2
$\begingroup$

I need to prove that for every infinite model $\mathfrak A$ of signature $\sigma$ exists model $\mathfrak B$ with attributes:

  1. $\mathfrak A \equiv \mathfrak B$.
  2. $\parallel \mathfrak B \parallel = \max\{\omega, \parallel\sigma\parallel\}$.
  3. Not every element of $\mathfrak B$ is an interpretation of constant symbol from $\sigma$.

I think that it looks very similar to downward Löwenheim — Skolem theorem, but I don't understand how to use third statement.

Do I need to use the Compactness Theorem somehow?

I have some ideas.

So what I made:

1)I extended signature $\Sigma = \sigma \cup \{b\}$

2)$C$ is a set of constants from $\sigma$ and $\parallel C \parallel = \lambda$

3)$T = Th(\mathfrak A) \cup \{b \not=c_i \mid c_i \in C , i \in \lambda\}$

4)Then I viewed finite subset $\bar{T} \subseteq T$ and $\bar{T} = \bar{Th(\mathfrak A)} \cup \{b \not=c_i \mid c_i \in C , i \le k \in \lambda\}$

5)Then we can construct the new model $\bar{\mathfrak A} = <A, \sigma^{\mathfrak A},b>$ and $\bar{\mathfrak A} \models \bar{T}$. $\nu (b) \in A\setminus \{\nu (c_i) \mid i\le k \}$. Where $\nu$ is an interpretation.

6)According to the Compactness Theorem $\exists \mathfrak M$ of signature $\Sigma$ and $\mathfrak M \models T$

7)Then we excepted $b$ from $\Sigma$ and we got a new model $\mathfrak N$ of signature $\sigma$ which $\mathfrak N \equiv \mathfrak A$

8)According to the Löwenheim — Skolem theorem $\exists \mathfrak B$ such that $\parallel \mathfrak B \parallel = \max\{\omega, \parallel \sigma \parallel \}$ and $\mathfrak B \equiv \mathfrak N \equiv \mathfrak A$

Could you check my proof, please?

$\endgroup$
  • $\begingroup$ The third statement is easy to make true by adding suitable axioms. For every finite subset $F$ of the constant symbols, add an axiom that says there is an $x$ different from all elements of $F$. $\endgroup$ – André Nicolas Dec 22 '13 at 10:26
  • $\begingroup$ Let $C$ be the set of constants from $\sigma$. $\lambda = \parallel C \parallel$. Let $T = Th(\mathfrak A)\cup\{c_\alpha \not=c_\beta\mid\alpha , \beta \in \lambda ; \alpha \not= \beta\} \cup \{\exists x(x \not= c_i)\mid i\in\lambda\}$. According to the Compactness Theorem, $\exists\mathfrak B: \mathfrak B\models T$. Can I prove somehow that $\mathfrak A \equiv \mathfrak B$? $\endgroup$ – Alexander Dec 22 '13 at 12:44
  • $\begingroup$ You just need to use the full Lowenheim-Skolem, or prov this modification of it. The theory $T$ is not right. (1) We don't want $c_\alpha\ne c_\beta$, that might make the theory inconsistent; (ii) To make the model as big as $\sigma$, you need to add enough new constant symbols, with axioms they are different, so exactly like yours except new symbols; (iii) It is not enough to say $x\ne c_i$ for individual old constant symbols. We need either the version I gave above, or more simply a new constant symbol $k$ with axioms that say it is different from all the old $c_i$. $\endgroup$ – André Nicolas Dec 22 '13 at 16:17
  • $\begingroup$ I add some ideas upper, can you check it, please? $\endgroup$ – Alexander Dec 23 '13 at 14:58
  • $\begingroup$ At a quick read it looks OK, the right tools are used. I should write out a solution, but have too many Christmas responsibilities. Much later! $\endgroup$ – André Nicolas Dec 23 '13 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.