Can models of set theory contain extra ordinals? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:46:56Z http://mathoverflow.net/feeds/question/16134 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16134/can-models-of-set-theory-contain-extra-ordinals Can models of set theory contain extra ordinals? Mike Shulman 2010-02-23T06:51:11Z 2010-03-01T09:33:01Z <p>In the paper "Complete topoi representing models of set theory" by Blass and Scedrov, they consider a general notion of Boolean-valued model of set theory, and one of the conditions they impose is that the model contain "no extra ordinals after those of V", i.e. that for all z in the model we have</p> <p>$$\Vert z \text{ is an ordinal} \Vert = \bigvee_{\alpha \text{ is an ordinal of } V} \Vert z=\check{\alpha}\Vert$$</p> <p>where $\Vert-\Vert$ denotes the truth function of the model valued in some complete Boolean algebra.</p> <p>My question is: do there exist models which <em>do</em> contain "extra ordinals" in this sense? I presume so, or they wouldn't have needed to impose this condition. What do such models look like?</p> <p>(By way of clarification, certainly if the starting model V is a set model in some larger universe, then one can find other set models in that larger universe which contain more ordinals. But I'm interested in just starting with a single model V and building models from it, which can be sets or proper classes.)</p> http://mathoverflow.net/questions/16134/can-models-of-set-theory-contain-extra-ordinals/16157#16157 Answer by Joel David Hamkins for Can models of set theory contain extra ordinals? Joel David Hamkins 2010-02-23T13:45:44Z 2010-02-23T14:22:27Z <p>I have two answers. </p> <p>First, the standard method of building B-valued models of set theory, where B is any complete Boolean algebra, always satisfies your condition.</p> <p>Suppose that B is any complete Boolean algebra, and denote the original set-theoretic universe by V. One constructs the B-valued universe V<sup>B</sup> by building up the collection of B-names by recursion, so that &tau; is a B-name, if it consists of pairs &lang;&sigma;,b&rang; where &sigma; is a previously constructed name and b &isin; B. One may impose a B-valued structure on the class V<sup>B</sup> of all B-names, by first defining it for atomic formulas by induction on names and then extending to all formulas by induction on formulas. This is the usual way to do forcing with Boolean-valued models, and V<sup>B</sup> is the Boolean-valued structure that results.</p> <p>The remarkable thing, providing the power of forcing, is that every ZFC axiom gets Boolean value 1 in V<sup>B</sup>. In particular, the assertion that any two ordinals are comparable will have Boolean value 1.</p> <p>Suppose that z is any B-name. If &beta; is an ordinal above the Levy rank of z, that is, the place in the V<sub>&beta;</sub> hierarchy where the name z first exists, then it is not difficult to see that [[ &beta; &isin; z ]] = [[ &beta; = z ]] = 0. It follows that [[ z &isin; &beta; ]] = 1. But this latter Boolean value is the same as V<sub>&alpha;&lt;&beta;</sub> [[ z = &alpha; ]]. Thus, we have proved your identity</p> <ul> <li>[[ z is an ordinal ]] = V<sub>&alpha; &isin; ORD</sub>[[ z = &alpha; ]],</li> </ul> <p>since all of the terms in this join are 0 beyond &beta; and below &beta; it is the expression we already observed.</p> <p>There is a subtle point about whether your condition actually expresses "no new ordinals" or not. Suppose that V<sup>B</sup> is the B-valued model we have constructed and let U be any ultrafilter on B. One may form the quotient model V<sup>B</sup>/U, and there is a Los theorem, showing that the quotient satisfies &phi; if and only if [[ &phi; ]] &isin; U. If U is not V-generic, for example, if U is in V and B is not atomic, then there will be names z such that [[ z is an ordinal ]] = 1, but [[ z = &alpha; ]] &notin; U for any ordinal &alpha; in V. One way of thinking about this is that V<sup>B</sup> knows that z is definitely an ordinal, and by your property, V<sub>&alpha; &isin; ORD</sub> [[ z = &alpha; ]] = 1, but V<sup>B</sup> doesn't know that z is any particular ordinal &alpha;. Thus, the ultrafilter U is able to squeeze between these two requirements, and in the quotient, z is a new ordinal. But this doesn't contradict your property.</p> <p>Now, second, I can give a negative example. It is implicit in your question that the B-valued model somehow includes V, since you refer to the V ordinals &alpha; inside the Boolean brackets. Suppose that V is the univese of all sets, and let j:V to M be any elementary embedding that is not an isomorphism. For example, perhaps M is the ultrapower of V by an ultrafilter (M may or may not be well-founded). In particular, not every ordinal of M has the form j(&alpha;) for an ordinal &alpha; of V. Since M is a model of ZFC, we may regard it as a 2-valued Boolean model, or as a B-valued model for any B, since 2 = {0, 1} is a subalgebra of B. But M has ordinals not of the form j(&alpha;) for any V ordinal &alpha;. If one identifies V with its image in M, then this would provide a counterexample to the desired property.</p> http://mathoverflow.net/questions/16134/can-models-of-set-theory-contain-extra-ordinals/16165#16165 Answer by François G. Dorais for Can models of set theory contain extra ordinals? François G. Dorais 2010-02-23T14:46:20Z 2010-02-23T19:02:16Z <p>As Joel pointed out, that requirement is true for internal Boolean valued models. It is also true for symmetric models, permutation models (Blass &amp; Scedrov allow atoms), and a variety of mixed constructions. Blass and Scedrov were simply isolating the key common features of these constructions.</p> <p>That said, there are ways to violate this condition. By an earlier requirement by Blass and Scedrov on canonical names, this would require the Boolean valued model to be an end-extension of V. For example, an old result of Keisler and Morley says that every countable models of ZF has an elementary end-extension. There is no internal way to construct such things in plain ZFC since that would easily violate Gödel's Theorem, but it is possible to have decent approximations assuming large cardinals.</p> <p>There is an interesting construction by Sy Friedman where he essentially forces with a poset of size Ord<sup>+</sup>. (You have to jump through several hoops to do this, see Chapter 5 of Sy Friedman's book <em>Fine Structure and Class Forcing</em>.) Let loose, this forcing would naturally violate the no new ordinals condition. However, Friedman is careful to cut down the model so that no new ordinals appear.</p> <hr> <p>Here are a few additional remarks on a potential construction that comes very close to what you want.</p> <p>Other than the fact that we just ran out of ordinals, there is no real reason to stop the usual construction of L at Ord. We can construct L<sub>Ord+1</sub> in the same way except that we can't replace definitions for classes by actual sets. Nevertheless, we can define elements of L<sub>Ord+1</sub> to be (Gödel codes for) one-variable formulas with ordinal parameters, identifying formulas that define the same class. This is tricky to do internally since we have no truth definition, but we can work around that using 0<sup>#</sup>, assuming it exists in V. Once we have L<sub>Ord+1</sub>, we can similarly construct L<sub>Ord+2</sub>. In fact, with large enough cardinals, we can keep going like this for quite a while. There is no reason we will ever hit a model of ZF in this way, but I don't think it's impossible. To accomplish this, you need to have L live inside a model V where all the relevant information is packed into a set. This is where Friedman (building on seminal work of Jensen and others) picks up. His methods in Chapter 5 suggest that you could pack all of the relevant information into a single real number. Unfortunately, this is also where I stop, opening Friedman's book at a random page should explain why...</p>