Book of Proof - Amazon S3

296 Solutions
11. Partition N into 8 countably infinite sets.
For each i ∈ {1,2,3,4,5,6,7,8}, let X i be those natural numbers that are congruent
to i modulo 8, that is,
X 1 = {1,9,17,25,33,...}
X 2 = {2,10,18,26,34,...}
X 3 = {3,11,19,27,35,...}
X 4 = {4,12,20,28,36,...}
X 5 = {5,13,21,29,37,...}
X 6 = {6,14,22,30,38,...}
X 7 = {7,15,13,31,39,...}
X 8 = {8,16,24,32,40,...}
13. If A = {X ⊂ N : X is finite }. Then |A| = ℵ 0 .
Proof. This is true. To show this we will describe how to arrange the items of
A in an infinite list X 1 , X 2 , X 3 , X 4 ,....
For each natural number n, let p n be the nth prime number, that is p 1 =
2, p 2 = 3, p 3 = 5, p 4 = 7, and so on. Now consider any element X ∈ A, so X =
{n 1 , n 2 , n 3 ,..., n k }, where k = |X| and n i ∈ N for each 1 ≤ i ≤ k. Define a function
f : A → N as follows: f (X) = p n1 p n2 ··· p nk . That is, we treat X ∈ A as an “index"
for the prime sequence and just map the entire set to the product of all the
primes with corresponding index. For example, take the set X = {1,2,3}. Then
f (X) = f ({1,2,3}) = p 1 p 2 p 3 = 2 · 3 · 5 = 30.
Note that f is an injection from A to N. Assume f (X) = f (Y ). Then, by definition
of the function, f (X) = p n1 p n2 ··· p nk . Similarly, f (Y ) = p m1 p m2 ··· p ms . By the
fundamental theorem of arithmetic, these are the prime decompositions of each
f (X) and f (Y ). Furthermore, the fundamental theorem guarantees that these
decompositions are unique. Hence {n 1 , n 2 ,..., n k } = {m 1 , m 2 ,..., m s } or X = Y .
This means each finite set X ⊆ N is associated with a unique natural number
f (X). Thus we can list the elements in X in A in increasing order of the
numbers f (X). The first several terms of this list would be
{1}, {2}, {3}, {1,2}, {4}, {1,3}, {5}, {6}, {1,4}, {2,3}, {7},...
It follows that A is countably infinite.
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Section 13.3 Exercises
1. Suppose B is an uncountable set and A is a set. Given that there is a surjective
function f : A → B, what can be said about the cardinality of A?