Homework 1 (revised) Solutions

Homework 1 (revised) Solutions 1. Textbook, 1.1.1, # 1.1.2 (p. 24) Let S be an ordered set. Let A be a non-empty finite subset. Then A is bounded and sup A, inf A A Solution. The hint was: Use induction, so induction will be used. The first thing is to rephrase the exercise in such a way as to be able to apply induction; transform what has to be proved into a property that can be satisfied by an integer n. The way to rephrase it is as follows: Let A be a subset of n elements of the ordered set A, n N, n 1. Then A is bounded and inf A, sup A A. Since this is our first proof, I will be do a painfully detailed exposition. The proof is by induction on n, the number of elements of A. The case n = 1. In this case A = {a}, a singleton set. Claim a is an upper bound of A. In fact, let x A. Then x = a and a a. Thus a is an upper bound of A, hence A is bounded above. Similarly one sees that a is a lower bound of A, hence A is bounded below. Thus A is bounded. This may actually a good moment to interrupt the proceedings at hand and prove the result of Exercise 3 (1.1.5 in the textbook), namely that if an upper bound of a set is in the set, that upper bound is the supremum. In fact, it already has one property of the supremum, the property of being an upper bound. If now b is any upper bound, it is larger than the upper bound in the set, since the upper bound in the set is in the set! This is essentially the proof, but let us be really organized and state it as a lemma, and give a more detailed proof. Lemma Let A be a subset of an ordered set S. Assume b is an upper bound of A and b A. Then b = sup A. Proof. Since b is an upper bound of A, it already has one of the two properties of the supremum. To see it has the other one, i.e., that it is less than or equal all upper bounds of A, let c be an upper bound of A. Since b A we have b c. Done. We get back to the proof at hand. We were considering a singleton set A = {a} and saw that a is both an upper and lower bound of A. Thus, by the Lemma, a = sup A = inf A, proving everything in case n = 1. Assume now the result proved for all subsets of n elements of S, for some n 1. Let A be a subset of S of n + 1 elements; say A = {a 1,..., a n+1 }. Let A = {a 1,..., a n }, which is a set of n elements hence, by the induction hypothesis it is bounded and inf A, sup A A. Without loss of generality we may assume that inf A = a 1 and sup A = a n if n 2; if n = 1 then

2 inf A = sup A = a 1. In all cases a n = sup A. By trichotomy either a n < a n+1 or a n+1 < a n. (The case a n = a n=1 is excluded; otherwise A would have n, not n + 1 elements.) If a n < a n+1 then a i < a n+1 for i = 1,... n (since a n = sup A); moreover, a n+1 = a n+1 so a n+1 is an upper bound of A. Since a n+1 A, we conclude that a n+1 = sup A. On the other hand, if a n+1 < a n, then a i a n for all i = 1,..., n + 1; so a n is an upper bound of A, hence it is the sup of A, being in A. In either case sup A A (incidentally, the existence of a sup implies the set is bounded above). Similarly one proves that inf A A; with inf A being either a 1 or a n+1. This concludes the proof of the statement of the exercise. A different proof, probably better, not using induction, can be found below as part of Exercise 4. 2. Textbook, 1.1.1, # 1.1.4 (p. 25) Let S be an ordered set. Let B S be bounded (above and below). Let A B be a nonempty subset. Suppose all the infs and sups exist. Show that inf B inf A sup A sup B. Solution. Let us give a name to the objects. I ll use Greek letters. Let α = inf A, β = inf B, γ = sup A, δ = sup B. For the first inequality; which is β α, let x A. Then x B and since β is a lower bound of B we have β x. Since x was an arbitrary element of A, this shows that β is a lower bound of A, thus β α since α is the greatest lower bound of A. For the second inequality, α γ it is essential to assume A is not empty. Then we are authorized to say 1 : Let x A. Then α x since α is a lower bound of A and x γ, since γ is an upper bound of A. Thus, by transitivity, α γ. For the final inequality, namely, γ δ, we proceed similarly as we did for the first. Let x A (what happens if there is no x in A? Yes, we have the hypothesis A, but this is not necessary for this last inequality). Then x B, hence x δ = sup B. This shows δ is an upper bound of A, hence δ γ, γ being the least upper bound of A. 3. Textbook, 1.1.1, # 1.1.5 (p. 25) Let S be an ordered set. Let A S and suppose b is an upper bound for A. Suppose b A. Show that b = supa. Solution. This was proved as Lemma in the proof of Exercise 1. 1 Why was saying x A legal when we said it earlier in this solution?

3 4. Textbook, 1.1.1, # 1.1.6 (p. 25) Let S be an ordered set. Let A S be a nonempty subset that is bounded above. Suppose sup A exists and sup A / A. Show that A contains a countably infinite subset. In particular, A is infinite. Solution. This exercise is essentially the same as exercise 1; if one takes into consideration that every infinite set contains a countably infinite one. Proof by contradiction: Assume A is finite. By Exercise 1, sup A A, contradicting the hypothesis that sup A / A. However, a better approach to both exercises 1 and 4 is to do 4 first. Here is a direct proof of Exercise 4. The hypothesis is that sup A exists and sup A / A. Also A is assumed to be non-empty. Since it nonempty, there is a A. Now we will define a one-to-one function f : N A. We proceed by induction. The first step is to set f(1) = a. Assume now f defined for 1, 2..., n, for some n 1 in such a way that f(i) < f(j) if 1 i < j n. This has been done for n = 1. Since f(n) A and sup A / A, we must have f(n) < sup A. Thus f(n) is not an upper bound of A (being < than the least upper bound), hence there is an element in A that is larger than f(n), set f(n + 1) equal to such an element. In more detail, there is a A, f(n) < a ; set f(n + 1) = a. By induction, f is defined for all n N. It is clearly one-to-one; if n, m N and n < m, then the construction shows f(n) < f(m), thus f(n) f(m). It follows that the subset f(n) of A is equipotent to N, hence countably infinite. Having proved this, Exercise 1 is an immediate consequence; if A is finite we must have sup A A (otherwise A would be infinite). 5. Textbook, 1.1.1, # 1.1.7 (p. 25) Find a (nonstandard) ordering of the set of natural numbers N such that there exists a nonempty proper subset A N and such that sup A exists in N, but sup A / A. Hint: Define a new ordering of N by m n if either 1 < m < n or m > 1 and n = 1. That is, take 1 away from the beginning of N and place it at the very end. Solution. The simplest example is given by the hint. Define the order relation as in the hint. We have to verify: It is an order relation; i.e., transitivity and trichotomy. For transitivity let k, m, n N, assume k m, m n. We see that neither k nor m can be 1 since x 1 for all x N, x 1. If n = 1, then k 1 by definition. If n 1, then k m, m n becomes k < m, m < n, hence k < n, hence k n. For trichotomy, let m, n N. Assuming m n and m n, we have to prove n m. If m 1 n, then is the same as <, and the result follows from the usual trichotomy law.

4 We can t have n = 1; if n = 1 then either m = n = 1 or m n = 1. Thus n 1. If m = 1, then by definition n m. Trichotomy holds in all cases. We have to find a set A so sup A exists but is not in A. The perhaps simplest example is A = {2, 3, 4,...}. We claim sup A = 1. In fact, every integer is now less than or equal 1, so 1 is clearly an upper bound not only of A but to all subsets of N. Having proved it is an upper bound we must see it is the smallest such bound. If 1 is not the least upper bound of A, there is n N such that n is an upper bound of A, n 1 (so n 1). The definition of A shows it contains all integers from 2 onwards, hence with n it contains n + 1 and n n + 1, contradicting that n is an upper bound of A. We are done, since 1 / A.. 6. Textbook, 1.1.1, # 1.1.9 (p. 25) Let S be an ordered set and A is a nonempty subset such that sup A exists. Suppose there is a B A such that whenever x A there is a y B such that x < y. Show that sup B exists and sup B = sup A. Solution. two things. We have to show sup A is also sup B. This implies showing (a) sup A is an upper bound of B (b) If b is an upper bound of B then b sup A. Let s get to work! (a) Since B A any upper bound of A is an upper bound of B. In fact, if x b for all x A, assume x B. Then x A and x b. Thus, since sup A is an upper bound of A, it is also an upper bound of B. (b) To prove this, we assume b is an upper bound of B. We ll be done once we can write sup A b. If everything else fails, try contradiction, so to arrive at a contradiction let us assume the conclusion is false; that is, let us assume b < sup A. There is an immediate consequence of a number being less than the least upper bound: Then b is NOT an upper bound of A, hence there is x A, x > b. Now the hypothesis kicks in to say that there is y B such that y > x. But then y > b, contradicting that b is an upper bound of B. We are done. 7. Textbook, 1.1.1, # 1.1.10 (p. 25) Let D be the ordered set of all possible words (not just English words, all strings of letters of arbitrary length) using the Latin alphabet, using only lower case letters. The order is the lexicographic order as in a dictionary (e.g. aaa < dog < door). Let A be the subset of D containing the words whose first letter is a (e.g. a ϵa, abcd ϵa). Show that A has a supremum and find what it is.

5 Solution. Here are some considerations, preceding the proof. A is certainly bounded above; any word beginning with a letter other than a is an upper bound of A. One can also see, as we will in the actual proof, that any word beginning with a can be dominated by another word beginning with a. This suggests that the supremum of A must be the first word not beginning with a, namely b. The proof is quite immediate: In the first place, every word beginning with a precedes b. Thus b is an upper bound of A. On the other hand every word < b must begin with a. Since it only has a finite number of letters, we can always add another letter at the end (say z ) to form a word that is larger lexicographically than that word, so NO word preceding b can be an upper bound of A. It follows that sup A = b.