Mathematics 220 Workshop Cardinality

Some harder problems on cardinality.

1. Does there exist a continuous bijective function f : R → R − {1}? Explain. Hint: Recall the Intermediate Value Theorem.

Solution: The answer is “No”. Suppose f : R → R − {1} is bijective. Then there exists a ∈ R, such that f(a) = 0, and there exists b ∈ R, such that f(b) = 2. Then, since f is continuous, by the Intermediate Value Theorem, there exists c ∈ (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) |A| = |A − B|.

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 ∈ A − B. Next, consider A − B ∪ {c1}. Note that B ∪ {c1} ⊆ A is countable, so by (1), A − B ∪ {c1} is uncountable, in particular not empty. So there is c2 ∈ A−B∪{c1}. Then, similarly, ∃c3 ∈ A−B∪{c1, c2}. Inductively, we can find cn ∈ A − B for all n ∈ N and ci 6= cj. Let C = {c1, . . . cn,... }. 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 ∈ A − (C ∪ B) h(x) = f(x) x ∈ C ∪ B.

We claim that h is bijective. To check injectivity, assume h(x) = h(y). Case 1: x, y ∈ A − (C ∪ B). Then h(x) = x, h(y) = y so x = y. Case 2: x, , y ∈ C ∪ B. Then f(x) = f(y). So x = y as f is injective. Case 3: x ∈ A − (C ∪ B), y ∈ C ∪ B. Then h(x) = x 6∈ C ∪ B, in particular, h(x) 6∈ C, and h(y) = f(y) ∈ 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 ∈ A−B. If y ∈ C, then take x = f −1(y), so h(x) = f(x) = y. If y 6∈ 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 |R − A| = |R|. (b) Prove that |I| = |R|, where I is the set of irrationals as usual. (c) Show that |[0, 1]| = |[0, 1)| = |(0, 1)|.

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 = {1} for the 1st ”=” and B = {0, 1} for the 2nd ”=”.

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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 = {0, 1}N 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 = {p1, p2,... }. 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’.

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