I need to prove that for every infinite model $\mathfrak A$ of signature $\sigma$ exists model $\mathfrak B$ with attributes:
- $\mathfrak A \equiv \mathfrak B$.
- $\parallel \mathfrak B \parallel = \max\{\omega, \parallel\sigma\parallel\}$.
- 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?