Section 2.6: Problem 3 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Establish the following facts:

(a) $\Sigma_{1}\subseteq\Sigma_{2}\Rightarrow\mbox{Mod}\Sigma_{2}\subseteq\mbox{Mod}\Sigma_{1}$ .

$\mathcal{K}_{1}\subseteq\mathcal{K}_{2}\Rightarrow\mbox{Th}\mathcal{K}_{2}\subseteq\mbox{Th}\mathcal{K}_{1}$ .

(b) $\Sigma\subseteq\mbox{Th}\mbox{Mod}\Sigma$ and $\mathcal{K}\subseteq\mbox{Mod}\mbox{Th}\mathcal{K}$ .

(c) $\mbox{Mod}\Sigma=\mbox{Mod}\mbox{Th}\mbox{Mod}\Sigma$ and $\mbox{Th}\mathcal{K}=\mbox{Th}\mbox{Mod}\mbox{Th}\mathcal{K}$ . (Part (c) follows from (a) and (b).)

(a) If $\mathfrak{A}$ is a model of $\Sigma_{2}$ , then it is a model of $\Sigma_{1}$ .

Similarly, if $\sigma$ is true in every structure of $\mathcal{K}_{2}$ , then $\sigma$ is true in every structure of $\mathcal{K}_{1}$ .

(b) Every sentence of $\Sigma$ is true in every model of $\Sigma$ , i.e. if $\sigma\in\Sigma$ , then $\sigma\in\mbox{Th}\mbox{Mod}\Sigma$ .

Every sentence true in every structure of $\mathcal{K}$ , is true in every structure of $\mathcal{K}$ , i.e. for every $\sigma\in\mbox{Th}\mathcal{K}$ , and for every $\mathfrak{A}\in\mathcal{K}$ , $\mathfrak{A}$ is a model of $\sigma$ , i.e. $\mathfrak{A}\in\mbox{Mod}\mbox{Th}\mathcal{K}$ .

(c) By (b), $\mbox{Mod}\Sigma\subseteq\mbox{Mod}\mbox{Th}\mbox{Mod}\Sigma$ , and $\Sigma\subseteq\mbox{Th}\mbox{Mod}\Sigma$ , so, by (a), $\mbox{Mod}\Sigma\supseteq\mbox{Mod}\mbox{Th}\mbox{Mod}\Sigma$ .

By (b), $\mbox{Th}\mathcal{K}\subseteq\mbox{Th}\mbox{Mod}\mbox{Th}\mathcal{K}$ , and $\mathcal{K}\subseteq\mbox{Mod}\mbox{Th}\mathcal{K}$ , so, by (a), $\mbox{Th}\mathcal{K}\supseteq\mbox{Th}\mbox{Mod}\mbox{Th}\mathcal{K}$ .

Note. The second equation of (c) holds in more general: $\Sigma=\mbox{Th}\mbox{Mod}\Sigma=\mbox{Cn}\Sigma\Leftrightarrow\Sigma\mbox{ is a theory}.$

Indeed, $\Rightarrow$ is immediate. To show $\Leftarrow$ , assume, according to (b), that for a theory $\Sigma$ , $\Sigma\subsetneq\mbox{Th}\mbox{Mod}\Sigma$ . Then, there is $\sigma\in\mbox{Th}\mbox{Mod}\Sigma-\Sigma$ . Then $\Sigma;\neg\sigma$ is consistent (otherwise, $\Sigma\vDash\sigma$ and $\sigma\in\Sigma$ ), and, hence, satisfiable. But then, there is a model of $\Sigma;\neg\sigma$ , which is a model of $\Sigma$ as well, such that $\sigma$ is false in the model, contradicting the assumption that $\sigma\in\mbox{Th}\mbox{Mod}\Sigma$ .

In particular, this equality means, that every theory $T$ is the theory of some class of structures, namely, $\mbox{Mod}T$ . This, however, does not mean that every theory is axiomatizable, as the definition of axiomatizability requires that $T=\mbox{Th}\mbox{Mod}\Sigma$ for some decidable $\Sigma$ .

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Section 2.6: Problem 3 Solution

Section 2.6: Problem 3 Solution