Similar to Exercise 8 of Section 2.5, and the method discussed in this section, we extend the language to include an infinitely countable number of new constant symbols $c_{0}$ , $c_{1}$ , $c_{2}$ , $\ldots$ . For every $n\ge1$ , consider the message $\sigma_{n}$ saying that $c_{1}<c_{0}\wedge\ldots\wedge c_{n}<c_{n-1}$ . For every $n\ge1$ , there is a model $\mathfrak{A}_{n}$ in the new language of the sentences $\mbox{Th}\mathfrak{A}\cup\{\sigma_{1},\ldots,\sigma_{n}\}$ . We just take $\mathfrak{A}$ , and let $c_{k}^{\mathfrak{A}_{n}}=n-k$ , for $k=0,\ldots,n$ . Hence, every finite subset of $\Sigma=\mbox{Th}\mathfrak{A}\cup\{\sigma_{1},\sigma_{2},\ldots\}$ is satisfiable. By the compactness theorem, there is a structure $\mathfrak{B}'$ that satisfies $\Sigma$ . The restriction $\mathfrak{B}$ of $\mathfrak{B}'$ to the initial language is a model of $\mbox{Th}\mathfrak{A}$ . Indeed, if $\mathfrak{A}$ is a model of a sentence $\sigma$ (which does not use constant symbols) then $\sigma\in\mbox{Th}\mathfrak{A}$ and $\Sigma\vdash\sigma$ . Then, there is a deduction of $\sigma$ from $\Sigma$ which does not use constant symbols (for example, we can proof this using rule EI), and, hence, $\mathfrak{B}$ proves $\sigma$ , implying $\sigma\in\mbox{Th}\mathfrak{B}$ . By Exercise 26(a) of Section 2.2, $\mathfrak{B}$ models $\mbox{Th}\mathfrak{A}$ implies $\mathfrak{B}\equiv\mathfrak{A}$ (another way is just to see that, similarly, if $\sigma\notin\mbox{Th}\mathfrak{A}$ , then $\neg\sigma\in\mbox{Th}\mathfrak{A}$ , $\neg\sigma\in\mbox{Th}\mathfrak{B}$ , $\sigma\notin\mbox{Th}\mathfrak{B}$ ).

To see that there is an infinite descending chain in $|\mathfrak{B}|$ , we note that $<a,b>\in<^{\mathfrak{B}}$ iff $<a,b>\in<^{\mathfrak{B}'}$ , and $\vDash_{\mathfrak{B}'}\sigma_{i}$ $\Rightarrow\vDash_{\mathfrak{B}'}c_{i+1}<c_{i}$ $\Rightarrow<c_{i+1,}^{\mathfrak{B}'}c_{i}^{\mathfrak{B}'}>\in<^{\mathfrak{B}'}$ .

Note. This is another way to argue the Example on page 152.