HW 2 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson 3.5 a) Show that |b| ≤ a if and only if −a ≤ b ≤ a. We have to prove both directions. First suppose that |b| ≤ a. Notice that since |b| ≥ 0, we must have a ≥ 0. Then there are two cases. If b is non-negative, then |b| = b and so b ≤ a. But b is non-negative, so −a ≤ 0 ≤ b ≤ a, so −a ≤ b ≤ a. If instead b is negative, then |b| = −b, so −b ≤ a. Multiplying by −1 and switching the sign, we get that b ≥ −a; i.e. that −a ≤ b. But also, b ≤ 0, so b ≤ a. Combining these, we see that −a ≤ b ≤ a. So in either case, −a ≤ b ≤ a. Conversely, suppose that −a ≤ b ≤ a. If b is positive, then |b| = b, so b ≤ a means that |b| ≤ a. If b is negative, then |b| = −b; we have −a ≤ b, so by multiplying by -1 and switching the direction, a ≥ −b, so a ≥ |b|. In either case, |b| ≤ a.  b) Prove that ||a| − |b|| ≤ |a − b| for all a, b ∈ R. By part a), this is equivalent to showing that |a| − |b| ≤ |a − b| for all a, b ∈ R, which is equivalent to showing that |a| ≤ |b| + |a − b| for all a, b ∈ R. So pick any a and b ∈ R, and let c = a − b. Then by the triangle inequality, we have |b + c| ≤ |b| + |c|. But b + c = a and c = a − b, so this statement is just that |a| ≤ |b| + |a − b|, which is equivalent to what we have been asked to prove. So ||a| − |b|| ≤ |a − b| for any a, b ∈ R.  3.6. a) Prove that |a + b + c| ≤ |a| + |b| + |c| for all a, b, c ∈ R. Pick a, b, and c ∈ R. Apply the triangle inequality to the numbers (a+b) and c to see that |a + b + c| ≤ |a + b| + |c|. But by the triangle inequality applied to a and b, the right hand side here is less than or equal to |a| + |b| + |c|. So |a + b + c| ≤ |a| + |b| + |c|.  b) Use induction to prove that for n real numbers a1 , a2 , . . ., an , we have |a1 + a2 + . . . + an | ≤ |a1 | + |a2 | + . . . + |an |. The basis for induction is n = 1, which is obvious (|a1 | ≤ |a1 |). For the induction step, we assume that Pn is true (Pn is the nth proposition, as usual). We want to show that |a1 +a2 +. . .+an +an+1 | ≤ |a1 |+|a2|+ . . .+|an |+|an+1|. By the triangle inequality applied to (a1 +. . .+an ) and an+1 , we know that |a1 +a2 +. . .+an +an+1 | ≤ |a1 +a2 +. . .+an |+|an+1 |. Applying Pn , the first term on the right-hand-side is ≤ |a1 | + . . . + |an |. So putting these together, we see that |a1 + a2 + . . . + an + an+1 | ≤ |a1 | + |a2 | + . . . + |an | + |an+1 |. So Pn+1 is true; by mathematical induction, this finishes the proof.  1

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3.8. Let a, b ∈ R. Show that if a ≤ b1 for every b1 > b, then a ≤ b. This was confusing for many people, so it helps to think about exactly what the hypothesis and conclusion are. The hypothesis is that a ≤ b1 for every b1 > b (not just for some b1 , then this would be false). The desired conclusion is that a ≤ b. We’ll do proof by contradiction. Assume for contradiction that a is not ≤ b. Then by an order property, a ≥ b; but of course a 6= b, so a > b. (Yes, this is obvious, so don’t worry if you omitted it). Now let b1 = 21 (a + b) be the average of a and b. We claim that b1 > b; indeed, a > b, so a/2 > b/2, so by adding b/2 to both sides, b1 > b. Similarly we see that a > b1 (subtract a/2 from both sides). But then, b1 > b and yet it is not true that a ≤ b1 . This contradicts the hypothesis, so our assumption is wrong; thus a ≤ b.  4.6. Let S be a nonempty bounded subset of R. a) Prove that inf S ≤ sup S. Since S is bounded and nonempty, both inf S and sup S exist. And since S is nonempty, there exists s ∈ S. By definition, we have inf S ≤ s and s ≤ sup S. So inf S ≤ s ≤ sup S. That’s it.  b) What can you say about S if inf S = sup S? Well, if inf S = sup S, then pick some s ∈ S (we’re still assuming S is nonempty and bounded). inf S ≤ s ≤ sup S, so inf S = sup S means that we must have equality in each case; that is, inf S = s = sup S. Since inf S is a lower bound for S, there can be no elements in S less than s. Since sup S is an upper bound for S, there can be no elements in S greater than s. Thus S has exactly one element - s itself!  4.7. Let S and T be nonempty bounded subsets of R. a) Prove that if S ⊆ T , then inf T ≤ inf S ≤ sup S ≤ sup T . Since S and T are bounded and nonempty, all the infs and sups exist, and by 4.6, inf S ≤ sup S. So it’s enough to show that inf T ≤ inf S and that sup S ≤ sup T . To show that inf T ≤ inf S, notice that inf T is a lower bound for T . If there were an element in S less than inf T , the fact that S ⊆ T would tell us that it would also be an element in T less than inf T , which is a contradiction. Hence inf T is a lower bound for S, and thus is less than or equal to the greatest lower bound, which is inf S. Similarly, to show that sup S ≤ sup T , notice that sup T is an upper bound for T . Since every element in S is also in T , sup T is also an upper bound for S. Thus it is greater than or equal to the least upper bound for S, which is sup S. So sup T ≥ sup S. This is enough. 

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b) Prove that sup(S ∪ T ) = max{sup S, sup T }. Do NOT assume that S ⊆ T . We will prove this by doing two things; first we will show that max{sup S, sup T } is an upper bound for S ∪ T , and then we will show that it is the least upper bound. To show that max{sup S, sup T } is an upper bound for S ∪ T , pick any element x ∈ S ∪ T . Then either x ∈ S, in which case x ≤ sup S, or x ∈ T , in which case x ≤ sup T . In either case, x ≤ max{sup S, sup T }. So max{sup S, sup T } is an upper bound for S ∪ T . To show that it is the least upper bound, suppose for contradiction that y is an upper bound for S ∪ T with y < max{sup S, sup T }. Then either y < sup S or y < sup T ; suppose that y < sup S. Then y is not an upper bound for S. So there is s ∈ S with s > y; but s ∈ S ∪ T , so this contradicts the fact that y is an upper bound for S ∪ T . Similarly, if y < sup T , y is not an upper bound for T , so there is t ∈ T with t > y; but t ∈ S ∪ T , and this is again a contradiction. Thus we get a contradiction in either case, and so max{sup S, sup T } is indeed the least upper bound for S ∪ T . Thus sup(S ∪ T ) = max{sup S, sup T }.  4.8. Let S and T be nonempty subsets of R with the following property: s ≤ t for all s ∈ S and t ∈ T . a) Observe that S is bounded above and T is bounded below. Well, T is nonempty, so there exists t ∈ T . By the property, t is an upper bound for S. So S is bounded above (and thus sup S exists). Similarly, S is nonempty, so there exists s ∈ S. By the property, s is a lower bound for T , so T is bounded below (and thus inf T exists).  b) Prove that sup S ≤ inf T . We’ll prove this by contradiction. Suppose for contradiction that sup S > inf T . Then inf T is less than the least upper bound for S, so in particular inf T is not an upper bound for S. So there exists s ∈ S with s > inf T . But now s > inf T , so s is not a lower bound for T . Thus there exists t ∈ T with t < s. However, since t ∈ T and s ∈ S, the property requires that s ≤ t. This is a contradiction, and so our assumption was wrong and sup S ≤ inf T .  c) Give an example of such sets S and T where S ∩ T is nonempty. Let S = [0, 1] and T = [1, 2]. Their intersection is {1}. d) Give an example of sets S and T where sup S = inf T and S ∩ T is empty.

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Let S = (0, 1) and T = (1, 2). The trick is that either S cannot contain its sup or T cannot contain its inf. 4.9. Complete the proof that inf S = − sup(−S) in Corollary 4.5 by proving (1) and (2). (1): Let s0 = sup(−S); we need to prove that −s0 ≤ s for all s ∈ S. Suppose for contradiction that there is s ∈ S with −s0 > s. Then we must have s0 < −s. But −s ∈ −S, so −s ≤ s0 , because s0 = sup(−S). This is a contradiction, and so −s0 ≤ s for all s ∈ S.  (2): We need to show that if t ≤ s for all s ∈ S, then t ≤ −s0 . Suppose for contradiction that t ≤ s for all s ∈ S but t > −s0 . Then let x ∈ −S; −x ∈ S, so t ≤ −x, so −t ≥ x. Thus −t is an upper bound for −S. However, t > −s0 , so −t < s0 , so −t is less than the least upper bound for −S. This is a contradiction, and so t ≤ s for all s ∈ S means that t ≤ −s0 .  4.13. Prove that the following are equivalent for real numbers a, b, c: a) |a − b| < c, b) b − c < a < b + c, c) a ∈ (b − c, b + c). Proof: To establish that these three statements are equivalent, we need to be able to start with any of them and prove any of the others; that is, we need to create a logical chain between them. In this case, we will prove that a) is true if and only if b) is true, and also that b) is true if and only if c) is true. This is enough. Honestly, to show b) is true if and only if c) is true, there’s nothing to do - it’s simply a matter of the definition of an open interval. This is completely obvious. (That’s a very dangerous word to use in analysis, but it’s true here!) So the real difficulty is to show that a) is true if and only if b) is true. It can be done directly, but I will use Exercise 3.5. First suppose a) is true; i.e. that |a − b| < c. Then of course |a − b| ≤ c. By Exercise 3.5, this means −c ≤ a − b ≤ c. Add b to all three terms in this inequality to see that b − c ≤ a ≤ b + c. How do we get ¡ signs instead of ≤ signs? Well, suppose a = b + c; then |a − b| = c, which is a contradiction. Similarly if a − b = −c, then |a − b| = c, contradiction. So we have strict inequality in each case, and b − c < a < b + c. Now suppose b) is true - i.e. b − c < a < b + c. Then of course b−c ≤ a ≤ b+c; subtracting b from all three terms gives −c ≤ a−b ≤ c, so by Exercise 3.5 we have |a − b| ≤ c. Suppose for contradiction that |a − b| = c; then if a ≥ b we have a − b = c, so a = b + c, contradiction.

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Similarly if b ≥ a we have b − a = c, so a = b − c, contradiction. Thus |a − b| = 6 c, and so this shows |a − b| < c. So a) is true if and only if b) is true if and only if c) is true. Thus all three statements are equivalent.  4.14. Let A and B be nonempty bounded subsets of R, and let S be the set of all sums a + b where a ∈ A and b ∈ B. a) Prove that sup S = sup A + sup B. First notice that all the sups exist since A and B are nonempty and bounded. Now we’d like to show that sup A + sup B is an upper bound for S, and moreover that it is the least upper bound. To show it’s an upper bound, take s ∈ S. There exist a ∈ A and b ∈ B such that s = a + b. a ≤ sup A and b ≤ sup B, so s = a + b ≤ sup A + sup B. Thus sup A + sup B is an upper bound for S. To show that it’s the least upper bound, suppose for contradiction that there exists t which is an upper bound for S but with t < sup A + sup B. Then t − sup B < sup A, so in particular t − sup B is not an upper bound for A. So there exists a ∈ A with t − sup B < a, or t < a + sup B, or t − a < sup B. Now, though, we see that t − a is not an upper bound for B, so there exists b ∈ B with t − a < b, or t < a + b. But a + b ∈ S, and t is an upper bound for S. This is a contradiction! So sup A + sup B = sup S.  b) This can be done by exactly mimicking the proof of a), switching all the inequalities, replacing greatest by least, et cetera. But there is an easier way. First we look at −S. If −s ∈ −S, then s ∈ S, so s = a + b for some a ∈ A and b ∈ B, so −s = −a − b; −a ∈ −A and −b ∈ −B, so −s is a sum of elements in −A and −B. Conversely if −a ∈ −A and −b ∈ −B, then a + b ∈ S, so −a − b ∈ S. Thus −S is precisely the set of all sums x + y where x ∈ −A and y ∈ −B. But now we may apply part a) to −S, −A, and −B to see that sup(−S) = sup(−A) + sup(−B). By Corollary 4.5, this tells us that inf S = inf A + inf B, and we’re done.