Unformatted text preview: elements of S1 with N, the set of natural numbers. Likewise, list the elements of S2 in their natural decreasing order to pair up the elements of this set with the numbers of N. To prove that S is countable, we need to pair up the elements of S with N. To do this, we pair the elements of S1 with the odd natural numbers and the elements of S2 with the even natural numbers: if an element s of S1 corresponds to n under the original pairing, under this new pairing it will correspond to 2n+1. If an element t of S2 corresponds to m, under the new pairing, let is correspond to 2m. This sets up a one
to
one correspondence between the elements of S and the elements of N, so S is countable. c) If W is any subset of the set Z of positive integers then infinitely many subsets of Z contain the same number of elements as W. This is FALSE as stated, because there is one exception. One corrected version that is true: If W is any non
empty subset of the set Z of positive integers then infinitely many subsets of Z contain the same number of elements as W. Proof. First, assume that W is a finite set. Then Z – M is an infinite set of integers. Let n be any element of W (n exists, since W is not empty). Then for any m ∈ Z – M let Wm be the (W – {n})∪{m}. Wm has the same number of elements as W, but each of these sets is different from the others, since m ∈ Wm but m ∈ Wn if m ≠ n. If W is infinite, W is countably infinite with the same number as N (by (b)). So W has the same number as Z – {n} for any n ∈ Z. And this is an infinite collection of subsets of Z. d) An uncountable subset may contain a countable subset. TRUE Proof. Since the statement says &qu...
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 Winter '11
 JamesKing
 Math, Sets, #, Max, 2m, Ω, Ψ, s00

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