Ultrapower
The intuitive idea behind ultrapower constructions (and ultraproduct constructions in general) is to take a sequence of already existing models and construct new ones from some combination of the already existing models. Ultrapower constructions are used in many major results involving elementary embeddings. A famous example is Scott's proof that the existence of a measurable cardinal implies $V\neq L$. Ultrapower embeddings are also used to characterize various large cardinal notions such as measurable, supercompact and certain formulations of rank into rank embeddings. Ultrapowers have a more concrete structure than general embeddings and are often easier to work with in proofs. Most of the results in this article can be found in [1].
Contents
General construction
The general construction of an ultrapower supposes given an index set $X$ set for a collection of (non-empty) models $M_i$ with $i\in X$ and an ultrafilter $U$ over $X$. The ultrafilter $U$ is used to define equivalence classes over the structure $\prod_{i\in X} M_i$, the collection of all functions $f$ with domain $X$ such that $f(i)\in M_i$ for each $i\in X$. When the $M_i$ are identical to one another, we form an ultrapower by "modding out" over the equivalence classes defined by $U$. In the general case where $M_i$ differs from $M_j$, we form a structure called the ultraproduct of $\langle M_i : i\in X\rangle$.
Two functions $f$ and $g$ are $U$-equivalent, denoted $f=_U g$, when the set of indices in $X$ where $f$ and $g$ agree is an element of the ultrafilter $U$ (intuitively, we think of $f$ and $g$ as disagreeing on a "small" subset of $X$). The $U$-equivalence class of $f$ is usually denoted $[f]$ and is the class of all functions $g\in \prod_{i\in X} M_i$ which are $U$-equivalent to $f$. When each $M_i$ happens to be the entire universe $V$, each $[f]$ is a proper class. To remedy this, we can use Scott's trick and only consider the members of $[f_U]$ of minimal rank to insure that $[f]$ is a set. The ultrapower (ultraproduct) is then denoted by $\text{Ult}_U(M) = M/U =\{[f]: f\in \prod_{i\in X} M_i\}$ with the membership relation defined by setting $[f]\in_U [g]$ when the set of all $i\in X$ such that $f(i)\in g(i)$ is in $U$.
Note that $U$ could be a principal ultrafilter over $X$ and in this case the ultraproduct is isomorphic to almost every $M_i$, so in this case nothing new or interesting is gained by considering the ultraproduct. However, even in the case where each $M_i$ is identical and $U$ is non-principal, we have the ultrapower isomorphic to each $M_i$ but the construction technique nevertheless yields interesting results. Typical ultrapower constructions concern the case $M_i=V$.
Formal definition
Given a collection of nonempty models $\langle M_i : i\in X \rangle$, we define the product of the collection $\langle M_i : i\in X \rangle$ as $$\prod_{i\in X}M_i = \{f:\text{dom}(f)=X \land (\forall i\in X)(f(i)\in M_i)\}$$
Given an ultrafilter $U$ on $X$, we then define the following relations on $\prod_{i\in X} M_i$: Let $f,g\in\prod_{i\in X} M_i$, then $$f =_U g \iff \{i\in X : f(i)=g(i)\}\in U$$ $$f \in_U g \iff \{i\in X : f(i)\in g(i)\}\in U$$
For each $f\in\prod_{i\in X} M_i$, we then define the equivalence class of $f$ in $=_U$ as follows: $$[f]=\{g: f=_U g \land \forall h(h=_U f \Rightarrow \text{rank}(g)\leq \text{rank}(h) \}$$
Every member of the equivalence class of $f$ has the same rank, therefore the equivalence class is always a set, even if $M_i = V$. We now define the ultraproduct of $\langle M_i : i\in X \rangle$ to be the model $\text{Ult}=(\text{Ult}_U\langle M_i : i\in X \rangle, \in_U)$ where: $$\text{Ult}_U\langle M_i : i\in X \rangle = \prod_{i\in X}M_i / U = \{[f]:f\in\prod_{i\in X}M_i\}$$
If there exists a model $M$ such that $M_i=M$ for all $i\in X$, then the ultraproduct is called the ultrapower of $M$, and is denoted $\text{Ult}_U(M)$.
Los' theorem
Los' theorem is the following statement: let $U$ be an ultrafilter on $X$ and $Ult$ be the ultraproduct model of some family of nonempty models $\langle M_i : i\in X \rangle$. Then, for every formula $\varphi(x_1,...,x_n)$ of set theory and $f_1,...,f_n \in \prod_{i\in X}M_i$, $$Ult\models\varphi([f_1],...,[f_n]) \iff \{i\in X : \varphi(f_1(i),...,f_n(i))\}\in U$$
In particular, an ultrapower $\text{Ult}=(\text{Ult}_U(M), \in_U)$ of a model $M$ is elementarily equivalent to $M$. This is a very important result: to see why, let $f_x(i)=x$ for all $x\in M$ and $i\in X$, and now let $j_U(x)=[f_x]$ for every $x\in M$. Then $j_U$ is an elementary embedding by Los' theorem, and is called the canonical ultrapower embedding $j_U:M\to\text{Ult}_U(M)$.
Properties of ultrapowers of the universe of sets
Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on some measurable cardinal $\kappa$ and $j_U:V\to\text{Ult}_U(V)$ be the canonical ultrapower embedding of the universe. Let $\text{Ult}=\text{Ult}_U(V)$ to simplify the notation. Then:
- $U\not\in\text{Ult}$
- $\text{Ult}^\kappa\subseteq\text{Ult}$
- $2^\kappa\leq(2^\kappa)^{\text{Ult}}<j_U(\kappa)<(2^\kappa)^+$
- If $\lambda>\kappa$ is a strong limit cardinal of cofinality $\neq\kappa$ then $j(\lambda)=\lambda$.
- If $\lambda$ is a limit ordinal of cofinality $\kappa$ then $j_U(\lambda)>lim_{\alpha\to\lambda}$ $j_U(\alpha)$, but if $\lambda$ has cofinality $\neq\kappa$, then $j_U(\lambda)=lim_{\alpha\to\lambda}$ $j_U(\alpha)$.
Also, the following statements are equivalent:
- $U$ is a normal measure
- For every $X\subseteq\kappa$, $X\in U$ if and only if $\kappa\in j_U(X)$.
- In $(\text{Ult}_U(V),\in_U)$, $\kappa=[d]$ where $d:\kappa\to\kappa$ is defined by $d(\alpha)=\alpha$ for every $\alpha<\kappa$.
Let $j:V\to M$ be a nontrivial elementary embedding of $V$ into some transitive model $M$ with critical point $\kappa$ (which is a measurable cardinal), also let $D=\{X\subseteq\kappa:\kappa\in j(X)\}$ be the canonical normal fine measure on $\kappa$. Then:
- There exists an elementary embedding $k:\text{Ult}\to M$ such that $k(j_D(x))=j(x)$ for every $x\in V$.
Iterated ultrapowers
Given a nonprincipal $\kappa$-complete ultrafilter $U$ on some measurable cardinal $\kappa$, we define the iterated ultrapowers the following way: $$(\text{Ult}^{(0)},E^{(0)})=(V,\in)$$ $$(\text{Ult}^{(\alpha+1)},E^{(\alpha+1)})=\text{Ult}_{U^{(\alpha)}}(\text{Ult}^{(\alpha)},E^{(\alpha)})$$ $$(\text{Ult}^{(\lambda)},E^{(\lambda)})=lim dir_{\alpha\to\lambda}\{(\text{Ult}^{(\alpha)},E^{(\alpha)}),i_{\alpha,\beta})$$ where $\lambda$ is a limit ordinal, $limdir$ denotes direct limit, $i_{\alpha,\beta} : \text{Ult}^{(\alpha)}\to \text{Ult}^{(\beta)}$ is an elementary embedding defined as follows: $$i_{\alpha,\alpha}(x)=j^{(\alpha)}(x)$$ $$i_{\alpha,\alpha+n}(x)=j^{(\alpha)}(j^{(\alpha+1)}(...(j^{(\alpha+n)}(x))...))$$ $$i_{\alpha,\lambda}(x)=lim_{\beta\to\lambda}i_{\alpha,\beta}(x)$$ and $j^{(\alpha)}:\text{Ult}^{(\alpha)}\to \text{Ult}^{(\alpha+1)}$ is the canonical ultrapower embedding from $\text{Ult}^{(\alpha)}$ to $\text{Ult}^{(\alpha+1)}$. Also, $U^{(\alpha)}=i_{0,\alpha}(U)$ and $\kappa^{(\alpha+1)}=i_{0,\alpha}(\kappa)$ where $\kappa=\kappa^{(0)}=crit(j^{(0)})$.
If $M$ is a transitive model of set theory and $U$ is (in $M$) a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, we can construct, within $M$, the iterated ultrapowers. Let us denote by $\text{Ult}^{(\alpha)}_U(M)$ the $\alpha$th iterated ultrapower, constructed in $M$.
Properties
- For every $\alpha$ the $\alpha$th iterated ultrapower $(\text{Ult}^{(\alpha)},E^{(\alpha)})$ is well-founded. This is due to $U$ being nonprincipal and $\kappa$-complete.
- The Factor Lemma: for every $\beta$, the iterated ultrapower $\text{Ult}^{(\beta)}_{U^{(\alpha)}}(\text{Ult}^{(\alpha)})$ is isomorphic to the iterated ultrapower $\text{Ult}^{(\alpha+\beta)}$.
- For every limit ordinal $\lambda$, $\text{Ult}^{(\lambda)}\subseteq \text{Ult}^{(\alpha)}$ for every $\alpha<\lambda$. Also, $\kappa^{(\lambda)}=lim_{\alpha\to\lambda}$ $\kappa^{(\alpha)}$.
- For every $\alpha$, $\beta$ such that $\alpha>\beta$, one has $\kappa^{(\alpha)}>\kappa^{(\beta)}$.
- If $\gamma<\kappa^{(\alpha)}$ then $i_{\alpha,\beta}(\gamma)=\gamma$ for all $\beta\geq\alpha$.
- If $X\subseteq\kappa^{(\alpha)}$ and $X\in \text{Ult}^{(\alpha)}$ then for all $\beta\geq\alpha$, one has $X=\kappa^{(\alpha)}\cap i_{\alpha,\beta}(X)$.
The representation lemma
References
- Jech, Thomas J. Set Theory. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex