1
$\begingroup$

That is, prove that there is no set $T$ of purely universal sentences such that for every structure $A$ over the signature $\{\leq\}$, $A$ is a dense linear order iff $A\models T$.

$\endgroup$

closed as off-topic by Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon Feb 2 at 20:09

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

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, Adrian Keister, mrtaurho, Pierre-Guy Plamondon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Given a model of such a theory, consider whether a substructure must be a model as well. $\endgroup$ – Jishin Noben Jan 26 at 9:36
3
$\begingroup$

Suppose there was such a set $T$. Clearly $(\mathbb{Q}, \le) \models T$, as it is a dense linear order.

$(\mathbb{Z}, \le)$ is a substructure of $(\mathbb{Q}, \le)$ so $T$ also holds in this, because purely universal sentences stay true on subsets: $(\mathbb{Z}, \le) \models T$.

But $(\mathbb{Z}, \le)$ is not a dense linear order, which contradicts the assumption on $T$.

$\endgroup$
  • $\begingroup$ Why the purely universal sentences are true on subsets? $\endgroup$ – Gonzalo Jan 26 at 11:44
  • $\begingroup$ @Gonzalo $\forall x \phi(x)$ holds for the larger set, so also for the smaller. Induct on the structure of the formula. $\endgroup$ – Henno Brandsma Jan 26 at 11:49
  • $\begingroup$ Thank you for the clarification $\endgroup$ – Gonzalo Jan 26 at 12:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.