Proof that $\bigcup_{i \epsilon I} A_{i}$ is an elementary substructure of $\mathfrak{U}$

Question

I'm currently studying model theory with the book "A Course in Model Theory" by Katrin Tent and Martin Ziegler. Currently I have written a proof for Exercise 2.1.1, and I would like to know if it's correct. The problem states:

Exercise 2.1.1. Let $\mathfrak{A}$ be an L-structure and ( $\mathfrak{A}_{i} )_{i \epsilon I}$ a chain of elementary substructures of $\mathfrak{A}$. Show that $\bigcup_{i \epsilon I} A_{i}$ is an elementary substructure of $\mathfrak{A}$.

My solution goes like this:

1. $\forall i, \forall a, (a \in A_{i} \Rightarrow a \in \bigcup_{i \epsilon I} A_{i})$
2. $\forall i, \mathfrak{A}_{i} \prec \mathfrak{A}$
3. $\forall i, \forall \phi, \forall a , ((\mathfrak{A}_{i} \prec \mathfrak{A}) \Leftrightarrow (\mathfrak{A}_{i} \models \phi (a) \Leftrightarrow \mathfrak{A} \models \phi (a) ))$

This first three are basically just definitions, the actual inferences are the following:

4. The hypothesis: $ \exists i, \exists a \in \bigcup_{i \epsilon I} A_{i}, ( \mathfrak{A} \not\models \phi(a) \Rightarrow \mathfrak{A}_{i} \nprec \mathfrak{A}) $
5. By Modus Ponens: $\exists i, \mathfrak{A}_{i} \nprec \mathfrak{A}$
6. $(\forall i, \mathfrak{A}_{i} \prec \mathfrak{A}) \wedge (\exists i, \mathfrak{A}_{i} \nprec \mathfrak{A}) \Rightarrow \bot$ (by 2 and 5)

And finally, by Modus Tollens on the contradictory hypothesis (4):

7. $\forall i, \forall a \in \bigcup_{i \epsilon I} A_{i}, \mathfrak{A} \models \phi(a)$

Which should satisfy Tarski's Test which according to the book states:

Let $\mathfrak{B}$ be an $L$-structure and $A$ a subset of $B$. Then $A$ is the universe of an elementary substructure if and only if every $L(A)$-forumla $\phi (x)$ which is satisfiable in $\mathfrak{B}$ can be satisfied by an element of $A$.

So again, does the proof seems correct? are there not any inferential jumps or plain just downright wrong inferences?

Answer 1

I don't understand the proof you've written. (5) does not follow from (4) by Modus Ponens, and (7) is not the negation of the hypothesis (4). Further, your proof concludes that $\phi(x)$ is satisfied by every element of $\bigcup_{i\in I}A_i$, and it should be clear that this is too strong. To apply Tarski's test, you need to consider a formula $\phi(x)$ which is satisfied by some element of $\mathfrak{A}$. But it's quite possible that it's satisfied by only one element!

Anyway, you have the right idea to apply Tarski's test. Let $\varphi(x,\bar{a})$ be an $L$-formula with parameters $\bar{a}$ from $\bigcup_{i\in I}A_i$, such that $\varphi(x,\bar{a})$ is satisfied in $\mathfrak{A}$, i.e., $\mathfrak{A}\models \exists x\, \varphi(x,\bar{a})$. You want to find some $b\in \bigcup_{i\in I} A_i$ such that $\mathfrak{A}\models \varphi(b,\bar{a})$. Then you're done by Tarski's test.

Hint: Use the assumption that $\mathfrak{A}_i\preceq \mathfrak{A}$ to pull satisfaction of $\exists x\, \varphi(x,\bar{a})$ down to $\mathfrak{A}_i$. Careful: not any $i\in I$ works. Why not? How can you pick the right $i$?