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Modern Analysis

EXAM 1. Friday, September 23, 2011

INSTRUCTIONS

•   You do not have to copy the statements of the exercises. But make sure you clearly indicate where youranswer to any given exercise begins, and where it ends.

•   The main thing I’d like to verify with this exam is whether you are able to write a proof correctly. So,please, write proofs correctly. Don’t improvise, don’t invent strange things, don’t fly off tangents. Don’t useundefined terms. What you write should be clear and coherent.

1. Let A, B  be sets of real numbers. Assume they are bounded below and non-empty. Prove

inf(A + B) = inf  A + inf  B.

Do not waste time proving separately that the set  A  + B  is non-empty or bounded below. You may acceptthat it is. The set A + B  is defined by  A + B = {a + b   :   a ∈ A, b ∈ B}.

Solution.   Let   x ∈   A +  B; then   x   =   a +  b   for some   a ∈   A,   b ∈   B. Because infima are lower bounds,a ≥   inf  A, b ≥   infB, thus   x   =   a +  b  ≥   inf  A  + inf  B. Since   x   was an arbitrry element of   A +  B , thisproves that inf  A  + inf  B   is a lower bound of   A +  B ; in particular, inf(A +  B) ≥   inf  A + inf  B. For theconverse inequality, let  > 0 be a positive number. Then inf  A < inf  A +  

2 thus there exists  a ∈ A  such that

a < inf  A +  

2. Similarly, there exists inf  A < inf  A +  

2 thus there exists  b ∈ B  such that b < inf  B +  

2. Then

inf(A + B) ≤ a + b <  inf  A +inf  B + ; since  > 0 is arbitrary, this implies inf(A + B) ≤ inf  A +inf  B. Havingproved inf(A + B) ≥ inf  A + inf  B  and inf(A + B) ≤ inf  A + inf  B, we are done.

2. Let {an}  be a sequence of real numbers. Assume {an}   is increasing and bounded above. Prove:

limn→∞

an = sup{

an   :   n ∈

N

}.

Solution.   Because the sequence is bounded above, the non empty set {an   :   n ∈ N}  is bounded above; letβ  be its supremum. Let  > 0 be given. Then  β − < β , thus β −  is not an upper bound of the set of whichβ  is the least upper bound, hence there is  x ∈ {an   :   n ∈ N}  such that β  − < x. Given the definition of theset,  x =  aN   for some  N  ∈  N, thus there exists  N  ∈  N  with  aN   > β  − . If  n > N , because the sequence isincreasing, we’ll have  an ≥ xN  > β − ; by the definition of  β ,  an ≤ β . Thus  β − < an ≤ β < β  + ; that is,|an − β | <  whenever  n > N .

3. Prove, using only the definition of limit, that the following sequences {an} converge to the given limit  L.

(a) {an} = { n2 + 5

3n2 − 2}, L =

 1

3.

Solution.

(b) {an} = {(−1)n

+ 2√ n√ n

  }, L = 2.

(c)   a1  = 1,  a2  = 4/2,  a3  = 5/3,  a4  = 9/4, and in general

an = n-th digit after the decimal point of the expansion of  π

n  ;   L = 0.

4. Let {an}, {bn}  be sequences of real numbers. Assume {an}, {bn}  both converge and  an ≤  bn   for all  n ∈  N.Prove: limn→∞an ≤ limn→∞ bn.

Hint:  You could give names to the limits, say  a  = limn→∞an,  b  = limn→∞ bn. You have to prove  a ≤ b; thealternative is  a > b. What happens if you take    = (a − b)/2?