Prompted by a really cool proof of the Hahn–Banach theorem that relies only on nonstandard-analysis and the ultrafilter lemma, I have decided to take some time to learn more about the non-standard analysis. This quest has lead me to this paper and the following theorem on transferred sets:
Finiteness Lemma For Transferred Sets.
Let $\phi:\mathbb R \hookrightarrow {}^*\mathbb R$ be any nonstandard model of the reals given by ultrafilter $\mathfrak U$. Then for any set $A \subseteq \mathbb R$ we have that $^*A \subseteq {}_*A$ implies that $A$ is finite.
In the interest of readability I will quickly define my notation for ultrapowers below:
For any indexing set $I$ and set $S$, pick a choice of product space $\prod_{i \in I}S$ with projectons $\left(\pi_i:\prod_{i \in I}S \to S\right)_{i \in I} $ and a proper free ultrafilter $\mathfrak U$ (here assuming the ultrafilter lemma). Let now $\epsilon: S \to \prod_{i \in I}S$ be given by $\pi_i \circ \epsilon= 1_S$ be given by universal property. Define the relation $\sim$ on $\prod_{i \in I}S$ by $x \sim y \iff \left\{i \in I \ \big | x_i = y_i\right\} \in \mathcal U$. It is a quick exercise to show that $\sim$ is an equivalence. Let then $\psi: \prod_{i \in I}S \twoheadrightarrow \prod_{i \in I}S/ \mathfrak U$ be a congruence of the form $\prod_{i \in I}S/\sim$. Further let $\phi = \psi \circ \epsilon$. It is a quick argument based on the fact that $\mathfrak U$ is a proper filter to show that $\phi$ is an injection. Define furthermore for all $A \subseteq S$ the following two sets $*A$ and ${}^*A$ called the inclusion and transferof $A$ into $\prod_{i \in I}S/ \mathfrak U$ respectively.
$$ _*A := \phi[A] = \left\{x \in \prod_{i \in I}S/ \mathfrak U \ \big| \ \exists a \in A: x = \phi(a) \right\} $$$$ {}^*A := \left\{x \in \prod_{i \in I}S/ \mathfrak U \ \big | \ \exists y \in \prod_{i \in I}S \ \left\{i \in I \ \big | y_i \in A\ \right\} \in \mathfrak U\right\} $$
Lastly we set $^* \mathbb R$ for a given $\mathfrak U$ ultrafilter on $\mathbb N$ to be $\prod_{n \in \mathbb N}\mathbb R / \mathfrak U$.
The proof of the lemma given in my source text is a hand wave, but I cobbled together a simple enough remark, given below, that implies the lemma when choice is assumed:
My Remark:
Let $I$ be a set with ultrafilter $\mathfrak U$ thereon. Further suppose that $\tau:I \hookrightarrow A$ is an injective function. Let then $\bar{\tau} \in \prod_{i \in I}S$ such that $\forall i \in I:\pi_i(\bar{\tau}) = \tau(i)$ and then $\beta := \psi(\bar{\tau}) \in {}^*A$. Suppose for contradiction that ${}^*A \subseteq *A$. thus we find some $a \in A$ such that $\phi(a) = \beta$. Set $T := \left\{i \in I \ \big | \ \bar{\tau}_i = a\right\}$, which lies in $\mathfrak U$ as $\phi(a) = \beta$. As $\tau$ is injective it follows that $T$ is either a singleton or empty. Either-way, as $T \in \mathfrak U$ this makes $\mathfrak U$ either non-proper or non-free - a contradiction! Thus ${}^*A \nsubseteq {}_*A$ and so $_*A \subsetneq {}^*A$.
Assuming choice, a set $A$ is infinite if and only if there is an injection from $\mathbb N$ into $A$. This means that for $A \subseteq \mathbb R$ and considering a choice $^*\mathbb R$ of hyperreals that ${}^* A \subseteq *A $ implies that $A$ is finite. Furthermore, this result holds so long as the indexing set $I$ is countable i.e. for all ultra powers with countable indexing set. However, if $I$ is uncountable then the argument of my proof does not hold as there exist nonfinite sets that cannot be injectively mapped to by $I$. This leads me to the following question:
My Question:
Is it true that for $S$ nonempty, indexing set $I$, an proper free ultrafilter $\mathfrak U \subseteq \mathcal P(I)$ that the ultrapower $\prod_{i \in I}S /\mathfrak U$ satisfies for any $A \subseteq S$ that ${}^*A \subseteq {}_*A$ only if $A$ is finite? Furthermore if true, is this provable without choice?
I have naively attempted to solve the problem by expanding upon the definition, but have hit a road block. See my working below:
My Thoughts:
Suppose that $x \in {}^*A \setminus {}_* A$ then we by the definition of ${}^*A$ and ${}_*A$ find some $\tau: I \to S$ such that $B := \left\{i \in I \ \big | \ \tau_i \in A\right\} \in \mathfrak U$ and $B_a := \left\{i \in I \ \big | \ \tau_i = a \right\} \not\in \mathfrak U $ for all $a \in A$. As $\emptyset \not\in \mathfrak U$ it follows that $\alpha := \tau_j \in A$ for some $j \in B$ and thus $A$ is nonempty. Set $C := I \setminus B$ and note $C \not\in \mathfrak U$. Note then as $\mathfrak U$ is an ultrafilter that $B_\alpha \cup C \not\in \mathfrak U$. Thus define $\tau^\prime: I \to A$ by $\tau^\prime(b) = \tau(b)$ for $b \in B$ and $\tau^\prime(c) = \alpha$ otherwise.
The above construction gives $I$ as a disjoint sum decomposition of the form $\bigsqcup_{a \in A}C_a$ for $C_a := {\tau^{\prime}}^{-1}[a] \not\in \mathfrak U$. Conversely such a decomposition $\bigsqcup_{a \in A}C_a$ gives a $\tau: I \to A$ by $\tau(i) = a \iff i \in C_a$ and hence an element $x \in {}^*A \setminus {}_* A$. This equivalce seems to show that if $A$ if sufficiently large in cardinality then ${}^*A \setminus {}_* A$ is nonempty. However this depends on the form of the ultrafilter $\mathfrak U$ and I don't know enough about the results for ultrafilters on uncountable sets to give a stronger result, including the question I am interested in. Does anyone have any thoughts?