Elementary equivalence using countable models

[Logic Cloud]

Homework Statement

Prove that $$(\mathbb{R}, <)$$ and $$(\mathbb{R} \backslash \{0\}, <)$$ are elementary equivalent using the fact that there exist countable models $$(A, <_0)$$ and $$(B, <_1)$$ which are elementary equivalent with $$(\mathbb{R}, <)$$ and $$(\mathbb{R} \backslash \{0\}, <)$$ respectively.

Homework Equations

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The Attempt at a Solution

Once you prove that the two countable models are elementary equivalent, the desired result follows immediately. So suppose the countable models are not elementary equivalent. Then there exists a sentence s such that s is satisfied by A and ~s is satisfied by B. I don't know how to proceed from here. I'll probably have to use the uncountable models in some way as well but I don't see how. Any help will be much appreciated.

 

[mighty2000]

Thank you for your question. To prove that (\mathbb{R}, <) and (\mathbb{R} \backslash \{0\}, <) are elementary equivalent, we can use the fact that there exist countable models (A, <_0) and (B, <_1) which are elementary equivalent with (\mathbb{R}, <) and (\mathbb{R} \backslash \{0\}, <) respectively. This means that any first-order sentence that is true in one of these models is also true in the other model.

To prove this, we can use the Lowenheim-Skolem theorem, which states that for any first-order theory T with an infinite model, there exists a countable model of T. Since both (\mathbb{R}, <) and (\mathbb{R} \backslash \{0\}, <) are infinite models of the first-order theory of ordered fields, we can apply the Lowenheim-Skolem theorem to obtain the countable models (A, <_0) and (B, <_1).

Now, to show that these countable models are elementary equivalent, we can use the fact that any first-order sentence is true in a countable model if and only if it is true in the corresponding uncountable model. This means that if s is a sentence that is satisfied by A, then it is also satisfied by (\mathbb{R}, <), and if ~s is satisfied by B, then it is also satisfied by (\mathbb{R} \backslash \{0\}, <). Since both of these models are uncountable, this means that s is true in (\mathbb{R}, <) and ~s is true in (\mathbb{R} \backslash \{0\}, <). Therefore, s is satisfied by both (\mathbb{R}, <) and (\mathbb{R} \backslash \{0\}, <), and thus they are elementary equivalent.

I hope this helps to answer your question. Please let me know if you have any further questions or need any clarification.