p> <p>If the length of the Solovay sequence is a successor ordinal, this means that there is a set of reals $A$ such that every set of reals is OD from $A$ and a real. In this case $\aleph(2^{\mathbb{R}})$ has the largest possible value, namely $\Theta(2^{\mathbb{R}})$. Given a surjection $F : \mathcal{P}(\mathbb{R}) \to Z$ for any set $Z$ we can construct an injection $Z \to \mathcal{P}(\mathbb{R})$. Let $G(z)$ be the set of pairs $(x,y)$ of reals such that $y$ is in the least $OD_{A,x}$ set of reals $B$ with $F(B) = z$ (if it exists.)</p> <p>On the other hand, if the length of the Solovay sequence is a limit ordinal then $\aleph(2^{\mathbb{R}})$ is equal to $\Theta$; that is, to $\Theta(\mathbb{R})$. We have $\aleph(2^{\mathbb{R}}) \ge \Theta$ on general grounds like you said in the question. But any function $F : \Theta \to \mathcal{P}(\mathbb{R})$ is OD from a set $A$ of reals because $V=L(\mathcal{P}(\mathbb{R}))$, so the range of $F$ consists of $OD_A$ sets of reals. The Solovay sequence has limit length, so the range of $F$ cannot be all of $\mathcal{P}(\mathbb{R})$.</p> http://mathoverflow.net/questions/102505/sets-of-reals-amenable-to-each-lx Sets of reals amenable to each L[x] Trevor Wilson 2012-07-18T00:55:26Z 2012-07-25T19:01:32Z <p>If $A$ is a set of reals such that $A \cap L[x] \in L[x]$ for each real $x$, is there a real $z$ such that $A \cap L[x] \in OD_z^{L[x]}$ for a cone of $x$?</p> <p>This can be proved under the Axiom of Determinacy using a game of roughly the same complexity as $A$.</p> <p>Just assuming boldface ${\bf \Delta}^1_2$ determinacy, if $A\cap L[x] \in OD^{L[x]}_z$ for a cone of $x$ then it can be shown that $A$ itself is $OD_z$. So perhaps one can find a counterexample by finding a homogeneous forcing that does not add reals but adds a set $A$ of reals with $A \cap L[x] \in L[x]$ for each real $x$.</p> <p>(The assumption of ${\bf \Delta}^1_2$ determinacy is necessary here because starting with $V=L$ it is possible to force a new set $A$ with $A\cap L[x] \in OD^{L[x]}$ for each real $x$ by a homogeneous forcing that does not add reals.)</p> http://mathoverflow.net/questions/95865/examples-of-conjectures-that-were-widely-believed-to-be-true-but-later-proved-fal/102950#102950 Answer by Trevor Wilson for Examples of conjectures that were widely believed to be true but later proved false Trevor Wilson 2012-07-23T17:59:18Z 2012-07-23T17:59:18Z <p>An example from set theory: My understanding is that it was once widely believed that all reals appearing in canonical inner models of large cardinals (at least up to supercompact cardinals) would be $\Delta^1_3$ in a countable ordinal. This is because it was assumed that linear iterations, the only kind known at the time, would suffice to compare such inner models. This assumption turned out to fail at the level of Woodin cardinals, far below supercompact cardinals. The resulting non-linear iterations (iteration trees) are a basic part of inner model theory today, whereas canonical inner models for supercompact cardinals are still far out of reach.</p> http://mathoverflow.net/questions/102846/whats-the-difference-between-zfcgrothendieck-zfcinaccessible-cardinals-and-ta/102849#102849 Answer by Trevor Wilson for What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory? Trevor Wilson 2012-07-22T01:10:38Z 2012-07-22T02:12:39Z <p>Yes. Assume ZFC. If there is a proper class of inaccessible cardinals, then Tarski's Axiom A holds because whenever $\kappa$ is inaccessible, the rank initial segment $V_\kappa$ of $V$ is a Tarski set. Conversely, if Tarski's Axiom A holds then for every set $x$ there is a Tarski set $y$ with $x \in y$. We will show that $|y|$ is an inaccessible cardinal greater than $|x|$, proving the existence of a proper class of inaccessible cardinals.</p> <p>To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, given $\zeta &lt; \kappa$ we take a subset $z$ of $y$ of size $\zeta$. We have $z \in y$ because $y$ contains its small subsets. Then we have $\mathcal{P}(z) \in y$ because $y$ is closed under the power set operation. Finally $\mathcal{P}(\mathcal{P}(z)) \subset y$ because $y$ contains all subsets of its elements. This shows that $2^{2^{\zeta}} \le \kappa$ and therefore that $2^{\zeta} &lt; \kappa$.</p> <p>To show that the cardinality $\kappa$ of $y$ is regular, notice that if $\kappa$ is singular then by the closure of $y$ under small subsets we can get a family of $\kappa^{cof(\kappa)}$ many distinct sets in $y$, contradicting the fact that $\kappa^{cof( \kappa)} > \kappa$ (which is an instance of Koenig's Theorem.)</p> http://mathoverflow.net/questions/102829/is-the-mostowski-collapse-natural/102832#102832 Answer by Trevor Wilson for Is the Mostowski collapse natural? Trevor Wilson 2012-07-21T18:06:15Z 2012-07-21T18:06:15Z <p>If I understand the definition correctly at <a href="https://en.wikipedia.org/wiki/Reflective_subcategory" rel="nofollow">https://en.wikipedia.org/wiki/Reflective_subcategory</a>, the question boils down to showing that every elementary embedding $f: B \to A$ between wellfounded models uniquely factors through the transitive collapse of $B$. This is true: It factors through the transitive collapse of $B$ because the transitive collapse map is an isomorphism. Uniqueness follows from the fact that isomorphic wellfounded models (in particular, $B$, its transitive collapse, and the range of $f$) are uniquely isomorphic.</p> http://mathoverflow.net/questions/102810/to-find-an-element-of-a-pi1-1-set/102820#102820 Answer by Trevor Wilson for To find an element of a $\Pi^1_1$ set Trevor Wilson 2012-07-21T14:14:59Z 2012-07-21T14:14:59Z <p>I can't access the full paper at the moment, but I'm pretty sure this is exactly the question addressed in the paper "A Note on the Kondo-Addison Theorem" by D. Guaspari </p> http://mathoverflow.net/questions/102376/measures-that-are-not-od Measures that are not OD Trevor Wilson 2012-07-16T17:55:56Z 2012-07-16T18:43:47Z <p>Is anything known about the consistency strength of the statement:</p> <p>"There is a normal measure (on a cardinal) that is not ordinal-definable"?</p> <p>In particular, is it consistent relative to the existence of a measurable cardinal? It looks like it's consistent relative to the existence of a supercompact cardinal. If $\kappa$ is supercompact then we can force to make it Laver indestructible.</p> <p>So assume that $\kappa$ is still $(\kappa+2)$-strong after we add $(2^{2^\kappa})^+$ many Cohen subsets of $\kappa^+$, more than the number of measures on $\kappa$ in $V$. Solovay proved that if $\kappa$ is $(\kappa+2)$-strong then for every set $X \in V_{\kappa+2}$ there is a normal measure on $\kappa$ whose ultrapower contains $X$. So letting $X$ range over the Cohen subsets of $\kappa^+$ that we added, a counting argument shows that we must get some normal measures on $\kappa$ that are not in $V$. Cohen forcing is homogeneous, so these measures cannot be ordinal-definable. I don't know how strong this kind of indestructibility is, or whether it's necessary.</p> <p>I am also interested to know anything about countably complete measures on any set that are not ordinal-definable from that set.</p> http://mathoverflow.net/questions/81455/models-of-ad-different-from-l-mathbbr/82434#82434 Answer by Trevor Wilson for Models of $AD$ different from $L(\mathbb{R})$ Trevor Wilson 2011-12-02T07:18:33Z 2011-12-02T07:18:33Z <p>To answer the first question, like Andres mentioned, larger models $L(\Gamma,\mathbb{R})$ of AD can behave quite differently from $L(\mathbb{R})$. For example they can satisfy AD$_{\mathbb{R}}$, the axiom of determinacy for Gale-Stewart games played on $\mathbb{R}$, which fails in $L(\mathbb{R})$. (This is because it implies the Axiom of Uniformization, i.e., that every binary relation on $\mathbb{R}$ contains a function with the same domain, whereas in a model such as $L(\mathbb{R})$ where every set is ordinal-definable from a real, the set of pairs $(x,y)$ such that $y$ is not ordinal-definable from $x$ cannot be uniformized.)</p> <p>To add to Andreas's answer to the second question, there is a $\Sigma_1$ statement in the parameter $\mathbb{R}$ that is true under ZFC but false under AD, namely the existence of an injection $\omega_1 \to \mathbb{R}$. (This is easily seen to be inconsistent with a countably complete nonprincipal ultrafilter on $\omega_1$.)</p> http://mathoverflow.net/questions/125108/stationary-many-subsets-of-kappa-whose-order-type-is-a-cardinal-and-whose-in/125114#125114 Comment by Trevor Wilson Trevor Wilson 2013-03-21T02:20:02Z 2013-03-21T02:20:02Z Thanks. By the way, I should probably share my motivation: I think it is interesting that the simultaneous failure of $\square(\kappa)$ and $\square_\kappa$ is strong whereas the failure of either on its own is weak. I have an application where it's not clear that $\neg \square(\kappa) \And \neg \square_\kappa$ is enough, however, so hopefully I can strengthen $\neg \square(\kappa)$ to &quot;$\kappa$ is weakly compact&quot; and strengthen $\neg \square_\kappa$ slightly to something like in the question, but still maintain the property that they are weaker on their own than they are together. http://mathoverflow.net/questions/125108/stationary-many-subsets-of-kappa-whose-order-type-is-a-cardinal-and-whose-in/125114#125114 Comment by Trevor Wilson Trevor Wilson 2013-03-21T02:02:55Z 2013-03-21T02:02:55Z ...whereas on the other hand if we don't know that $\kappa$ is weakly compact I don't see any way to get more than two Mahlo cardinals out of it, so I thought maybe it could be forced from two Mahlo cardinals somehow (the first being $\kappa$ and the second becoming $\kappa^+$.) I don't know if this is plausible though. http://mathoverflow.net/questions/125108/stationary-many-subsets-of-kappa-whose-order-type-is-a-cardinal-and-whose-in/125114#125114 Comment by Trevor Wilson Trevor Wilson 2013-03-21T02:02:02Z 2013-03-21T02:02:02Z Yes, that is definitely worth mentioning. Ideally, I was hoping for something that implied the stationarity of the set in the question without also implying that $\kappa$ is weakly compact. This is because if we add the additional assumption that $\kappa$ is weakly compact, then $\square(\kappa)$ and $\square_\kappa$ both fail, and the lower bound for this coincides with the current state of the art in inner model theory (I think)... http://mathoverflow.net/questions/120463/homogeneous-namba-like-forcing Comment by Trevor Wilson Trevor Wilson 2013-02-15T02:29:53Z 2013-02-15T02:29:53Z @alephomega When I use that phrase I usually mean &quot;almost homogeneous forcing&quot;, which means that for any two conditions $p$ and $q$ in the forcing poset $\mathbb{P}$ there is an automorphism $\pi$ of $\mathbb{P}$ such that $\pi(p)$ is compatible with $q$. The only consequence of this that I am interested in is that the theory of the forcing extension with parameters from the ground model does not depend on the choice of generic filter. http://mathoverflow.net/questions/116827/definition-of-limit-and-zfc Comment by Trevor Wilson Trevor Wilson 2012-12-19T22:39:45Z 2012-12-19T22:39:45Z I don't see how &quot;in other words...&quot; applies. It seems like the first sentence just asks whether every function $f: \mathbb{R} \to \mathbb{R}$ is continuous. (The answer is no.) http://mathoverflow.net/questions/112139/cardinality-of-group-of-order-preserving-functions-from-r-to-r Comment by Trevor Wilson Trevor Wilson 2012-11-12T04:30:34Z 2012-11-12T04:30:34Z A tangential remark: your question seems to suggest that the only uncountable cardinal less than $\beth_2$ is $\beth_1$, but this is not a consequence of ZFC. http://mathoverflow.net/questions/110871/can-measures-be-added-by-forcing/110898#110898 Comment by Trevor Wilson Trevor Wilson 2012-10-28T14:06:15Z 2012-10-28T14:06:15Z Hi Joel, thank you for your answer. I would also be interested in hearing about the issue Fran&#231;ois raises, either in seeing even one example of a new measure that does not lift an old measure, or the stronger property you mention. By the way, I think there may be a typo in the answer above immediately after &quot;We may factor the forcing as&quot;. http://mathoverflow.net/questions/109925/proof-of-ad-every-set-of-reals-is-suslin-implies-ad-mathbbr Comment by Trevor Wilson Trevor Wilson 2012-10-17T21:37:49Z 2012-10-17T21:37:49Z @alephomega It looks like Jackson just mentions that the equivalence between AD$_\mathbb{R}$ and &quot;every set is Suslin&quot; was proved by Woodin. http://mathoverflow.net/questions/109925/proof-of-ad-every-set-of-reals-is-suslin-implies-ad-mathbbr Comment by Trevor Wilson Trevor Wilson 2012-10-17T16:57:35Z 2012-10-17T16:57:35Z Hi Andres, it looks like he just says it is due independently to Martin and Woodin, and is unpublished. http://mathoverflow.net/questions/109925/proof-of-ad-every-set-of-reals-is-suslin-implies-ad-mathbbr Comment by Trevor Wilson Trevor Wilson 2012-10-17T16:36:39Z 2012-10-17T16:36:39Z Sorry about the formatting. I cannot seem to get a subscript \mathbb{R} in the body of the question. http://mathoverflow.net/questions/108707/banach-tarski-vs-von-neumann Comment by Trevor Wilson Trevor Wilson 2012-10-03T17:04:53Z 2012-10-03T17:04:53Z @FeldmannDenis Brackets usually denote the commutator $[g, h] = g^{−1}h^{−1}gh.$ http://mathoverflow.net/questions/107803/terminology-for-generalized-relations Comment by Trevor Wilson Trevor Wilson 2012-09-21T21:17:18Z 2012-09-21T21:17:18Z In the case that the range is contained in $[0,1]$ it would be called a fuzzy relation, I think. http://mathoverflow.net/questions/107785/in-zf-when-is-a-disjoint-union-of-metrizable-spaces-metrizable Comment by Trevor Wilson Trevor Wilson 2012-09-21T20:17:51Z 2012-09-21T20:17:51Z @Andrej I think we are supposed to forget how the space is embedded in $\mathbb{R}^2$ http://mathoverflow.net/questions/107441/does-every-nonempty-definable-finite-set-have-a-definable-member Comment by Trevor Wilson Trevor Wilson 2012-09-18T21:54:25Z 2012-09-18T21:54:25Z @Garabed I assume you are talking about the set of square roots of $-1$ in $\mathbb{C}$. Whether this pair has a definable member depends on how you define $\mathbb{C}$. In the usual construction of $\mathbb{C}$ by pairs of real numbers, both imaginary units are definable. However, using Fran&#231;ois's answer below, one can show that it is consistent that there is a definable field isomorphic to $\mathbb{C}$ with no definable imaginary unit. See my answer to <a href="http://math.stackexchange.com/q/181464/39378" rel="nofollow">math.stackexchange.com/q/181464/39378</a> for a proof. http://mathoverflow.net/questions/107441/does-every-nonempty-definable-finite-set-have-a-definable-member/107474#107474 Comment by Trevor Wilson Trevor Wilson 2012-09-18T15:35:42Z 2012-09-18T15:35:42Z Oops, I read that wrong. Never mind!