June 3rd, 2019 SWMS-2018-Follow up Worksheet 1 Infimum and Supremum
1. LetA R. Define: ⊂ (a) What is meant by sayingA is a bounded subset ofR? (b) Negate the above statement using logical notation. (c) What is meant by sayingα = sup(A) andβ = inf(A)? Solution: See Class Board Photos.[Page 1-6] 2. Find the infimum and supremum of the sets (a)B= 2,3,4 { } (b)S= 1 :n N 50 + 1 :n N { n ∈ }∪{ n ∈ } ( 1)n (c)A= 1 − :n N { − n ∈ } Solution: See Class Board Photos.[Page 7-10]
3. LetA= p Q:p 2 <2 . Show thatA is bounded above but does not have a least upper bound in { ∈ } Q Solution: See Principles of Mathematical Analysis by W. Rudin, Page 11,12
4. LetA andB be two subsets ofR, which are both bounded below. Letu = inf(A) andv = inf(B). Find inf(C), in terms of u, v, when (a)C= a+b:a A, b B , { ∈ ∈ } (b)C=A B, ∪ (c)C=A 10 . ∪{ } Solution: For 4(a) See Class Board Photos.[Page 11-12] For 4(b) Ans: min u, v and 4(c) Ans: min u, 10 { } { } 5. Letn 1 andx > 0. Prove the Bernoulli inequality: ≥ i n n (1 +x ) 1+ x i ≥ i i=1 i=1 � � 6. Ifa R such that 0 a<� for every�> 0, then show thata = 0. ∈ ≤ Solution: We will prove this by contradiction. Supposea= 0. Then, since it is given thata 0, a > 0. By taking�=a, we conclude thata 1 June 3rd, 2019 SWMS-2018-Follow up Practice Problems Set 1 1. LetA andB be two subsets ofR, which are both bounded below. Letu = sup(A) andv = sup(B). Find sup(C), in terms of u, v, when (a)C= a+b:a A, b B , { ∈ ∈ } (b)C=A B, (What ifC=A B?) ∪ ∩ (c)C=A 10 . ∪{− } Solution: See Related Problem for 1(a) 2. LetA andB be two non-empty subsets ofR. Define what is meant by a functionf:A B and when is it called one-one and onto. Provide examples off that are (and are not) one-one→ and (or) onto whenA,B arefinite or countable or uncountable. Solution: : Check E.B. Folland, Real Analysis 2 June 3rd, 2019 SWMS-2018-Follow up Worksheet 2 Sequences 1. In each of the cases below decide if x ∞ converges or not: { n}n=1 1 (a)x n = n 2 (b)x n = √n n n 2n − − (c)x n = n! , (d)x = nbn, forb (0, 1) n ∈ Solution: See Class Board Photos.[Page 14-18] 2. Leta n be a bounded sequence of real numbers. Lets = sup a n :n 1 ands a n :n N . Show { ≥ } �∈{ ∈ } that there is a subsequence ofa n that increases tos. Solution: Note that s = sup an :n 1 ands a n :n N . (1) { ≥ } �∈{ ∈ } From (1) it is immediate that there existsn 1 > 0 such that s 1 0 such that 2 2 { {| − i | ≤ ≤ 1} 2 s � < a < s. − 2 n2 By definition of� 2 we have that 1 n > n , a > a , ands < a < s. 2 1 n2 n1 − 2 n2 Thus inductively (why ?) we can construct for eachk 1 ≥ 1 nk > nk 1, ank > ank 1 , ands < ank < s. − − − k By constructiona is a subsequence ofa , it is increasing, anda s ask . nk n nk → →∞ 3. Suppose a n n 1 is a sequence of real numbers such that { } ≥ lim a n+1 a n = 0. n →∞ | − | Does this necessarily imply that the sequence is a Cauchy sequence. Solution: Answer is No. See Class Board Photos, Page 2 (upside down) 1 4. Consider the y n ∞ , such thaty 1 > 1 andy n+1 := 2 forn 2. Show thaty n converges. { }n=1 − yn ≥ 5. (Finding Roots of a number) Leta> 0 and chooses 1 > √a. Define 1 a sn+1 := (sn + ) 2 sn forn N. ∈ (a) Show thats n is monotonically decreasing and limn sn = √a. → 2 (b) Ifz =s √a then show thatz < zn . n n − n+1 2√a 2 f(s n 1) (c) Letf(x) =x a. Show thats n =s n 1 − . f �(sn 1) − − − − (d) Draw graph off witha = 4 and plot the sequences n for a few steps whens 0 = 5. Solution: See Class Board Photos. 3 June 3rd, 2019 SWMS-2018-Follow up Practice Problems Set 2 1. Decide if x n n∞=1either converges to a real number or diverges to or diverges to or none of { } 1 ∞ −∞ 2 the above whenx n =n n 2. Letx R, x n ∞ be a sequence of real numbers. Show that the following are equivalent: ∈ { }n=1 (a) �> 0, there is anN N such that x n x <� for alln N. ∀ ∈ | − | ≥ (b) � (0, 1), there is anN N such that x n x <� for alln N. ∀ ∈ ∈ | − | ≥ (c) LetC> 0, �> 0, there is anN N such that x n x C� for alln>N. ∀ ∈ | − |≤ 4 June 3rd, 2019 SWMS-2018-Follow up Worksheet 3 1. LetA be a non-empty set of real numbers which is bounded below. Let A := x R: x A . Show that inf(A) = sup( A). − { ∈ − ∈ } − − 2. If z,w,zi C fori=1,2, . . . , n then show that ∈ z +z + +z z + z + + z | 1 2 ··· n|≤| 1| | 2| ··· | n| and z w z w || |−| ||≤| − | Solution: We will prove thefirst inequality by induction onn. Let (n) be the statement: P n n zi zi � � ≤ | | �i=1 � i=1 �� � � � � (1) is true asz 1 z 1. We have seen that� (2)� is true. Let (k) be true for somek N such that kP 2. We will show≤ that (k + 1) is true. P P ∈ ≥ P k+1 k k k k+1 zi = zi +z k+1 zi + z k+1 zi + z k+1 = zi � � �� � � ≤ � � | |≤ � | |� | | | | �i=1 � � i=1 � �i=1 � i=1 i=1 �� � � � � �� � � � This proves� (k +� 1).� We have used both� � (2) and� (k) above. (1) is true, (2) is true, and �P � � � �P � P P P (k) = (k + 1) k N 1 . Hence by the principle of mathematical induction, (n) is true P ⇒P ∀ ∈ −{ } P for alln N. ∈ Now we will prove the second inequality. For any a, b R, ∈ a+ ib = a2 +b 2 = ( a)2 + ( b)2 = (a+ ib) | | − − |− | � � Hence z 1 = z 1 z C. In particular, z w = w z . Now, | | |− |∀ ∈ | − | | − | z = (z w)+w z w + w | | | − |≤| − | | | z w z w | |−| |≤| − | Interchangingw andz, w z w z = z w | |−| |≤| − | | − | Since w and z are reals, z w equals at least one of z w and w z , both of which are less than| | or equal| | to z w ||. Hence,|−| || | |−| | | |−| | | − | z w z w || |−| ||≤| − | 3. LetA andB be bounded nonempty subsets ofR, and letA+B := a+b:a A, b B . Prove that sup(A+B) = sup(A) + sup(B). { ∈ ∈ } Solution: We willfirst prove that sup(A+B) = sup(A) + sup(B). Letα = sup(A) andβ = sup(B). Asα a a A andβ b b B, we conclude thatα+β a+b a A, b B. Hence α+β≥ x∀ x∈ (A+B).≥α+∀β∈ is an upper bound ofA+B≥. ∀ ∈ ∈ ≥ ∀ ∈ Letγ<α+β. To prove thatα+β is the supremum ofA+B, we need show thatγ cannot be an upper bound ofA+B, i.e. we need to show that there existsx A+B such thatx>γ. Let ∈ p=(α+β γ)/2. Note thatp> 0. Asα p is not an upper bound ofA, a � A a � >α p. − − ∃ ∈ � − 5 Asβ p is not an upper bound ofB, b � B b � >β p. Now,a � +b � A+B. Also, − ∃ ∈ � − ∈ a� +b � >(α p) + (β p) =γ. So we have foundx=a � +b � A+B such thatx>γ. Henceα+β is the supremum− ofA−+B. sup(A+B) = sup(A) + sup(B). ∈ For anyX R, define X := x:x X . So, for anyy R,y X ( y) ( X). ⊆ − {− ∈ } ∈ ∈ ⇐⇒ − ∈ − ( A) + ( B) = x+y:x ( A), y ( B) − − { ∈ − ∈ − } = ( x) + ( y):x A, y B { − − ∈ ∈ } = (x+y):x A, y B {− ∈ ∈ } = z:z (A+B) {− ∈ } = (A+B) − We know that ifX R is non-empty and bounded below, inf(X)= sup( X) by below Claim. 1Hence, we also have⊆ − − inf(A+B) = sup( (A+B)) − − = sup(( A) + ( B)) − − − = (sup( A) + sup( B)) − − − = sup( A) sup( B) − − − − = inf(A) + inf(B) This completes the proof. Claim: LetA be a non-empty set of real numbers which is bounded below. Let A := x R: x A . Show that inf(A) = sup( A). − { ∈ − ∈ } − − Solution: : A is non-empty and bounded below, so it has a greatest lower bound. Letα = inf(A). Asα is a lower bound ofA,α 4. In each of the cases below decide if x n n∞=1either converges to a real number or diverges to or diverges to or none of the above.{ } ∞ −∞ 2n2+1 (a)x n = 3n2 1 − ( 1)n (b)x n =n − n! (c)x n = nn 6 June 3rd, 2019 SWMS-2018-Follow up Practice Problems Set 3 1. Decide if x n n∞=1either converges to a real number or diverges to or diverges to or none of { } n n n 1 ∞ −∞ the above whenx n = (a1 +a 2 +a 3 ) n , witha 1, a2, a3 >0 2. Let a ∞ be a bounded sequence. Then show that it has a convergent subsequence. { n}n=1 3. Let x n ∞ be a sequence of real numbersR and suppose thatx n x. { }n=1 → (a) Let m, n N, show thatx m+n x asm . ∈ → →∞ l k m k (b) Let m, l N,p:R R such thatp(x) = k=0 pkx , andq:R R 0 ,q(x) = k=0 qkx , ∈ → → \{ } p(x) withp k R, q k R fork=1,2, . . . , n. Show that ifr:R R defined byr(x) = then ∈ ∈ � → �q(x) r(x ) r(x). n → (c) Show that x ∞ also converges {| n |}n=1 7 June 3rd, 2019 SWMS-2018-Follow up Homework 1 Solution 1. LetA andB be two subsets ofR, which are both bounded below. Letu = inf(A) andv = inf(B). Find inf(C), in terms of u, v, whenC= ab:a A, b B andA,B [0, ), { ∈ ∈ } ⊂ ∞ Solution: See Night Board discussion[Page 1] 2. Let x n n∞=1be a sequence of real numbers that is not bounded. Then show that there is either a subsequence{ } that diverges to or a subsequence that diverges to . ∞ −∞ Solution: See Night Board discussion[Page 2] nα 3. Decide if x n ∞ converges or not whenx n = n withα R, p > 0. { }n=1 (1+p) ∈ Solution: See Night Board discussion[Page 3] 4. Lety 1, y2 R be given, and define recursively forn 1, ∈ ≥ 1 2 y = y + y n+2 3 n 3 n+1 for alln N. Decide ify n converges and if it does thenfind its limiting value. ∈ 8