I want to know if there are fairly simple combinatorial necessary conditions for when a direct limit of ultrapowers of $V$ is well-founded similar to $\sigma$-completeness. By combinatorial, I mean that these conditions are conditions on the ultrafilters instead of the elementary embeddings they produce.


For simplicity, we shall formulate the question with regards to the following ultrapower construction. Assume that $X$ is a set and $(P\_{n})\_{n}$ is a sequence of partitions of $X$ such that $P\_{n+1}$ refines $P\_{n}$ for all $n$ and whenever $x,y\in X,x\neq y$ there is some $n$ where $x,y$ belong to different blocks of the partition $P\_{n}$. Furthermore, assume that whenever $R\_{n}\in P\_{n}$ and $R\_{n+1}\subseteq R\_{n}$ for all $n$ then $\bigcap\_{n}R\_{n}$ is non-empty. Let $B=\bigcup\_{n\in\omega}\{\bigcup\mathcal{R}|\mathcal{R}\subseteq P\_{n}\}$. Then $B$ is a Boolean algebra. Let $U$ be an ultrafilter on $B$. We shall use the ultrafilter $U$ to construct an ultrapower.


Suppose $\mathcal{A}$ is a structure. Then let $\mathcal{A}^{(P\_{n})\_{n}}$ be the collection of all functions $f:X\rightarrow\mathcal{A}$ where there is some natural number $n$ where $P\_{n}\preceq\{f^{-1}[\{a\}]|a\in\mathcal{A}\}$ (the ordering $\preceq$ is refinement of partitions). Now let $\simeq\_{U}$ be the equivalence relation on $\mathcal{A}^{(P\_{n})\_{n}}$ where $f\simeq\_{U}g$ if and only if $\{x\in X|f(x)=g(x)\}\in U$. One can define the fundamental operations, constants, and relations on $\mathcal{A}^{(P\_{n})\_{n}}/\simeq\_{U}$ as one does with the classical ultrapower construction. Let the structure $\mathcal{A}^{(P\_{n})\_{n}}/\simeq\_{U}$ be denoted by
$\mathcal{A}^{(P\_{n})\_{n}}/U$.


Is there a fairly simple combinatorial characterization of when the ultrapower $V^{(P\_{n})\_{n}}/U$ well-founded? For instance, in order for $V^{(P\_{n})\_{n}}/U$ to be well-founded, the ultrafilters $\{\mathcal{R}\subseteq P\_{n}|\bigcup\mathcal{R}\in U\}$ must be $\sigma$-complete. On the other hand, if whenever $R\_{n}\in\mathcal{U}$ for all $n$, we have $\bigcap\_{n}R\_{n}\neq\emptyset$, then the ultrapower $V^{(P\_{n})\_{n}}/U$ is well-founded. Even though I just listed necessary conditions and sufficient conditions for when an ultrapower is well-founded, I do not know of any conditions that are both necessary and sufficient. Can someone give a reference or a proof of combinatorial necessary and sufficient conditions for when such an ultrapower is well-founded? If $V^{(P\_{n})\_{n}}/U$ is well-founded, then do we necessarily have $\bigcap\_{n}R\_{n}\neq\emptyset$ whenever $R\_{n}\in U$ for all $n$? 


$\textrm{This last question was answered affirmatively by Joel David Hamkins}$


$\textbf{Remark}$ 


Although I phrased this question in terms of a direct limit of a countable sequence of ultrapowers, this characterization also holds for any kind of direct limit of ultrapowers since any possible infinite descending sequence in an ultrapower of a well-founded set must be in a countable direct limit of ultrapowers. Furthermore, the condition that $\bigcap\_{n}R\_{n}\neq\emptyset$ whenever $R\_{n}\in P\_{n},R\_{n+1}\subseteq R\_{n}$ for all $ n$ does not restrict the types of ultrapowers that one can form since one can always extend the set $X$ to a larger set $\hat{X}$ so that $\bigcap\_{n}R\_{n}\neq\emptyset$ whenever $R\_{n}\in P\_{n},R\_{n+1}\subseteq R\_{n}$. For instance, give $X$ the uniformity where the uniform covers are generated by the uniform partitions $P\_{n}$. Then the appropriate set $\hat{X}$ is the completion of the uniform space $X$. Equivalently, one can let $\hat{X}=\varprojlim\_{n}P\_{n}.$ By replacing $X$ with $\hat{X}$ one obtains an isomorphic ultrapower but where $\bigcap\_{n}R\_{n}\neq\emptyset$ whenever $R\_{n}\in P\_{n},R\_{n+1}\subseteq R\_{n}$ for all $n$.



Is there a simple combinatorial characterization for when a direct limit of ultrapowers of $V$ is well-founded?
The answer is yes.


**Theorem.** The direct limit ultrapower you describe is
well-founded if and only if $\bigcap\_n A\_n\neq \emptyset$ whenever
$A\_n\in U$ for all $n$.


Proof. You've already noted the converse direction, since any
instance of ill-foundedness amounts to $[f\_{n+1}]\in\_U [f\_n]$,
which gives measure one sets $A\_n=\{x\mid f\_{n+1}(x)\in
f\_n(x)\}\in U$, with empty intersection.


For the forward implication, suppose that we have $A\_n\in U$ with $\bigcap\_n
A\_n=\emptyset$. Each $A\_n$ lives on some level of the sequence of partitions,
but let us assume for convenience (by padding if necessary) that
$A\_n$ respects the partition $P\_n$. We may also assume that
$A\_{n+1}\subset A\_n$.


There is a tree structure $T$ in this situation, whose nodes are
the elements of any of the partitions $P\_n$ that are in $A\_n$, ordered by inclusion. That is, if $b\in P\_n$ and $b\subset A\_n$, then we place 
$b\in T$, and we say that $b$ is a parent to any of the corresponding $c\in
P\_{n+1}$ with $c\subset b$ and $c\subset A\_{n+1}$, if there are
any such $c$.


The key insight to make is that this tree is well-founded. Any
infinite descending sequence of nodes in the tree amounts to a
descending sequence $R\_n\in P\_n$ of partition elements, and you
had stated that $\bigcap\_n R\_n\neq\emptyset$ in any such case. But
any element of that intersection would be in $\bigcap\_n A\_n$,
since $R\_n\subset A\_n$. So there can be no such infinite descending paths through the tree, and so the tree is well-founded. 


Because of this, we may define an ordinal ranking
function on the tree. Namely, the rank of any $b\in T$ is the supremum
of the $\text{rank}(c)+1$ for all $c\in T$ below $b$. This is well-defined by recursion on the tree. Note that the 
minimal nodes in $T$ get rank $0$, and the rank of any child node
is strictly smaller than the rank of its parent.


Consider the ranking function of the nodes of $T$ that happen to
be in $P\_n$. Let $f\_n(x)=\text{rank}(b)$ whenever $x\in b\in T\cap
P\_n$ be the corresponding rank function. This is an ordinal-valued
function with support in $P\_n$. But since the rank of a child node
is strictly less than the rank of a parent, we see that
$f\_{n+1}(x)<f\_n(x)$, whenever $x\in c\in T\cap P\_{n+1}$. Thus,
$[f\_{n+1}]<\_U [f\_n]$, and so we have found ill-foundedness in the
direct limit ultrapower. QED


If one doesn't insist that $\bigcap\_n R\_n\neq\emptyset$ when $R\_n$ are nested elements of the partition, then one can make a counterexample with principal ultrafilters concentrating on those partition elements.


