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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 (Jech, 2003).
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 $M_i=V$.
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 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 $j_U:M\to\text{Ult}_U(M)$.
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:
Also, the following statements are equivalent:
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:
Definition of the ultrapower axiom needed here.
The ultrapower axiom ($\text{UA}$) has many significant consequences. Assume there is a strongly compact cardinal, then according to [1]:
$\text{UA}$ holds in all known inner models, but none of them contains a strongly compact cardinal, let alone a supercompact. It is currently unknown whether $\text{UA}$ is consistent with the existence of a supercompact or of a strongly compact. If $\text{UA}$ is consistent with a strongly compact, then supercompactness and strong compactness are equiconsistent: if there is a model of with a strongly compact, then there is a model with a supercompact.
Given a nonprincipal $\kappa$-complete ultrafilter $U$ on some measurable cardinal $\kappa$, we define the iterated ultrapowers the following way: \((\\text{Ult}^{(0)},E^{(0)})=(V,\\in)\) \((\\text{Ult}^{(\\alpha+1)},E^{(\\alpha+1)})=\\text{Ult}\_{U^{(\\alpha)}}(\\text{Ult}^{(\\alpha)},E^{(\\alpha)})\) \((\\text{Ult}^{(\\lambda)},E^{(\\lambda)})=\\text{lim dir}\_{\\alpha\\to\\lambda}\\{(\\text{Ult}^{(\\alpha)},E^{(\\alpha)}),i\_{\\alpha,\\beta})\) 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: \(i\_{\\alpha,\\alpha}(x)=j^{(\\alpha)}(x)\) \(i\_{\\alpha,\\alpha+n}(x)=j^{(\\alpha)}(j^{(\\alpha+1)}(...(j^{(\\alpha+n)}(x))...))\) \(i\_{\\alpha,\\lambda}(x)=\\mathrm{lim}\_{\\beta\\to\\lambda}i\_{\\alpha,\\beta}(x)\) 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 $\kappa=\kappa^{(0)}=\text{crit}(j^{(0)})$.
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 $M$.
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, $\kappa^{(\lambda)}=\text{lim}_{\alpha\to\lambda}$ $\kappa^{(\alpha)}$.
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 $X=\kappa^{(\alpha)}\cap i_{\alpha,\beta}(X)$.