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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:

  1. 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}) $
  2. By Modus Ponens: $\exists i, \mathfrak{A}_{i} \nprec \mathfrak{A}$
  3. $(\forall i, \mathfrak{A}_{i} \prec \mathfrak{A}) \wedge (\exists i, \mathfrak{A}_{i} \nprec \mathfrak{A}) \Rightarrow \bot$ (by 2 and 5)

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

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

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?

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  • $\begingroup$ Welcome to MSE. I suggest that you add the solution-verification tag to your question. $\endgroup$ – José Carlos Santos Feb 19 at 7:14
  • $\begingroup$ Sorry about that @JoséCarlosSantos , just edited the tag. $\endgroup$ – Vagoltof Feb 19 at 7:17
  • $\begingroup$ I just made a bunch of edits to your question. Most importantly: (1) use $\in$ (\in) instead of $\epsilon$ (\epsilon) for set membership, (2) the symbol $\mathfrak{A}$ is actually a mathfrak A, not a mathfrak U (the U looks like $\mathfrak{U}$). This is why we use $A$ for the domain of $\mathfrak{A}$, not $U$. $\endgroup$ – Alex Kruckman Feb 20 at 19:56
  • $\begingroup$ Sorry about that @AlexKruckman kind of a newbie with the Latex notation, and about te mathfrak A I really never realized it was actually an A, but thanks a lot for taking the bother of correcting my "proof"! $\endgroup$ – Vagoltof Feb 20 at 20:41
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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 $i$?

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  • $\begingroup$ I appreciate your response @AlexKruckman, and certainly, the fact of how "strong" the conclusion was should have been an alert, of which I will now and onwards be aware of, otherwise, I will work heeding your hint. I would like to ask some things regarding steps (4), (5) and (7) though, first regarding (4) if I had stablished the condition " $\exists \phi \in \mathfrak A_{i}$ " even if this is the incorrect route, would this have made the hypothesis valid? $\endgroup$ – Vagoltof Feb 21 at 4:38
  • $\begingroup$ Secondly in (4) I added the material condition (as it seemed intuitively correct given the prior correction) just so I could apply MP in 5, but it felt out of bounds (quite arbitrary), would it be valid to conclude by MP something that is entailed by an hypothesis withou stablishing a material conditional in the hypothesis itself? And regarding (7), I intended just to negate the condition of the conditional as in Modus Tollens, and just inverted the quantifiers and the sentence, isn't that the process of negation? Sorry if I have overstepped with these extra questions. $\endgroup$ – Vagoltof Feb 21 at 4:41
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    $\begingroup$ @Vagoltof You seem to be trying to give the proof in a very formal style, which actually makes it quite hard to understand the argument. You wrote "if I had stablished the condition `$\exists \phi \in \mathfrak{A}_i$' ... would this have made the hypothesis valid?" This makes no sense at all to me - there must be a typo, since $\phi$ is a formula not an element of $\mathfrak{A}_i$ - and the hypothesis (4) is something you're assuming for contradiction, right? What does it mean for it to be "valid"? $\endgroup$ – Alex Kruckman Feb 21 at 4:56
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    $\begingroup$ You then write "would it be valid to conclude by MP something that is entailed by an hypothesis without establishing a material conditional in the hypothesis itself". Again, I'm not sure what this means, but you seem to be asking a question about the logical form of the proof itself, not its content. I would strongly recommend that you get some more practice with proof writing and clear up any confusions you have about basic logic before you try to dive into a more advanced and abstract topic like model theory. $\endgroup$ – Alex Kruckman Feb 21 at 5:00
  • $\begingroup$ Well @AlexKruckman, I actually come from a logic-philosophy background, so while I have seen in, for example, Tent and Ziegler book more "informal" proofs, I have a specific need for proofs to be as logically unambiguous as possible, and for each step of an inference be formally stablished. Regarding the typo, yes, as a matter of fact I meant ' $\exists \phi \in Th(\mathfrak A_{i})$' ; by 'valid' in this case I mean, that by adding the correction, what the conditional implies can be correctly inferred ($\mathfrak A_{i} \not \prec \mathfrak A$). $\endgroup$ – Vagoltof Feb 21 at 5:12

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