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Math 323 – Real Anlaysis – Homework 1 Solutions
1.2.2 (b) True: Since ∩∞ i=1Ai ⊂ A 1 and A 1 is finite, the intersection must be finite. Next, we must show that ∩∞ i=1Ai 6 = ∅. In other words, we need to show that there exists some x ∈ An, for all n ∈ N. Suppose there is no such x then we will show that there is some set AN which is empty. (In other words, we argue by contrapositive.) Since A 1 is finite, we can number its elements: A 1 = {x 1 , x 2 , ..., xm}. For each k, the element xk 6 ∈ ∩∞ i=1Ai, and thus, there is some Nk such that xk ∈/ An for n ≥ Nk. But now, if N ≥ Nk for each k = 1, 2 , ..., m then AN ∩ A 1 = ∅. Since AN ⊂ A 1 this means that AN = ∅.
1.2.2 (c) False: Take B = ∅ then A ∩ (B ∪ C) = A ∩ C and (A ∩ B) ∪ C = C. But for any choice of A and C such that A 6 ⊃ C, A ∩ C 6 = C.
1.2.2 (d) True: This is the Associative Law for Sets. The following string of ”if and only if” (iff) statements constitutes the proof:
x ∈ A ∩ (B ∩ C) iff x ∈ A and x ∈ (B ∩ C) iff x ∈ A and (x ∈ B and x ∈ C) iff (x ∈ A and x ∈ B) and x ∈ C iff (x ∈ A ∩ B) and x ∈ C iff x ∈ (A ∩ B) ∩ C.
1.2.2. (e) True: This is the Distributive Law for Sets. The following string of ”if and only if” statements is the proof.
x ∈ A ∩ (B ∪ C) iff x ∈ A and x ∈ (B ∪ C) iff x ∈ A and (x ∈ B or x ∈ C) iff (x ∈ A and x ∈ B) or (x ∈ and x ∈ C) iff (x ∈ A ∩ B) or (x ∈ A ∩ C) iff x ∈ (A ∩ B) ∪ (A ∩ C).
1.2.3 (a) (A ∩ B)c^ ⊂ Ac^ ∪ Bc
Proof: If x ∈ (A ∩ B)c^ then x 6 ∈ A ∩ B then x 6 ∈ A or x 6 ∈ B then x ∈ Ac^ or x ∈ Bc then x ∈ Ac^ ∪ Bc
1.2.3 (b) (A ∩ B)c^ ⊃ Ac^ ∪ Bc
Proof: If x ∈ Ac^ ∪ Bc^ then x ∈ Ac^ or x ∈ Bc^. then x 6 ∈ A or x 6 ∈ B then x 6 ∈ A ∩ B then x ∈ (A ∩ B)c.
By part (a) the reverse inclusion is true and thus (A ∩ B)c^ = Ac^ ∪ Bc.
1.2.3 (c) The following string of if and only statements constitutes the proof: (A ∪ B)c^ = Ac^ ∩ Bc
Proof: If x ∈ (A ∪ B)c^ iff x 6 ∈ A ∪ B. iff x 6 ∈ A and x 6 ∈ B iff x ∈ Ac^ and x ∈ Bc iff x ∈ Ac^ ∩ Bc
Thus, (A ∪ B)c^ = Ac^ ∩ Bc.
1.2.5 (a) |a − b| ≤ |a| + |b|.
Proof: Making the substitution c = −b and applying the usual triangle inequality gives |a − b| = |a + c| ≤ |a| + |c| = |a| + | − b| = |a| + |b|.
1.2.5 (b) ||a| − |b|| ≤ |a − b|.
Proof: Let c = a − b, then a = c + b and by the usual triangle inequality
|a| = |c + b| ≤ |c| + |b| = |a − b| + |b|
and subtracting |b| from both sides gives |a| − |b| ≤ |a − b|. By a symmetric argument |b| − |a| ≤ |b − a| and thus ||a| − |b|| ≤ |a − b|.
1.3.2 (a) Definition: A real number t is the greatest lower bound or infimum of a set A ⊂ R if the following two conditions are satisfied:
(a) t is a lower bound for A; (b) if c is any lower bound for A, then t ≥ c.
1.3.2 (b) Assume t ∈ R is a lower bound for the set A ∈ R. Then t = inf A if and only if, for every choice of ≤ > 0 , there exists an element a ∈ A satisfying s + ≤ > a.
1.3.9 (d) True: Similar to 1.3.5 (a).
1.3.9 (e) False: Take A = { 1 } and B = (0, 1). Then sup A = sup B but there is no element in B that is an upper bound of A.
