User norman lewis perlmutter - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:27:13Z http://mathoverflow.net/feeds/user/3183 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77632/do-indiscernibility-embeddings-exist-for-an-initial-segment-of-an-inner-model-of Do indiscernibility embeddings exist for an initial segment of an inner model of many measurable cardinals? Norman Lewis Perlmutter 2011-10-09T23:36:39Z 2012-02-17T00:10:11Z <p><strong>Background</strong></p> <p>I am interested in elementary embeddings from a model of set theory into itself. One way of producing such elementary embeddings is when the model is generated by indiscernibles; this idea is very closely related to the existence of sharps. Jech and Kanamori discuss $0^\#$ and $0^\dagger$ in detail but don't tell me much about other sharps. More advanced resources are difficult to understand without a lot of background knowledge.</p> <p><strong>Hypotheses</strong></p> <p>Let $\theta$ be an inaccessible cardinal, and suppose that some set $A$ of measurable cardinals below $\theta$ is a stationary subset of $\theta$. For each $\kappa \in A$, let $\mu_\kappa$ be a normal measure on $\kappa$, and let $\mathcal{U} = \{ \langle \kappa, \mu_\kappa \rangle : \kappa \in A \}$. Let $L[\mathcal{U}]_\theta$ denote those elements of $L[\mathcal{U}]$ of rank less than $\theta$.</p> <p><strong>Question statement</strong></p> <p>Do there exist large cardinal assumptions which imply the existence of a closed unbounded set of ordinal indiscernibles for $L[\mathcal{U}]_{\theta}$ such that every order-preserving map of these indiscernibles extends to an elementary embedding $j:L[\mathcal{U}]_\theta \to L[\mathcal{U}]_\theta \, \, ?$</p> <p><strong>Remarks</strong></p> <p>The large cardinal assumptions may be on $\theta$, the elements of $A$, or some other large cardinal. The values of $\theta$, $A$, and the $\mu_\kappa$ may be chosen in whatever way you like subject to the hypotheses above -- I just want this to work in some example, not in every example. </p> <p>In <i>The Core Model</i>, Dodd mentions double mice, a generalization of $0^\dagger$. Maybe some version of these can be used to answer the question affirmatively, but I know nothing about them.</p> http://mathoverflow.net/questions/86595/the-consistency-of-martins-axiom/86739#86739 Answer by Norman Lewis Perlmutter for The consistency of Martin's Axiom Norman Lewis Perlmutter 2012-01-26T17:45:35Z 2012-01-26T17:45:35Z <p>The part that you are having trouble with essentially boils down to understanding the construction of a Booolean Algebra from a partial order.</p> <p>I don't have Jech with me right now, so this answer is off the top of my head. I will make corrections later if necessary.</p> <p>When one defines the Boolean algebra $B(P)$ corresponding to a separative partial order $P$, the members of the Boolean algebra are just taken to be arbitrary formal Boolean sums of antichains of the partial order. (In case $P$ is not separative, one must first take the separative quotient, but many common forcing notions are actually separative, so for simplicity assume $P$ is separative when first trying to understand this.)</p> <p>A Boolean sum corresponds to a logical disjunction -- that is, if $p$ forces $\phi$ and $q$ forces $\psi$, then $p+q$ forces $\phi$ or $\psi$. But we take arbitrary Boolean sums -- so infinite sums are allowed in addition to finite ones.</p> <p>One then shows that the collection of formal Boolean sums of antichains of $P$ satisfies the axioms for a Boolean algebra, which involves checking a lot of details. </p> <p>So the number of antichains in $P$ equals the number of elements in $B(P)$. In the case that $P$ satisfies the countable chain condition (which should really be called the countable antichain condition in this context, but is not for historical reasons), then every antichain of $P$ is at most countable, and so the size of the Boolean algebra equals the number of at most countable antichains in $P$ (this includes the finite antichains, not just the countably infinite ones, but the number of countably infinite antichains is the same as the number of at most countably infinite antichains.)</p> <p>But the question in your mind might be, why does it suffice to just consider formal Boolean sums of antichains, rather than of arbitrary subsets of P? After all, when we're working in the Boolean algebra, we want to be able to take Boolean sums of arbitrary subsets of $B(P)$ that may not be antichains. The idea is the same as the reason why considering only antichains suffices to define nice names. Suppose a subset $X$ of $P$ is not an antichain. Let $q \in X$ be stronger than some other element of $X$. Let $Y = X - {q}$. Then the Boolean sum of $Y$ is forcing equivalent to the Boolean sum of $X$. Repeat this process recursively until we get an antichain whose Boolean sum is equivalent to that of $X$. So we see that considering only antichains suffices in the construction of the Boolean algebra.</p> http://mathoverflow.net/questions/73251/characterizing-elementary-embeddings-of-l-and-l-alpha-under-0 Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0# Norman Lewis Perlmutter 2011-08-19T23:38:05Z 2011-08-21T19:20:46Z <p>Suppose 0# exists.</p> <p>It is clear that every order preserving map from the indiscernibles to the indiscernibles gives an elementary embedding from $L$ to $L$. Furthermore, following lemmas 18.7 and 18.8 of Jech, if $\alpha$ is an <i> <strike>infinite </strike> infinite limit</i> ordinal, an increasing map from alpha to beta gives an elementary embedding from $L_{i_\alpha}$ to $L_{i_\beta}$, where $i_\alpha$ is the $\alpha$-th indiscernible. This is because $L_{i_\alpha}$ equals the Skolem hull in itself of the first $\alpha$ indiscernibles. However, I am not clear on the following points.</p> <p>1) Is it the case that for a <i><strike>finite</strike> successor</i> ordinal, n, $L_{i_n}$ is necessarily equal to the Skolem hull in $L_{i_n}$ of the first n indiscernibles? Jech only proves this result for infinite ordinals. </p> <p>2) Is it possible that there could be an elementary embedding from $L$ to $L$, or from $L_{i_\alpha}$ to $L_{i_\beta}$ ($\alpha, \beta$ may be finite or infinite), that does not always map indiscernibles to indiscernibles? This sounds weird, but I'm not convinced it's impossible. As far as I know, there's no formula in $L$ that defines "$\alpha$ is a Silver indiscernible." (In fact there is no such formula -- see Andreas Blass's comment below.)</p> http://mathoverflow.net/questions/64280/a-question-about-aleph-1-closed-forcing-notions/73254#73254 Answer by Norman Lewis Perlmutter for A question about $\aleph_1-$closed forcing notions. Norman Lewis Perlmutter 2011-08-20T00:40:49Z 2011-08-20T00:40:49Z <p>I can't answer your question, but I can give a simpler sounding formulation that might be helpful. Analyze the question in two cases.</p> <p>Case 1: The continuum hypothesis holds <br> In this case, the statement is false, because any $&lt;\aleph_1$ -closed forcing of size $\aleph_1$ cannot collapse cardinals. The forcing to add a cohen subset to $\aleph_1$ is a nontrivial example of such a forcing. </p> <p>Case 2: The continuum hypothesis fails <br> In this case, it is a theorem that every $&lt;\aleph_1$-closed forcing notion which collapses a cardinal collapses the continuum. (See <a href="http://mathoverflow.net/questions/11633/is-it-possible-for-countably-closed-forcing-to-collapse-aleph-2-to-aleph-1%20%22this%20question%22" rel="nofollow">this question</a>, referenced by Joel in the comments.) But every such forcing is equivalent to the canonical collapse forcing to collapse the continuum to $\aleph_1$. The most general version of this latter theorem that I know of (although the degree of generality makes it hard to follow) can be found in Handbook of Boolean Algebras, Volume 2, Corollary 1.15. </p> <p>So really, your question boils down to whether every $&lt;\aleph_1$-closed forcing of size continuum is isomorphic to Coll$(\aleph_1, c)$ This sounds strange to me, but I can't prove it's false, and if Foreman is entertaining it, who am I to judge it?</p> http://mathoverflow.net/questions/72807/sigma-n-version-of-hod $\Sigma_n$ version of HOD Norman Lewis Perlmutter 2011-08-13T02:42:20Z 2011-08-13T04:14:41Z <p>Fix a natural number, $n \geq 1$. Consider the class, M, of all sets hereditarily ordinal-definable using some $\Sigma_n$ formula. Since there is a universal $\Sigma_n$ formula, M is definable. Is M necessarily a model of ZF? It seems to me that it is closed under Godel operations and almost universal for the same reasons that HOD is, and therefore a model of ZF. But I feel like I'm missing something, since I've never heard anything about this model. </p> <p>If it is a model of ZF, where can I learn more about it? Has anybody done any research about it? How does it relate to HOD?</p> <p>Note: I require $n \geq 1$ because the formula witnessing that HOD is almost universal is $\Sigma_1$.</p> http://mathoverflow.net/questions/47172/completeness-of-the-club-filter-without-ac Completeness of the club filter without AC Norman Lewis Perlmutter 2010-11-24T01:26:02Z 2010-11-25T06:59:40Z <p>Let $\kappa$ be a regular cardinal. If I understand correctly, the proof that the intersection of $&lt;\kappa$ many club subsets of $\kappa$ is a club does not require AC. However, the proof that the club filter on $\kappa$ is $\kappa$-complete does ostensibly require AC because given a sequence of sets in the club filter, we need to pick out a sequence of clubs that they contain in order to show that their intersection is in the club filter. So I'd like to know, in the absence of AC, how strong is the assertion that the club filter on $\kappa$ is $\kappa$-complete? Does it imply $AC_{\kappa}$, or is it weaker? If weaker, then how much weaker?</p> <p>Is there any well-known model of ZF without AC in which there is a proper class or better of regular cardinals $\kappa$ such that the club filter on $\kappa$ is $\kappa$-complete?</p> http://mathoverflow.net/questions/11505/closure-of-separative-quotients/14948#14948 Answer by Norman Lewis Perlmutter for closure of separative quotients Norman Lewis Perlmutter 2010-02-10T21:22:50Z 2010-02-10T21:22:50Z <p>Stevo Todorcevic answered this question for me at the MAMLS conference in honor of Richard Laver last weekend in Boulder, CO. Apparently, the answer is that examples of forcings that are closed, whose separative quotients are not closed, come up frequently, with one particular example being forcings involving semi-selective coideals studied by Ilija and Farah. </p> http://mathoverflow.net/questions/11505/closure-of-separative-quotients closure of separative quotients Norman Lewis Perlmutter 2010-01-12T05:09:12Z 2010-02-10T21:22:50Z <p>Does there exist a partial order, nontrivial for forcing, that is countably closed, but whose separative quotient is not countably closed? Supposing the answer is yes, then is there a partial order, nontrivial for forcing, that is countably closed, but is not forcing equivalent to any countably closed separative partial order? </p> <p>For those of you unfamiliar with the separative quotient of a partial order, it is defined as follows. Two elements of a partial order are compatible iff there is some element below both of them. We form the separative quotient of a partial order by taking equivalence classes: x is equivalent to y when x and y are compatible with the exact same things. We then define a new partial order for the separative quotient -- $x \leq y$ iff everything compatible with x is compatible with y.</p> <p>A partial order is said to be separative if whenever $x \nleq y$, there is $z \leq x$ such that z is incompatible with y. The separative quotient of any partial order is separative. </p> <p>Some of the ways, order-theoretically speaking, that two partial orders can be forcing equivalent are</p> <p>(1) They are isomorphic, or more generally, (2) A dense subset of one of them is isomorphic to a dense subset of the other. </p> http://mathoverflow.net/questions/11633/is-it-possible-for-countably-closed-forcing-to-collapse-aleph-2-to-aleph-1 Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum? Norman Lewis Perlmutter 2010-01-13T06:39:02Z 2010-02-10T20:55:37Z <p>Suppose the continuum is larger than $\aleph_2$. Does there exist a countably closed notion of forcing that collapses $\aleph_2$ to $\aleph_1$, but does not collapse the continuum to $\aleph_1$? Moreover, does there exist such a forcing notion that is separative and has size continuum? It is known (see below) that the canonical collapse Coll$(\aleph_1, \aleph_2)$ collapses the continuum. Trying something like the canonical collapse relativized to some inner model will fail to answer the question, because this forcing will not be countably closed in V. </p> <p><br> Background information:</p> <p>This question came up as a result of my studies of the following theorem. </p> <p>Let $\kappa &lt; \theta$ be cardinals, with $\kappa$ regular and $\theta^{&lt;\kappa} = \theta$. Then any forcing of size $\theta^{&lt;\kappa}$ which is separative and $&lt;\kappa$ closed and which collapses $\theta$ to $\kappa$ is forcing equivalent to the canonical collapse forcing Coll$(\kappa, \theta)$.</p> <p>I want to know whether this theorem still holds in the case where $\theta^{&lt;\kappa} = \theta$ fails. The question above is the simplest possible such case.</p> <p>The reason why Coll$(\aleph_1, \aleph_2)$ collapses the continuum (when CH fails) is that we can think of $\aleph_1$ as $\aleph_1$ many $\aleph_0$-blocks. Consider only the elements of Coll$(\aleph_1, \aleph_2)$ such that on each $\aleph_0$ block, they are either fully defined or fully undefined. This is a dense set in Coll$(\aleph_1, \aleph_2)$, and it's isomorphic to Coll$(\aleph_1, \aleph_2^{\aleph_0})$ = Coll$(\aleph_1, \bf{c})$.</p> http://mathoverflow.net/questions/11633/is-it-possible-for-countably-closed-forcing-to-collapse-aleph-2-to-aleph-1/14946#14946 Answer by Norman Lewis Perlmutter for Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum? Norman Lewis Perlmutter 2010-02-10T20:55:37Z 2010-02-10T20:55:37Z <p>I got the answer from Stevo Todorcevic last weekend at the MAMLS conference in honor of Richard Laver in Boulder, CO. He told me that it is an unpublished result of his that any semi-proper forcing which collapses $\aleph_2$ collapses the continuum. As countably closed forcing is semi-proper, the answer to my question is no. Stevo sketched a proof for me, but I do not remember it well enough to reproduce it here. </p> http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? Norman Lewis Perlmutter 2010-02-02T01:46:33Z 2010-02-07T03:28:53Z <p>It is a theorem of Solovay that any stationary subset of a regular cardinal, $\kappa$ can be decomposed into a disjoint union of $\kappa$ many disjoint stationary sets. As far as I know, the proof requires the axiom of choice. But is there some way to get a model, for instance a canonical inner model, in which ZF + $\neg$C holds and Solovay's Theorem fails?</p> <p>I am interested in this problem because Solovay's theorem can be used to prove the Kunen inconsistency, that is, that there is no elementary embedding j:V -->V, where j is allowed to be any class, under GBC. The Kunen inconsistency may be viewed as an upper bound on the hierarchy of large cardinals. Without choice, no one has yet proven the Kunen inconsistency (although it can be proven without choice if we restrict ourselves to definable j). So if there is hope of proving Solovay's Theorem without choice, we could use this to prove the Kunen inconsistency without choice.</p> http://mathoverflow.net/questions/13609/models-of-zfc-set-theory-getting-started/13775#13775 Answer by Norman Lewis Perlmutter for Models of ZFC Set Theory - Getting Started Norman Lewis Perlmutter 2010-02-02T02:27:13Z 2010-02-02T02:27:13Z <p>According to Godel's incompleteness theorem, ZFC cannot prove its own consistency. Therefore, it is relatively consistent with ZFC that there are not any set models of ZFC. In this case, there is still a proper class model of ZFC, namely the von Neumann universe, V, itself, among others (i.e. L, forcing extensions of V). However, the fact that V is a model of ZFC cannot be proven formally within ZFC. Indeed, truth in V cannot be defined in V due to a result of Tarski.</p> <p>If we allow for some stronger axioms, then we can get set models of ZFC. For instance, if there exists an inaccessible cardinal, $\kappa$, then $V_\kappa$ is a set model of ZFC.</p> http://mathoverflow.net/questions/8976/ordinals-that-are-not-sets/11638#11638 Answer by Norman Lewis Perlmutter for Ordinals that are not sets Norman Lewis Perlmutter 2010-01-13T07:11:02Z 2010-01-13T07:11:02Z <p>According to Kanamori, Reinhardt worked on extending the ordinals beyond the height of the universe. Kanamori talks about this in <em>The Higher Infinite</em>, page 313, where he cites the following reference from his bibliography.</p> <p>Reinhardt, William N. Remarks on reflection principles, large cardinals, and elementary embeddings, pages 189-205</p> <p>Jech, Thomas J. (ed.), <em>Axiomatic Set Theory</em>. Proceedings of Symposia in Pure Mathematics vol. 13, part 2. Providence, American Mathematical Society 1974. </p> http://mathoverflow.net/questions/8874/what-are-some-slogans-that-express-mathematical-tricks/11634#11634 Answer by Norman Lewis Perlmutter for What are some slogans that express mathematical tricks? Norman Lewis Perlmutter 2010-01-13T06:40:28Z 2010-01-13T06:40:28Z <p>I forget who this is attributed to, but someone said something like "A technique is a trick used twice."</p> http://mathoverflow.net/questions/7284/how-to-select-a-journal/7396#7396 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2013-05-31T21:39:28Z 2013-05-31T21:39:28Z I don't agree with Sam's comment. It is probably true for research grants and for applications to research-oriented jobs. But if you are applying to work at a liberal arts colleges, they may care that you are doing good research, even if nobody in the department can understand it. http://mathoverflow.net/questions/77632/do-indiscernibility-embeddings-exist-for-an-initial-segment-of-an-inner-model-of/88674#88674 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2012-02-17T09:08:11Z 2012-02-17T09:08:11Z Thanks for the answer. http://mathoverflow.net/questions/86595/the-consistency-of-martins-axiom/86739#86739 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2012-01-27T05:10:38Z 2012-01-27T05:10:38Z Yes, thanks for the correction. http://mathoverflow.net/questions/45257/when-can-we-detect-forcing Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2011-10-13T00:18:08Z 2011-10-13T00:18:08Z In the definition of consistent, do you mean to say &quot;forcing extension&quot; rather than &quot;elementary extension&quot;? http://mathoverflow.net/questions/77632/do-indiscernibility-embeddings-exist-for-an-initial-segment-of-an-inner-model-of Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2011-10-12T15:20:42Z 2011-10-12T15:20:42Z Thanks, Andreas. What is the intuition behind why this should work? Of course we can get indiscernibles from that assumption, but why should they be a club, and why should they generate the model? If I replaced $\mathcal{U}$ with an arbitrary subset of $V_\theta$, would it still work? http://mathoverflow.net/questions/5353/how-to-respond-to-i-was-never-much-good-at-maths-at-school/5493#5493 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2011-10-09T19:49:14Z 2011-10-09T19:49:14Z As a large cardinal theorist, I can gleefully answer &quot;yes&quot; to that one. http://mathoverflow.net/questions/73251/characterizing-elementary-embeddings-of-l-and-l-alpha-under-0/73292#73292 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2011-08-21T18:24:04Z 2011-08-21T18:24:04Z Ah, yes, infinite limit ordinals. Oops. http://mathoverflow.net/questions/67786/bijection-of-proper-classes/68122#68122 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2011-08-20T17:17:57Z 2011-08-20T17:17:57Z Is there an example of a set model of set theory with classes, in which the classes are, externally, not all of the same size? http://mathoverflow.net/questions/47172/completeness-of-the-club-filter-without-ac/47178#47178 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-11-24T02:00:03Z 2010-11-24T02:00:03Z As for comment (2), it is interesting, thanks. http://mathoverflow.net/questions/47172/completeness-of-the-club-filter-without-ac/47178#47178 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-11-24T01:57:13Z 2010-11-24T01:57:13Z Thanks for comment (1). I don't know so much about ZF without AC. No regular cardinals above $\omega_1$ -- that's kind of disturbing. I'm not sure whether my modification works better than what I had previously. http://mathoverflow.net/questions/11505/closure-of-separative-quotients/14948#14948 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-02-11T17:01:26Z 2010-02-11T17:01:26Z Yes, it must have been Ilijas Farah. I have a handwritten name in my notes, so it was hard to transcribe. I didn't get any details, though, and was unable to find the work online after a quick search. http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails/14455#14455 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-02-10T20:51:34Z 2010-02-10T20:51:34Z There is certainly not a known proof without choice -- otherwise the open question as to whether the Kunen inconsistency can be proven in ZF would be solved. http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails/13783#13783 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-02-04T05:17:37Z 2010-02-04T05:17:37Z Actually, I realized that since the Kunen inconsistency just needs Solovay's Theorem up high, your answer may not fully answer my question . . . but it's a great answer anyway. http://mathoverflow.net/questions/13770/model-of-zf-negc-in-which-solovays-theorem-on-stationary-sets-fails/13783#13783 Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-02-03T16:42:09Z 2010-02-03T16:42:09Z Thanks -- this fully answers my question. What does $\Theta$ refer to in the second paragraph? http://mathoverflow.net/questions/11505/closure-of-separative-quotients Comment by Norman Lewis Perlmutter Norman Lewis Perlmutter 2010-01-13T05:09:14Z 2010-01-13T05:09:14Z Thanks. I thought about adding the forcing tag, but it only had one other occurrence so far, and the site suggests not creating new tags.