Ultrapowers, and Models of Set theory. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:03:28Z http://mathoverflow.net/feeds/question/52901 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52901/ultrapowers-and-models-of-set-theory Ultrapowers, and Models of Set theory. Michael Blackmon 2011-01-23T03:21:30Z 2011-01-23T06:38:16Z <p>Let $V$ be the set-theoretic universe, and suppose that $U$ is some ultrafilter over $\kappa\gt\omega.$ Then, we can go through the motions and produce the ultrapower $M = V^{\kappa}/U$. </p> <p>Now, the existence of an $\omega_1$-complete ultrafilter (and the existence of the transitive collapse of $M$) is subject to (as I currently understand it) the existence of a measurable cardinal. So my questions are:</p> <ol> <li>in the absence of a strong large cardinal hypothesis (like the existence of a measurable cardinal) what can we say about the ultrapower M? </li> <li>Is it possible to recover or produce some kind of embedding (which will not be a full elementary embedding) of $V$ into $M$, which preserves enough structure to work within $M$?</li> </ol> http://mathoverflow.net/questions/52901/ultrapowers-and-models-of-set-theory/52907#52907 Answer by Andres Caicedo for Ultrapowers, and Models of Set theory. Andres Caicedo 2011-01-23T04:15:27Z 2011-01-23T06:38:16Z <p>Michael, </p> <ul> <li>the existence of a well-founded $M$ and an embedding $j:V\to M$ different from the identity </li> </ul> <p>is <em>equivalent</em> to </p> <ul> <li>measurability, </li> </ul> <p>which is equivalent to </p> <ul> <li>the existence of an $\omega_1$-complete non-principal ultrafilter, </li> </ul> <p>which is equivalent to </p> <ul> <li>the existence of a non-principal ultrafilter ${\mathcal U}$ such that the model $M$ obtained by taking the ultrapower by ${\mathcal U}$ is an $\omega$-model, i.e., the natural number of $M$ are isomorphic to the natural numbers.</li> </ul> <p>This means that we really cannot say anything special along these lines about the ultrapower unless we are in the presence of measurable cardinals. (On the other hand, the ultrapower $M$ by an ultrafilter ${\mathcal U}$ on a set $\kappa$ is always an elementary map, so $M$ is elementarily equivalent to $V$ and, for each $V$-cardinal $\lambda$, the sets $\lambda^\kappa/{\mathcal U}$ are still linearly ordered (by dominance a.e.), even if ill-founded.)</p> <p>Of course, if we relax conditions then we can actually say a great deal. For example, the existence of $0^\sharp$, a significantly weaker assumption than a measurable, ensures an embedding from $L$ to itself, or equivalently, an "$L$-ultrafilter" $U$ such that the <em>internal</em> ultrapower of $L$ by $U$ is well-founded. </p> <p>Many large cardinal notions weaker than measurability can also be described in terms of elementary embeddings, though typically these are embeddings where the source and the target are (well-founded) sets rather than proper classes. The best known example of a large cardinal notion admitting such a characterization is weak compactness:</p> <blockquote> <p>$\kappa$ is weakly compact iff $\kappa$ is strongly inaccessible and for every transitive $M$ such that $\kappa\in M$, ${}^{&lt;\kappa}M\subseteq M$, $|M|=\kappa$, and $M$ models enough set theory there exists an elementary embedding $k:M\to N$ where $N$ is a transitive set and ${\rm cp}(k)=\kappa$.</p> </blockquote> <p>The existence of elementary or "partially elementary" embeddings between transitive sets is in general something that requires no assumptions beyond ZFC. For example, you obtain such embeddings by inverting transitive collapses of elementary hulls. Also, such embeddings guide the construction of square sequences in $L$.</p> <p>If you relax the condition that the embedding and the ultrafilter live in $V$ and allow them to exist in some forcing extension $V[G]$, then (consistently) we can have well-founded $j:V\to M$ where the critical point is $\omega_1$, for example. Also, we can this way obtain embeddings where $M$ is an $\omega$-model, but not well-founded.</p> <p>This is actually useful. Silver's original proof that SCH holds if it holds at singulars of cofinality $\omega$ proceeded by comparing $V$ and an $M$ as in the last paragraph (this idea led to the Galvin-Hajnal results on exponentiation that, in turn, led to pcf theory). The technique of generic embeddings has since proved to be very fruitful.</p> <p>I recommend that you take a look at the articles by James Cummings and Matthew Foreman in the Handbook of Set Theory for <em>much</em> more on this subject.</p> <p>That being said, the absence of measurability or "significant" large cardinals does not mean that different ultrafilters give us virtually indistinguishable ultrapowers $M$. In fact, Shelah's impressively fruitful pcf theory begins by looking at the cofinality (in $V$) of the ordered sets $\lambda^\kappa/{\mathcal U}$, and there is still much to explore here.</p> http://mathoverflow.net/questions/52901/ultrapowers-and-models-of-set-theory/52910#52910 Answer by Jason for Ultrapowers, and Models of Set theory. Jason 2011-01-23T04:46:27Z 2011-01-23T04:46:27Z <p>(1) Even in the absence of a large cardinal hypothesis, Łoś's theorem still applies so we still have $V \models \varphi(x_1, \ldots, x_n)$ if and only if $V^{\kappa}/U \models \varphi(c_{x_1}, \ldots, c_{x_n})$ where $c_{x_i}$ is the constant function assuming $x_i$ on every value in $\kappa$. In particular, we still will have an elementary embedding between the two structures. However, the problem with working with this structure directly is that if the ultrapower is nonprincipal, then there will even be nonstandard elements of $\omega^{V_{\kappa}/U}$ so none of the usual absoluteness arguments will carry over. Therefore, we really want $\in_U$ to be a well-founded relation on V_{\kappa}/U so we could compose the map with the transitive collapse to get an elementary embedding $j: V \rightarrow N$ for some proper inner model $N$ (i.e., transitive and containing all ordinals).</p> <p>As I was typing up this response, I see that Andres already replied with basically everything I was going to say about (2). Let me just add that part of the strength of having a nontrivial embedding $j: V \rightarrow N$ with critical point $\kappa$ ($\kappa$ is the least ordinal such that $j(\kappa) > \kappa$) is that we could always use such an embedding to get another elementary embedding $j: V \rightarrow N^{*}$ with critical point $\kappa$ where $N^*$ is transitive and contains all of its $\kappa$ sequences. We cannot even get $\omega$ closure for $M$ when $M$ contains a nonstandard $\omega$. </p>