Download Rudin Exercise 2 on p. 43 A complex number z is said to be

Rudin Exercise 2 on p. 43
A complex number z is said to be algebraic if there are integers a0 ; : : : ; an , not all zero such that
a0 z n + a1 z n
1
+
an 1 z + zn = 0:
Prove that the set of all algebraic numbers is countable. Hint: For every positive integer N there are
only …nitely many equations with
n + ja0 j + ja1 j +
+ jan j = N:
Hints. Let P denote the set of complex polynomials with integer coe¢ cients minus the (trivial)
zero polynomial. De…ne f : P ! Z+ by
f a0 z n + a1 z n
Let PN = f
1
(N ). Then P =
1
+
[
N 2Z+
an 1 z + zn = n + ja0 j + ja1 j +
+ jan j :
PN (is a disjoint union). Let A denote the set of algebraic
numbers. For N 2 Z+ , let AN denote the set of complex zeroes of polynomials in PN , i.e.,
Then since P =
[
N 2Z+
AN = fz 2 C j P (z) = 0 for some P 2 PN g :
PN we have
A=
[
N 2Z+
AN :
Now each PN is a …nite set (see Rudin’s hint above). Moreover, each P 2 P has only a …nite number
of zeroes (by the Fundamental Theorem of Algebra).
Note that, as I mentioned brie‡y in class, Theorem 2.12 generalizes easily to the statement:
Let fEn g, n = 1; 2; 3; : : : ; be a sequence of at most countable sets, and put
S=
1
[
En :
n=1
Then S is at most countable.
Finally, give a very simple reason why the set A is in…nite. (Consider linear, i.e., degree one,
polynomials.)
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