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Math 310: Introduction to Mathematical Reasoning Summer 2006 Homework 6 - Solutions
- Suppose that ∆n → X is a surjection. Prove, by induction on n, that X is a finite set and |X| ≤ n.
Proof. Consider the base case, n = 1. Suppose for some set X we have a surjection f : ∆ 1 → X. Then X = {f (1)} and f is in fact a bijection, so X is finite and |X| = 1 ≤ 1. Now suppose that, for some k, for any set X, if we have a surjection from ∆k to X then X is finite and |X| ≤ k. We wish to prove that, for any set X, if we have a surjection f : ∆k+1 → X then X is finite and |X| ≤ k + 1. Intuitively, we want to think of X as arising from a set with at most k elements by adding one new element. We will make this rigorous. Let y = f (k + 1) ∈ X. Then X = (X − {y}) ∪ {y} and this union is disjoint. There are two cases to consider. Either f (i) 6 = y for any i ≤ k, or for some i ≤ k, y = f (k + 1) = f (i). If f (i) 6 = y for any i ≤ k, then f (i) ∈ (X − {y}) for all i ≤ k, and so we can obtain a new function g : ∆k → (X − {y}) by letting g(i) = f (i). This function is a surjection since f was a surjection, and we can apply our inductive hypothesis to conclude that (X − {y}) is finite and |(X − {y})| ≤ k, so |X| = |(X − {y}) ∪ {y}| = |(X − {y})| + |{y}| = |(X − {y})| + 1 ≤ k + 1, as desired. If f (i) = y for some i ≤ k, then we can define a new function g : ∆k → X by g(i) = f (i), and g will still be a surjection. We can apply our inductive hypothesis to conclude that X is finite and |X| ≤ k < k + 1, as desired. By induction, our proposition holds for all n.
- Suppose that X and Y are finite sets such that X ⊆ Y. Prove that
(a) X = Y ⇔ |X| = |Y |, (b) X ⊂ Y ⇔ |X| < |Y |.
(Recall that this means X ⊆ Y and X 6 = Y .)
Proof. This is easy.
- Prove that if A is a finite set and B is denumerable then A ∪ B is denumerable.
Proof. First, we observe that A ∪ B = (A − B) ∪ B, and A − B and B are disjoint. Since A is finite and A − B ⊆ A then A − B is finite. If A − B is empty, then A ∪ B = B is denumerable, and we are done. If A − B is non-empty, we can find for some natural number n a bijection f : { 1 , 2 , ..., n} → A − B. Since B is denumerable, we can find a bijection g : N → B. Now we construct a function h : N → (A − B) ∪ B = A ∪ B, by letting h(i) = f (i) for 1 ≤ i ≤ n, and h(i) = g(i − n) for i ≥ n + 1. Suppose that for some i and j, h(i) = h(j) = b. If i ≥ n + 1 then b ∈ B, so we must also have j ≥ n + 1 since for j ≤ n we have h(j) = f (j) ∈ A − B. But then h(i) = g(i − n) = h(j) = g(j − n) and since g is injective, i − n = j − n so i = j. Similarly if i ≤ n then j ≤ n and i = j. Hence h is an injection. But h is clearly a surjection, so h is a bijection and hence A ∪ B is denumerable.
- Prove that if {An|n ∈ N} is a denumerable set of pairwise disjoint denumerable sets then the union
n∈N
An = {x|x ∈ An for some n ∈ N}
is also denumerable.
(We say the sets An are pairwise disjoint if i 6 = j implies Ai ∩ Aj = ∅.)
Proof. Let A = {x|x ∈ An for some n ∈ N} be the set we are considering. Since each set An is denumerable, we have bijections fn : N → An for each n. Define a function f : N × N → A by f (n, m) = fn(m) ∈ An ⊂ A. We will show that this function is a bijection. First, suppose that f (n 1 , m 1 ) = f (n 2 , m 2 ) = a; then by definition fn 1 (m 1 ) = fn 2 (m 2 ) = a. But fn 1 (m 1 ) ∈ An 1 and fn 2 (m 2 ) ∈ An 2 , so a ∈ An 1 ∩ An 2. Since these sets are pairwise disjoint, we must have n 1 = n 2. But then since fn 1 is an injection, m 1 = m 2 and hence (n 1 , m 1 ) = (n 2 , m 2 ) and f is an injection. Now, for any a ∈ A, a ∈ An for some n, so since fn is a surjection there is some m ∈ N with fn(m) = a. But then a = fn(m) = f (n, m), so f is surjective. Hence f is a bijection. Since N × N is denumerable, and f : N × N → A is a bijection, then A is also denumerable.
- Prove that if X and Y are finite sets with |X| < |Y | there does not exist a surjection X → Y.
Proof. We will prove the contrapositive, that if X and Y are finite and there does exist a surjection f : X → Y then |Y | ≤ |X|. Since X is finite, for some natural number n there is a bijection φ : ∆n → X. Then the function f ◦ φ : ∆n → Y is a composition of two surjective functions and hence is surjective. By the first problem on this homework set, |Y | ≤ n = |X|, as desired.
