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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.) 1