Mathematics 220 Workshop Cardinality Some harder problems on cardinality. 1. Does there exist a continuous bijective function f : R ! R − f1g? Explain. Hint: Recall the Intermediate Value Theorem. Solution: The answer is \No". Suppose f : R ! R − f1g is bijective. Then there exists a 2 R, such that f(a) = 0, and there exists b 2 R, such that f(b) = 2. Then, since f is continuous, by the Intermediate Value Theorem, there exists c 2 (a; b), such that f(c) = 1. Thus, the range of such a continuous function has to contain 1. 2. Let A be any uncountable set, and let B ⊆ A be a countable subset of A. Prove that (1) A − B is uncountable, and (2) jAj = jA − Bj. Solution: (1) If A − B is countable, then as B is a subset of A, A = (A − B) [ B, and it follows that A is the union of two countable sets, therefore A is countable. Contradiction. (2) Since B is a proper subset of A by (1), there exists a c1 2 A − B. Next, consider A − B [ fc1g. Note that B [ fc1g ⊆ A is countable, so by (1), A − B [ fc1g is uncountable, in particular not empty. So there is c2 2 A−B[fc1g. Then, similarly, 9c3 2 A−B[fc1; c2g. Inductively, we can find cn 2 A − B for all n 2 N and ci 6= cj. Let C = fc1; : : : cn;::: g. Both C and C [ B are countably infinite, and therefore there exists a bijective function f : C [ B ! C. Now, define h : A ! A − B by: ( x; x 2 A − (C [ B) h(x) = f(x) x 2 C [ B: We claim that h is bijective. To check injectivity, assume h(x) = h(y). Case 1: x; y 2 A − (C [ B). Then h(x) = x; h(y) = y so x = y. Case 2: x; ; y 2 C [ B. Then f(x) = f(y). So x = y as f is injective. Case 3: x 2 A − (C [ B); y 2 C [ B. Then h(x) = x 62 C [ B, in particular, h(x) 62 C, and h(y) = f(y) 2 C. So we cannot have h(x) = h(y) in this case. We conclude that h is injective. To check surjectivity of h, let y 2 A−B. If y 2 C, then take x = f −1(y), so h(x) = f(x) = y. If y 62 C, then take x = y, so h(x) = h(y) = y. 3. (a) Prove that if A is a countable subset of real numbers, then jR − Aj = jRj. (b) Prove that jIj = jRj, where I is the set of irrationals as usual. (c) Show that j[0; 1]j = j[0; 1)j = j(0; 1)j. Solution: (a) It follows from Question 2. (b) It follows from (a) by taking A = Q. (c) Apply Question 2 to A = [0; 1];B = f1g for the 1st "=" and B = f0; 1g for the 2nd "=". Page 1 of 2 Mathematics 220 Workshop Cardinality Note: In the proof of Question 2, we constructed a bijection h from a set A to a proper subset of A. We can apply this general construction to the special cases above if you want to see (or are asked for) a bijection. 4. Let P = f0; 1gN be the set of all possible sequences of 0's and 1's. Prove that P is uncountable. Solution: Suppose P was countable. This means, we would have been able to list all the elements of P as P = fp1; p2;::: g. By definition of P , each element pi is a sequence of 0s and 1s. Let us make a new sequence a = a1a2a3 ::: , defined by ak = 0 if the sequence pk has a 1 in the kth place, and ak = 1 if the sequence pk has a 0 in the kth place. This new sequence is also an element of P , and a cannot coincide with any of the sequences pk, because a differs from pk at the kth term. We have arrived at the contradiction. Note that this is very similar to the proof of uncountability of R that we discussed in class. This is Cantor's `diagonal argument'. Page 2 of 2.