Section 2.6: Problem 2 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

Let $T_{1}$ and $T_{2}$ be theories (in the same language) such that (i) $T_{1}\subseteq T_{2}$ , (ii) $T_{1}$ is complete, and (iii) $T_{2}$ is satisfiable. Show that $T_{1}=T_{2}$ .

If $\sigma\in T_{2}$ , then $\neg\sigma\notin T_{2}$ , hence, $\neg\sigma\notin T_{1}$ , and $\sigma\in T_{1}$ .

Note. Another way to look at the problem, as suggested in the commentary to the book by the author, is to recall Exercise 8 of Section 2.2, which says that for a set of sentences $\Sigma$ such that for every sentence $\sigma$ , $\Sigma\vDash\sigma$ or $\Sigma\vDash\neg\sigma$ , and every model $\mathfrak{A}$ of $\Sigma$ , $\vDash_{\mathfrak{A}}\sigma$ iff $\Sigma\vDash\sigma$ . In our case, we just note that every model $\mathfrak{A}$ of $T_{2}$ is a model of $T_{1}$ , and, hence, by the result of the exercise, if $\sigma\in T_{2}$ then $T_{2}\vDash\sigma$ , then $\vDash_{\mathfrak{A}}\sigma$ for every model $\mathfrak{A}$ of $T_{2}$ , then $T_{1}\vDash\sigma$ , and $\sigma\in T_{1}$ .

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Section 2.6: Problem 2 Solution

Section 2.6: Problem 2 Solution