18.100pset1

# 18.100pset1 - 18.100B/C Fall 2008 Solutions to Homework 1 1...

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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 1 1. No, the given properties do not define an order if S has more than one element. As a counter-example, note that the relation defined by x y if and only if x = y satisfies all three conditions, but that the resulting relation ≺ is empty and so certainly not an order. If we replaced the first axiom with a statement such as • For all x,y ∈ S , x y or y x , the problem statement would be true. The axioms given in the problem define what is known as a partial order , while Definition 1.5 defines a total or linear order. One example of a nontrivial partial order is the division order on N , defined by m n if and only if m divides n . It’s easy to check that every integer divides itself, that if two integers divide each other they are equal, and that the divides relation has the transitivity property. However, there exist nonequal integers neither of which divides the other (e.g., 2 and 3). 2.(a) For a rational number p , define I p = { [ p,q ]; q ≥ p,q ∈ Q } . There is a natural bijection between I p and the set Q ≥ p = { q ∈ Q ; q ≥ p } ⊂ Q , so I p is at most countable. Then by the corollary to Theorem 2.12, I = S p ∈ Q I p is at most countable. Since it is clearly infinite, it is in fact countable. (b) Suppose for sake of contradiction that P is countable. Then there is a bijection f : N → P . Let D = { n ∈ N ; n 6∈ f ( n ) } . Then for any n ∈ N , n ∈ f ( n ) if and only if n 6∈ D . Since f is a bijection there must be some m ∈ N such that f ( m ) = D , but then by the preceding sentence m ∈ D if and only if m 6∈ D , a contradiction. Thus our assumption was false and P is not in fact countable. Alternatively, note that there is a natural bijection between P and sequences ( a 1 ,a 2 ,... ) whose elements are all 0 or 1: for a set S ∈ P , the associated sequence is defined by a i = 1 if and only if i ∈ S . But by Theorem 2.14, the set of such sequences is uncountable, so P is as well. Note that under this bijection, the set ¯ D corresponds to the anti-diagonal set discussed in class and used in the proof of Theorem 2.14....
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18.100pset1 - 18.100B/C Fall 2008 Solutions to Homework 1 1...

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