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# Analysis MATH 131A

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This 24 page Class Notes was uploaded by Kaylin Wehner on Friday September 4, 2015. The Class Notes belongs to MATH 131A at University of California - Los Angeles taught by Staff in Fall. Since its upload, it has received 106 views. For similar materials see /class/177815/math-131a-university-of-california-los-angeles in Mathematics (M) at University of California - Los Angeles.

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Date Created: 09/04/15

Math 131A1 Spring 2004 Handout 2 Some pointers on the logical conventions of mathematics It is probably hardest to get used to the way mathematicians use the word implies77 or the symbol The idea is that you want to know if the implication is true or not just on the basis of whether the ingredients are true or false without any more thinking 0 Any true statement any true statement eg 1 l 2 there are in nitely many primes Fermat7s last conjecture 0 Any false statment any true statement because in particular you want to be able to say that l 2 l X 0 2 X 0 is a correct deduction 0 Any false statement any false statement because in particular you want to be able to say that 12 ll 2l is a correct deduction o The following is false truth false H The most common logical errors made by beginners 0 They think that or77 is exclusive thus although they know that S means less than or equal to77 they think it is wrong77 to write 3 S 3 because they know that actually 3 377 0 They think that you cannot prove P Q if you already know that is ase 0 They think that if P Q is true then Q is true 0 When asked to prove P Q they instead prove Q P 0 They get equality say of numbers mixed up with 42gt logical equivalence used for propositions 2 Some correct illustrations of logic 0 The proposition 6 lt 7 or 4 lt 577 is true 0 The proposition 1 2 0 X l 0 X 277 is true 0 The proposition For any real number 1 12 3 zz2 13 is true 0 The proposition For any real number I zz2 13 12 377 is false 3 It is important to be able to take the negations of statements in order to prove things by contradiction Here is the general scheme donlt worry about the last two at this point Math 131a Midterm 2 Lecture 2 Spring 2008 Name Instructions 0 There are 4 problems Make sure you are not missing any pages 0 Give complete convincing and clear answers or points will be deducted o No calculators books or notes are allowed 0 Answer the questions in the spaces provided on the question sheets If you run out of room for an answer continue on the back of the page Question Points Score 1 10 2 10 3 10 4 10 Total 40 1 10 points Let A be a nonempty subset of R that is bounded above and let Oz sup A If 04 A7 prove that 04 is a limit point of A Solution We suppose that 04 is not a limit point of A and show that 04 31 sup A Since 04 is not a limit point of A7 there exists a 6 gt 0 such that 04 7 604 6 A C a Since 04 A7 we thus have 04 7 604 6 H A Q We may assume that 04 is an upper bound for 04 since otherwise it certainly can7t be the least upper bound and were done But7 since 04 is an upper bound for A and Oz 7 604 6 A 0 we have that 04 7 6 is an upper bound for A7 and so 04 is not the least upper bound for A 2 10 points Let 1511 be a sequence of real numbers which converges to a point a E R Consider the sequence 5321 de ned a1 an 8 Show that 53311 converges to a Solution Let E gt 0 Since lirnH00 ak 17 we may nd an 711 such that ak 7 1 lt 62 for k 2 711 Choose 712 large enough so that alianL am fed712 lt 62 Then for n gt rnaxnlng7 we have a1an snicd 7a 71 7 a17aania i n lt a17aamiaam1iaania 7 n n lt a17aamia7 7 712 n ltE 76 2 3 10 points Let E C R and let F be the set of limit points of E Show that F is closed Solution We need to show that F0 is open Suppose z 6 F0 Then z is not a limit point for E so there exists a 6 gt 0 such that N5x E C We claim that N5x C F0 and hence F0 is open Suppose y E N5z and y 31 x Let 6 rninly 7l757 ly 7zl and consider N5y Since 6 3 lg 7 zl we have z N5 y Since 6 S 6 7 lg 7 zl it follows from the triangle inequality that N5y C N5 Thus N5 y E Q and so y is not a limit point for E7 which means that y 6 F0 4 10 points Show you are only allowed to use the de nitions and the triangle inequality that every compact subset of R is closed Solution Suppose K C R is compact We need to show that K0 is open Suppose z 6 K0 Since z K7 we have ly 7 ml gt 0 for each y 6 K7 and so N yixiyyd is an open cover 2 of K Since K is compact7 there is a nite subcover N y c 7317 Niywxi Set 2 2 6 minly1mllyquotml and consider N5z lfy E K then lt lywzl2 for some 239 Thus S lyiiyllyizl lt lyiixlQnLlyizl and so lyisl gt lyiislQ 2 6 Hence N5x C K0 Math 131a Handout 6 Our completeness axiom If S is a nonempty subset of R and S is bounded above ie7 S S b for some b7 then S has a least upper bound bo sup 5 You fomulate the corresponding result for nonempty sets that are bounded below Here are theorems about sequences and their limts that you should be able to prove including the relevant de nitions lf In is a convergent sequence7 then it must be bounded Proof Suppose that In 7gt L Choose no such that n 2 no In 7 L lt 1 Then 7 L S In7L lt 1 implies that lt L 1 Let M max I1 lIno1llLl 1 We have that for all n S M lf In 7gt L and In 0 and L f 07 then there is a constant c gt 0 such that In 2 c for all n Proof Suppose rst that In gt 0 and L gt 0 Choose no such that n 2 no In 7 L lt L2 Then L 7 In S In7L lt L2 implies that In gt L7L2 L2 Let c min IhIg7 7Ino17 LQ It follows that In 2 c for all n For the general case note that 7gt L and use the positive result lfIn 7gt L and for all n7 In 2 07 then L 2 0 Proof Suppose that L lt 0 Then let 5 7L We may choose no such that In 7 L lt 5 Then InO 7 L lt 5 InO lt L 5 07 contradicting Inc 2 0A The usual limit theorems such as In 7gt L and yn 7gt M implies Inyn 7gt L M lf In is an increasing sequence7 and In S 127 then In 7 ho sup You should be able to state and prove the corresponding result for de creasing sequences Proof Given 5 gt 07 we have that ho 7 5 lt bo implies that ho 7 5 is not an upper bound for 7 hence there exists an no with bo 7 5 lt Inc It follows that ifn 2 no7 then bo75 lt InO S In S no and thus In 7 bo lt 5 If 0 f S Q R and bo sup 5 then there is a sequence In E S such that In 7 no You should be able to state and prove the corresponding result for the in mum Proof Given n E N bo 7 ln is not an upper bound for 5 hence we may choose an In E S such that ho 7 ln lt In S no It follows that In 7 bo lt ln7 and thus In 7 ho Math 131a Midterm 1 Lecture 2 Spring 2008 Name Instructions 0 There are 4 problems Make sure you are not missing any pages 0 Give complete convincing and clear answers or points will be deducted o No calculators books or notes are allowed 0 Answer the questions in the spaces provided on the question sheets If you run out of room for an answer continue on the back of the page Question Points Score 1 10 2 10 3 10 4 10 Total 40 1 10 points Let f A a B and g B a A be functions satisfying 9 o z for all z E A Prove that f is one to one Must f be onto B Justify your answer Solution First we prove that f is one to one Suppose 12 E A and fx1 zz We need to show that 1 2 Since fz1 zz we have 9 o f1 gfz1 gfz2 gof But by our assumption that gof x this implies that 1 gof 1 9 O f2 2 It is not necessarily the case that f is onto B For example we may take A 1 B 12 f 11 and g 1121 Then f does not map A onto B but 9 0 fx z for every x E A 2 10 points Let J be a nonempty subset of R that has the following properties i J is bounded ii supJ E J iii inf J g J iv lf Ly E J with z lt y then t E J for every t satisfying z lt t lt y Prove that J is the interval inf Jsup J Solution We need to show that a J C inf Jsup J and b that inf Jsup J C J For a let t E J Since ian is a lower bound for J and supJ is an upper bound for J we have ian S t S supJ Since ian J we also have ian lt t S supJ and so t E inf Jsup J For b let t E inf Jsup J If t sup J we have t E J by ii and so we may assume that inf J lt t lt sup J Since t gt inf J and inf J is the greatest lower bound we see that t is not a lower bound for J and so there is an x E J with z lt t Since t lt supJ and supJ is the least upper bound we see that t is not an upper bound for J and so there is a y E J with y gt t Thus z lt t lt y and so by property iv t E J 3 10 points Let A and B be nonempty subsets of the positive real numbers which are bounded above Prove that supa b a 6 Ab E B sup A sup B Solution We need to show that supA supB is an upper bound for a b a 6 Al 6 B7 and ii if 04 lt supA supB then 04 is not an upper bound for a b a 6 Al 6 B For i7 let a E A and b E B Then a S supA and b S supB7 so ab S supAb S supA sup B For ii7 suppose 04 lt supA supB and let 6 supA supB 7 04 gt 0 Since supA is the least upper bound7 we may nd a E A with a gt supA 7 64 Since sup B is the least upper bound7 we may nd b E B with b gt sup B764 Then ab gt supAsupB762 gt supA supB 7 E Q So 04 is not an upper bound for a b a 6 Al 6 B 4 10 points For each positive integer k let Nk 17xk 1k E N be the set of ordered k tuplets of positive integers Prove that for every positive integer k Nk is countable You are allowed to use7 without proof7 the fact that N gtlt N is countable You are also allowed to use7 without proof7 the fact that if f A a B and g B a C are both bijective7 then 9 o f A a C is bijective Hint Try using induction on k Solution By induction it suf ces to show that N1 is countable and that ii if Nk is countable then Nki l is countable For i7 it is clear that f N a N1 de ned by fn is a bijection7 and so N1 is countable For ii assume that Nk is countable Then7 there exists a bijective g N a Nk Consider the function f N gtlt N a N19 l de ned by fnm 717 7yk1 where 12 7yk1 We claim that f is bijective To see that f is onto7 let 11 7yk1 E Nk Since 9 is onto7 there exists m E N with gm 12 7yk17 and so fy1m 11 7yk1 To see that f is one to one7 suppose fn1m1 y17yk1 fn2m2 Then gm1 yg7 7yk1 9m2 Since 9 is one to one7 this implies that m1 m2 But we also have 711 yl 712 and so 7117711 7127712 Since N gtlt N is countable7 there exists a bijective h N a N gtlt N Since f is bijective and h is bijective7 f o h N a Nki l is bijective7 and so Nki l is countable Math 131a Midterm 1 Lecture 2 Spring 2008 Name Instructions 0 There are 4 problems Make sure you are not missing any pages 0 Give complete convincing and clear answers or points will be deducted o No calculators books or notes are allowed 0 Answer the questions in the spaces provided on the question sheets If you run out of room for an answer continue on the back of the page Question Points Score 1 10 2 10 3 10 4 10 Total 40 1 10 points Let f A a B and g B a A be functions satisfying 9 o z for all z E A Prove that f is one to one Must f be onto B Justify your answer 2 10 points Let J be a nonempty subset of R that has the following properties i J is bounded ii supJ E J iii inf J g J iv If Ly E J with z lt y then t E J for every t satisfying z lt t lt y Prove that J is the interval inf J7 sup J 3 10 points Let A and B be nonempty subsets of the positive real numbers which are bounded above Prove that supa b a 6 Ab E B sup A sup B 4 10 points For each positive integer k let Nk 17xk 1k E N be the set of ordered k tuplets of positive integers Prove that for every positive integer k Nk is countable You are allowed to use7 without proof7 the fact that N gtlt N is countable You are also allowed to use7 without proof7 the fact that if f A a B and g B a C are both bijective7 then 9 o f A a C is bijective Hint Try using induction on k Math 131a Midterm 1 Lecture 2 Spring 2008 Name Instructions 0 There are 4 problems Make sure you are not missing any pages 0 Give complete convincing and clear answers or points will be deducted o No calculators books or notes are allowed 0 Answer the questions in the spaces provided on the question sheets If you run out of room for an answer continue on the back of the page Question Points Score 1 10 2 10 3 10 4 10 Total 40 1 10 points Let f A a B and g B a A be functions satisfying 9 o z for all z E A Prove that f is one to one Must f be onto B Justify your answer Solution First we prove that f is one to one Suppose 12 E A and fx1 zz We need to show that 1 2 Since fz1 zz we have 9 o f1 gfz1 gfz2 gof But by our assumption that gof x this implies that 1 gof 1 9 O f2 2 It is not necessarily the case that f is onto B For example we may take A 1 B 12 f 11 and g 1121 Then f does not map A onto B but 9 0 fx z for every x E A 2 10 points Let J be a nonempty subset of R that has the following properties i J is bounded ii supJ E J iii inf J g J iv lf Ly E J with z lt y then t E J for every t satisfying z lt t lt y Prove that J is the interval inf J sup J Solution We need to show that a J C inf Jsup J and b that inf Jsup J C J For a let t E J Since ian is a lower bound for J and supJ is an upper bound for J we have ian S t S supJ Since ian J we also have ian lt t S supJ and so t E inf Jsup J For b let t E inf Jsup J If t sup J we have t E J by ii and so we may assume that inf J lt t lt sup J Since t gt inf J and inf J is the greatest lower bound we see that t is not a lower bound for J and so there is an x E J with z lt t Since t lt supJ and supJ is the least upper bound we see that t is not an upper bound for J and so there is a y E J with y gt t Thus z lt t lt y and so by property iv t E J 3 10 points Let A and B be nonempty subsets of the positive real numbers which are bounded above Prove that supa b a 6 Ab E B sup A sup B Solution We need to show that supA supB is an upper bound for a b a 6 Al 6 B7 and ii if 04 lt supA supB then 04 is not an upper bound for a b a 6 Al 6 B For i7 let a E A and b E B Then a S supA and b S supB7 so ab S supAb S supA sup B For ii7 suppose 04 lt supA supB and let 6 supA supB 7 04 gt 0 Since supA is the least upper bound7 we may nd a E A with a gt supA 7 64 Since sup B is the least upper bound7 we may nd b E B with b gt sup B764 Then ab gt supAsupB762 gt supA supB 7 E Q So 04 is not an upper bound for a b a 6 Al 6 B 4 10 points For each positive integer k let Nk 17xk 1k E N be the set of ordered k tuplets of positive integers Prove that for every positive integer k Nk is countable You are allowed to use7 without proof7 the fact that N gtlt N is countable You are also allowed to use7 without proof7 the fact that if f A a B and g B a C are both bijective7 then 9 o f A a C is bijective Hint Try using induction on k Solution By induction it suf ces to show that N1 is countable and that ii if Nk is countable then Nki l is countable For i7 it is clear that f N a N1 de ned by fn is a bijection7 and so N1 is countable For ii assume that Nk is countable Then7 there exists a bijective g N a Nk Consider the function f N gtlt N a N19 l de ned by fnm 717 7yk1 where 12 7yk1 We claim that f is bijective To see that f is onto7 let 11 7yk1 E Nk Since 9 is onto7 there exists m E N with gm 12 7yk17 and so fy1m 11 7yk1 To see that f is one to one7 suppose fn1m1 y17yk1 fn2m2 Then gm1 yg7 7yk1 9m2 Since 9 is one to one7 this implies that m1 m2 But we also have 711 yl 712 and so 7117711 7127712 Since N gtlt N is countable7 there exists a bijective h N a N gtlt N Since f is bijective and h is bijective7 f o h N a Nki l is bijective7 and so Nki l is countable Math 131A1 Handout 2 Some pointers on the logical conventions of mathematics It is probably hardest to get used to the way mathematicians use the word implies77 or the symbol The idea is that you want to know if the implication is true or not just on the basis of whether the ingredients are true or false without any more thinking 0 Any true statement any true statement eg 11 2 there are in nitely many primes Fermat7s last conjecture 0 Any false statment any true statement because in particular you want to be able to say that 1 2 1 X 0 2 X 0 is a correct deduction 0 Any false statement any false statement because in particular you want to be able to say that 12 11 21 is a correct deduction o The following is false truth false H The most common logical errors made by beginners 0 They think that or77 is exclusive thus although they know that S means less than or equal to77 they think it is wrong 7 to write 3 S 3 because they know that actually 3 377 0 They think that you cannot prove P Q if you already know that P is false 0 They think that if P Q is true then Q is true 0 When asked to prove P Q they instead prove Q P 0 They get equality say of numbers mixed up with 42gt logical equivalence used for propositions 2 Some correct illustrations of logic 0 The proposition 6 lt 7 or 4 lt 577 is true 0 The proposition 1 2 0 X 1 0 X 277 is true 0 The proposition For any real number 1 12 3 zz2 13 is true 0 The proposition For any real number I zz2 13 1 2 377 is false 3 It is important to be able to take the negations of statements in order to prove things by contradiction Here is the general scheme donlt worry about the last two at this point Math 131a Handout 6 Our completeness axiom If S is a nonempty subset of R and S is bounded above ie7 S S b for some b7 then S has a least upper bound bo sup 5 You fomulate the corresponding result for nonempty sets that are bounded below Here are theorems about sequences and their limts that you should be able to prove including the relevant de nitions lf In is a convergent sequence7 then it must be bounded Proof Suppose that In 7gt L Choose no such that n 2 no In 7 L lt 1 Then 7 L S In7L lt 1 implies that lt L 1 Let M max I1 lIno1llLl 1 We have that for all n S M lf In 7gt L and In 0 and L f 07 then there is a constant c gt 0 such that In 2 c for all n Proof Suppose rst that In gt 0 and L gt 0 Choose no such that n 2 no In 7 L lt L2 Then L 7 In S In7L lt L2 implies that In gt L7L2 L2 Let c min IhIg7 7Ino17 LQ It follows that In 2 c for all n For the general case note that 7gt L and use the positive result lfIn 7gt L and for all n7 In 2 07 then L 2 0 Proof Suppose that L lt 0 Then let 5 7L We may choose no such that In 7 L lt 5 Then InO 7 L lt 5 InO lt L 5 07 contradicting Inc 2 0A The usual limit theorems such as In 7gt L and yn 7gt M implies Inyn 7gt L M lf In is an increasing sequence7 and In S 127 then In 7 ho sup You should be able to state and prove the corresponding result for de creasing sequences Proof Given 5 gt 07 we have that ho 7 5 lt bo implies that ho 7 5 is not an upper bound for 7 hence there exists an no with bo 7 5 lt Inc It follows that ifn 2 no7 then bo75 lt InO S In S no and thus In 7 bo lt 5 If 0 f S Q R and bo sup 5 then there is a sequence In E S such that In 7 no You should be able to state and prove the corresponding result for the in mum Proof Given n E N bo 7 ln is not an upper bound for 5 hence we may choose an In E S such that ho 7 ln lt In S no It follows that In 7 bo lt ln7 and thus In 7 ho

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