# 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?

## 1 Answer

Yes, you're right. In fact we often can prove strict containment:

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, $x\in\bigcup \mathbb{A}^*\setminus \bigcup_{n\in\mathbb{N}}A_n^*$. $\quad\Box$

This argument works in much greater generality, of course. Basically, any time you have an appropriately nontrivial family of sets of reals, the union of the nonstandard versions of the individual sets in the sequence will be strictly smaller than the union of the nonstandard version of the whole sequence.