# Is the union of non-standard analogs of a family of sets a proper subset of the non-standard analog of the union of those sets?

The book on formal logic I'm using for self-study builds the foundation for non-standard analysis (to illustrate the usage of non-standard models) using the following construction.

Consider the real line $\mathbb{R}$ as a model of a very rich signature $\sigma$ which has a symbol for every $n$-ary function $\mathbb{R}^n \rightarrow \mathbb{R}$ and a symbol for every $n$-ary predicate on $\mathbb{R}$. Then there exists a non-trivial elementary extension $*\mathbb{R}$ (either by the Löwenheim–Skolem theorem, or directly by augmenting it with some fresh constant $\hat c$ and $\mathfrak{c}$ formulas of the form $\{ c < \hat c \mid c \in \mathbb{R} \}$ and then applying the compactness theorem).

Thus, for every $A \subset R$ we can consider the corresponding predicate (which exists by construction of $\sigma$) and its interpretation $*A \subset *\mathbb{R}$ and call that the non-standard analog of $A$.

Now consider some family $\{ A_\alpha \subset \mathbb{R} \}_{\alpha \in \mathcal{A}}$ indexed by some $\mathcal{A}$ and the union of its members $A = \cup_\alpha A_\alpha$. What can we say about $*A$ and $\cup_\alpha *A_\alpha$?

Clearly, if $\mathcal{A}$ is finite, then $*A = \cup_\alpha *A_\alpha$ (it's easy to show this by writing down a formula relating $A$ and $A_\alpha$ and using elementary equivalence).

What happens if $\mathcal{A}$ is infinite? The above doesn't work then (since we can't write an infinite formula). Moreover, it doesn't hold: consider $\mathcal{A} = \mathbb{N}$, and $A_i = \{ i \}$. Then $*A_i = \{ i \}$ as well and $\cup_i *A_i = \mathbb{N}$, but $*A = *\mathbb{N}$ can be shown to have more elements than $\mathbb{N}$ (as for every infinite set and its non-standard analog). Thus suggests that for infinite $\mathcal{A}$: $\cup_\alpha *A_\alpha \subset *(\cup_\alpha A_\alpha)$, but I was unable to prove that. Is that true?

On a second approach this looks more solvable. Let's just show that $\forall x : x \in \cup_\alpha *A_\alpha \Rightarrow x \in *A$. So, consider any such $x$. If $x \in \cup_\alpha *A_\alpha$, then (by definition of set union) this means there exists $\alpha_0$ such that $x \in *A_{\alpha_0}$. But $A_{\alpha_0} \subset A$, so by elementary equivalence $*A_{\alpha_0} \subset *A$, which means $x \in *A$, as required.

Does this look reasonable?

If $A_0\subsetneq A_1\subsetneq A_2\subsetneq ...$ is an infinite strictly increasing sequence of sets and $\mathbb{A}=(A_i)_{i\in\mathbb{N}}$ (note that we can code $\mathbb{A}$ as a single set of reals), then $\bigcup\mathbb{A}^*$ is strictly greater than $A_0^*\cup A_1^*\cup...$.
Proof: It's not hard to show that $\mathbb{A}^*$ is a $\mathbb{N}^*$-indexed sequence $(B_\alpha)_{\alpha\in\mathbb{N}^*}$ where $B_n=A_n^*$ for all standard $n$. Pick $\alpha\in\mathbb{N}^*\setminus\mathbb{N}$; by transfer, we have some $x\in B_\alpha$ which is not in any of the $B_\gamma$s for $\gamma<\alpha$. In particular, $x\in\bigcup \mathbb{A}^*\setminus \bigcup_{n\in\mathbb{N}}A_n^*$. $\quad\Box$