# Section 2.2: Problem 12 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 be the structure . (The language is assumed to have equality and the parameters , , and . is the structure whose universe is the set of real numbers and such that and are the usual addition and multiplication operations.)
(a) Give a formula that defines in the interval .
(b) Give a formula that defines in the set .
(c) Show that any finite union of intervals, the endpoints of which are algebraic, is definable in . (The converse is also true; these are the only definable sets in the structure. But we will not prove this fact.)
(a) ( iff for some , ).
(b) ( iff for some , and for all , ).
(c) Suppose we have a finite union of intervals and defines , then defines the union. So, we need only to show that one interval with algebraic endpoints is definable in .
Suppose an interval has endpoints and such that and are defined by and . Then, let be in which each variable is substituted by , and be in which each variable is substituted by , and we have defines the closed interval . We can further require in addition that and/or do define , and . For infinite intervals we can skip those part that restrict from that end. Therefore, we need only to show that for any algebraic number , the set is definable in .
First we show that for any rational number , is definable in . Indeed, defines in the set .
Now, if an algebraic number is a root of equation with integer coefficients, then let be an interval with rational endpoints such that is the only root of that lies within , and be a wff that defines in the set (by this point we have already shown that such a set is definable in ). Then, defines in the set .