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Let $L$ be a first order language and $M$ an interpretation of $L$. If $A$ and $B$ are two substructures of $M$ and their intersection $C=A\cap B\ne \emptyset$, then is it the case that every sentence (in language $L$) true for both $A$ and $B$ is also true for $C$?

By "intersection" of $A$ and $B$, I mean the smaller substructure whose domain is the intersection of the domain of $A$ and $B$.

I came to this conjecture when I learned that there was a smallest standard submodel in a universe of some set theories (such as $ZF$) if there was a standard submodel in it. The smallest standard submodel is the intersection of all the standard submodels. But is the intersection of any two standard submodels still a standard submodel? If so, considered every standard submodel is nothing more than a substructure of the universe for which a group of sentences, i.e. the axioms of some set theory, are true, can we further generalize it into the conjecture I mentioned above?

I haven't found a proof or disproof for this so far (actually I am not very good at it). Can anyone prove it or give a counterexample?

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The question in the title is trivial. Thanks for answers from Andrej Bauer and 喻良. But whether the intersection of two standard submodels of $ZF$ is still a stanadard submodel? (which is my earliest concern and let's only focus on $ZF$ or $ZFC$ here)

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closed as off-topic by Andrés E. Caicedo, Andrej Bauer, Gro-Tsen, Andreas Thom, Pace Nielsen Jul 25 at 0:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Andrés E. Caicedo, Andrej Bauer, Andreas Thom, Pace Nielsen
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ My answer to this question gives an example of two transitive models of ZFC whose intersection does not satisfy ZF: mathoverflow.net/questions/297756/… $\endgroup$ – Gabe Goldberg Jul 24 at 13:59
  • $\begingroup$ This link is very helpful. I’ll reed it again after I learn more about forcing. $\endgroup$ – Zhiwei Jul 24 at 23:51
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If $\mathcal C$ is any substructure of a structure $\mathcal A$, then $\mathcal C$ is the intersection of $\mathcal A$ and an isomorphic copy of it, $\mathcal A'$, inside a larger structure $\mathcal M$. (Take an isomorphic copy of $\mathcal A$ and combine it with $\mathcal A$, identifying corresponding elements of $\mathcal C$.) So a sentence is preserved by intersections if and only if it's preserved by substructures. And that's well-known to be the case only for sentences equivalent to universal sentences.

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  • $\begingroup$ The way you demonstrate that a sentence is preserved by intersections implies it's preserved by substructures is impressive! And thank you for this informative answer! $\endgroup$ – Zhiwei Jul 24 at 23:50
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Of course not. For example, consider the empty language. A model is just a set. Take $M = \{1, 2, 3\}$, $A = \{1, 3\}$ and $B = \{2, 3\}$. Both $A$ and $B$ satisfy $\exists x y . x \neq y$, but $A \cap B$ does not.

Supplemental: The OP comments that the simple answer makes the original question look stupid. I disagree, no question is stupid. In fact, very often a seemingly stupid question can lead to interesting things.

So we have discovered that the answer to the question is negative. But now, rather than hiding in shame, we should drill some more, by asking trickier questions. For instance, can we put some further restrictions, either on the language, the statements, or the substructures, so that the answer becomes positive?

For example, suppose our language only has function symbols (on relation symbols, the only primitive relation is $=$). If the statement $\phi$ under consideration contains no $\exists$ and $\lor$, is its validity preserved by substructures? What if we have a statement with only $\forall$, $\land$ and $=$ (so we drop implications as well)?

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    $\begingroup$ The question does specify that the intersection is nonempty. But of course you could just add $0$ to all the sets and have the sentence say “there are two distinct elements”. $\endgroup$ – Gabe Conant Jul 24 at 12:03
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    $\begingroup$ @GabeConant: ah, I miseed that point. I fixed the answer to fit the question as stated. $\endgroup$ – Andrej Bauer Jul 24 at 12:18
  • $\begingroup$ Yes! It's quite simple! This answer just makes my question look stupid. $\endgroup$ – Zhiwei Jul 24 at 12:19
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    $\begingroup$ As I tell my students, there are no stupid questions. Actually, let me make a comment about that in my answer. $\endgroup$ – Andrej Bauer Jul 24 at 12:21
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    $\begingroup$ The OP comments that the simple answer makes the original question look stupid. The "Of course not" is also condescending and could very well contribute to OP's impression. $\endgroup$ – Najib Idrissi Jul 24 at 14:15
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Let $L$ be the language of multiplication, let $M=\mathbf{R},\ A=[0,1), B=(0,1]$. Then $\exists x(xx=x)$ is true in $A$ and in $B$ but not in $A \cap B$.

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Of course not. For example, if $g$ and $h$ are mutually Cohen reals, then $L[g]\cap L[h]=L$. But both $L[g]$ and $L[h]$ satisfy $V\neq L$.

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  • $\begingroup$ Thanks for your counterexample! I can stop my useless struggling on that problem now. But I need more time to digest Cohen reals. $\endgroup$ – Zhiwei Jul 24 at 11:52
  • $\begingroup$ Cohen forcing does not really matter here. The point is that $g$ and $h$ form a ``minimal pair". Most mutually reals have such property. $\endgroup$ – 喻 良 Jul 24 at 11:55
  • $\begingroup$ Hi, I'm still wondering whether the intersection of any two standard submodels is still a standard submodel. Could you please explain this for me? $\endgroup$ – Zhiwei Jul 24 at 12:07
  • $\begingroup$ Of course $L$ is the least proper class model of $ZF$. If you want a standard set model, you need an inaccessible cardinal. $\endgroup$ – 喻 良 Jul 24 at 12:11

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