Math 312 Intro to Real Analysis Homework 2 Solutions Stephen G Simpson Wednesday February 4 2009 The assignment consists of Exercises 4 1 4 3 4 7 4 14 5 3 in the Ross textbook Each problem counts 10 points 4 1 The sups of these sets are found in Exercise 4 3 In each case if is the sup of the set then 1 and 2 are three different upper bounds of the set If the set has no sup then it has no upper bound 4 3 The sups are a 1 b 1 c 7 d e 1 f 0 g 3 h NO SUP i 1 j 1 k NO SUP l 2 m 2 n 2 o 0 p 10 q 16 r 1 s 1 2 t 2 u NO SUP v 1 2 w 1 2 4 7 Let S and T be nonempty bounded subsets of R The completeness property of R implies that sup S and sup T exist a Assume S T In other words every x belonging to S also belongs to T i Since is an upper bound of T every x belonging to S is In other words is an upper bound of S Since is the least upper bound of S it follows that In other words sup S sup T ii A similar argument shows that inf T inf S iii Since S is nonempty let x be an element of S By definition of inf and sup we have inf S x sup S for all such x Combining our results we have inf T inf S sup S sup T Q E D b S T is again a nonempty bounded subset of R so let sup S T i Since S S T it follows1 that Similarly since T S T it follows that Combining these two inequalities we have max ii Conversely given 0 we know that hence is not an upper bound of S T Therefore let z be such that z and z belongs to S T If z belongs to S then 1 by part a applied to S and S T 1 z sup S Similarly if z belongs to T then z sup T In either case we have max Since this inequality holds for all 0 it follows that max Combining these two results we see that max In other words sup S T max sup S sup T Q E D 4 14 Let A and B be nonempty bounded subsets of R Let S A B a b a in A b in B Let sup A sup B and sup A B a Let 0 be given Since 2 sup A we can find a in A such that 2 a Similarly we can find b in B such that 2 b Let c a b Then 2 2 a b c and c belongs to A B It follows that sup A B Since this holds for all 0 if follows that b Conversely given c in A B we can find a in A and b in B such that c a b Then a sup A and b sup B hence c a b Thus is an upper bound of A B Since is the least upper bound of A B it follows that Combining these two results we see that In other words sup A B sup A sup B Similarly it can be shown that inf A B inf A inf B 5 3 For the unbounded sets in 4 1 we have h inf 2 sup k inf 0 sup l inf sup 2 o inf sup 0 t inf sup 2 u inf 0 sup All of the other sets in 4 1 are bounded 2