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Countable Sets Countable Sets Definition 6 Let A be a set. It is countably infinite if N ∼ A and is countable if it is finite or countably infinite. If it is not countable it is uncountable . Note that if A ∼ B then A is countable iff B is countable. Proposition 10 Every infinite subset of N is countably infinite. Proof 1 Let A ⊆ N, A not finite. Define f : N → A as follows: f(0) = min A, f (n + 1) = min( A\{ f(0) , f (1) ,...,f (n)}) where for ∅ 6= X ⊆ N, min X denotes the least element of X. f is defined for all n ∈ N since A is not finite, it is clearly injective and also surjective since any a ∈ A must be f(n) for some n ≤ a. Corollary 1 Every subset of N is countable. Proposition 11 Let A be a set. The following are equivalent. (i) A is countable. (ii) A ¹ N. (iii) A = ∅ or N º∗ A. Proof (i) = ⇒ (iii) Assume A is countable and not equal to ∅. Then there is f : Nn ∼ A for some n ∈ N, n > 0 or f : N ∼ A. In the latter case clearly f : N º∗ A and in the former case we can obtain h : N º∗ A by choosing a point a ∈ A and defining f(m) if m < n, h(m) = ½ a if m ≥ n. (iii) = ⇒ (ii) If A is empty then trivially A ¹ N. If f : N º∗ A then h : A → N defined by h(a) = min {n | f(n) = a} is an injection, so A ¹ N. (ii) = ⇒ (i) If A ¹ N then A ∼ B for some B ⊆ N. B is countable by Corollary 1 and hence A is countable. Proposition 12 Let A ¹ B. (i) If B is countable then so is A. (ii) If A is uncountable then so is B. 1This proof (like some other ones in this section) gives the idea of why the result should hold. A rigorous proof will follow from from a theorem on definition by recursion proved later on. 11 Proof (i) Using Propositions 11 and 4(ii), if B is countable then B ¹ N and hence since A ¹ B, we also have A ¹ N, so A is countable. (ii) follows from (i). Note It follows that any subset of a countable set is countable (by Propositions 4(i) and 12). Proposition 13 Let A º∗ B. (i) If A is countable then so is B. (ii) If B is uncountable then so is A. Proof Exercise. By a list we mean a sequence of objects x0, x 1, x 2,... , or x0, x 1, x 2,...,x n. We write {x0, x 1, x 2,... } or {x0, x 1, x 2,... x n} for the sets containing (just) the objects in these lists. An empty list does not feature any object. Proposition 14 Let A be a set. A is countable iff its elements can be arranged in a finite or infinite list, perhaps with repetitions. If the set is nonempty we can assume that the list is infinite (even if the set is finite). Proof This follows from Proposition 11. For the empty set we have the empty list, and otherwise countability of A is equivalent to the existence of a surjective function f : N º∗ A, which in turn is equivalent to the existence of a list featuring each element of A at least once (via the correspondence of f and the list f(0) , f (1) , f (2) ,... ). Proposition 15 If A and B are countable sets then A ∪ B and A × B are countable. Proof Assume A and B are countable. If one of them is empty then A ∪ B is equal to the other one, and A × B is empty, so the proposition holds. If they are not empty then by proposition 14 A = {a0, a 1, a 2,... }, B = {b0, b 1, b 2,... } so the elements of A ∪ B can be arranged in a list: A ∪ B = {a0, b 0.a 1, b 1, a 2, b 2,... } and hence A ∪ B is countable by Proposition 14 again. 12 The elements of A × B can be arranged in an array: ha0, b 0i h a0, b 1i h a0, b 2i h a0, b 3i h a0, b 4i . ha1, b 0i h a1, b 1i h a1, b 2i h a1, b 3i h a1, b 4i . ha2, b 0i h a2, b 1i h a2, b 2i h a2, b 3i h a2, b 4i . ha3, b 0i h a3, b 1i h a3, b 2i h a3, b 3i h a3, b 4i . .. and subsequently, following the diagonals, listed: A × B = {ha0, b 0i, ha0, b 1i, ha1, b 0i, ha0, b 2i, ha1, b 1i, ha2, b 0i, ha0, b 3i, ha1, b 2i,... } z }| { z }| { z }| { By Proposition 14, A × B is countable. ¥ Note It follows by induction that if Ak for k ∈ { 1, . n } are countable then n their union k=1 Ak is countable and their Cartesian product A1 ×A2 ×. ×An is countable.S In what follows we will often use the correspondence of countable sets and lists (Proposition 14) without further mention. ∞ Proposition 16 Let An for n ∈ N be countable sets. Then n=1 An is a countable set. S 1 Proof If some of the An are empty then we can just leave them out. If there are only finitely many non-empty sets left then the result follows by the above remark. Otherwise assume An are already the non-empty ones and let A0 = {a00 , a 01 , a 02 , a 03 ,... } A1 = {a10 , a 11 , a 12 , a 13 ,... } A2 = {a20 , a 21 , a 22 , a 23 ,... } . Then following the diagonals again ∞ An = {a00 , a 01 , a 10 , a 02 , a 11 , a 20 ,... }. ¥ n[=1 Note that the above proof requires us to choose a list for each An, simultaneously. 1This is the only proof in the naive part of the course that employs what is now known as the Axiom of Choice. Cantor often used it. 13 Theorem 1 The set N×N and the sets Z of integers and Q of rational numbers are all countably infinite sets. Proof For N × N countability is an immediate consequence of Proposition 15. Z is equal to Z− ∪ N, where Z− are the negative integers. We have Z− ¹ N via the injection f(k) = −k so Z− is countable by Proposition 11 and consequently so is Z (by Proposition 15). Finally, the function f : Z × N → Q: 0 if n = 0 f(m, n ) = m ½ n otherwise. is a surjection so since Z is countable by the above and Z × N is countable by Proposition 15, Q is countable by Proposition 13. We have N ¹ N × N, N ¹ Z and N ¹ Q (via the injections f1(n) = hn, 0i, f2(n) = n, f3(n) = n respectively), so none of the sets is finite by Propositions 7 and 8. Hence they are all countably infinite. ¥ Note. Cantor knew the above facts some years before 1873. By that time, he has been trying for some years to prove also that the set of real numbers is countably infinite . Some correspondence with Richard Dedekind helped him to switch focus and he then quickly proved that R is not countable. Later, other sets were shown to be uncountable. Theorem 2 (i) The set {0, 1}N is uncountable. (ii) The set R of real numbers is uncountable. Proof (i) Clearly {0, 1}N is not empty since e.g the function f : N → { 0, 1} defined by f(n) = 0 for all n ∈ N belongs to it, so assuming that {0, 1}N is count- N able means that we can arrange its elements in a list: {0, 1} = {f0, f 1, f 2, f 3,... } (using Proposition 14). Define a function g : N → { 0, 1} as follows: 0 if fn(n) = 1 , g(n) = ½ 1 if fn(n) = 0 . Then g is a function from N to {0, 1} but g differs from each fn, namely in the value it assigns to n: g(n) 6= fn(n). This is a contradiction because N N f0, f 1, f 2, f 3 . is a list of all elements of {0, 1} . We must conclude that {0, 1} is not countable. (ii) The above is an example of a diagonal argument due to Cantor (1891). To show that the set R is uncountable, we give Cantor’s original proof of it (1874). However, see Example 2.5 for the better known diagonal argument proof. 14 Lemma 1 Suppose that a set A contains at least 2 elements, it is linearly densely ordered 1 and has the following property: For any partition of A into two nonempty sets X and Y such that every element of X is less than every element of Y, there is c ∈ A so that every point less than c is in X and every point greater than c is in Y. Then A is not countable. The lemma is proved below, so since the set R satisfies the above conditions, it is uncountable. ¥ Proof of Lemma 1 Assume to the contrary that A is countable, A = {a0, a 1, a 2, a 3,... } (by Proposition 14). Define x0, x 1, x 2,... and y0, y 1, y 2,... as follows: Choose x0, y 0 ∈ A such that x0 < y 0. Choose xn+1 ∈ A to be am where m = min {k | xn < a k < y n}.
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