# 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
.