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Tutorial-3 - Analysis II Tutorial Solution sketches...

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Unformatted text preview: Analysis II Tutorial Solution sketches Alexandre Tomberg * 1 January 27, 2011 1.1 Let X be a set and A,B X . 1. Prove that ( A B ) A B . 2. Find an example of X,A,B s.t. ( A B ) 6 = A B . 3. Find a example of a sequence of sets { A n } 1 s.t. [ n =1 A n [ n =1 A n ! properly . Solution. 1. Let us compute: ( A B ) = cl( A B ) cl( A B ) c = (cl A cl B ) cl( A c B c ) = ( cl A cl( A c B c ) ) ( cl B cl( A c B c ) ) ( cl A cl( A c X ) ) ( cl B cl( X B c ) ) = A B. 2. Consider X = R ,A = (0 , 1) ,B = (1 / 2 , 2). Then A = { , 1 } ; B = { 1 / 2 , 2 } ; ( A B ) = { , 2 } . 3. Take X = R ,A n = { r n } , where { r 1 ,...,r n ... } is an enumeration of Q . Then A n = { r n } ; [ n =1 A n = Q ; [ n =1 A n ! = Q = R . * Please report any errors and typos to atomberg@math.mcgill.ca. All tutorial solutions in one pdf can be found at http://www.cs.mcgill.ca/ ~ atombe2/AnalysisTutorial.pdf . 1 1.2 Let [ a,b ] be an interval and X = C ([ a,b ]) the vector space of all continuous real-valued functions on [ a,b ] equipped with the norm k f k = max a t b | f ( t ) | . Let F be a fixed continuous function such that F ( t ) > 0 for all t [ a,b ] and let S = { f X : | f ( t ) | < F ( t ) for all t [ a,b ] } . 1. Prove that S is open. 2. Find cl( S ). 3. Find S . Proof. 1. To prove that S is open, we need to show that ( f S ) ( > 0) ( D ( f, ) = { g X : k f- g k < } S ) . So, let f S be given, and consider the function ( t ) : [ a,b ] R , defined by ( t ) = F ( t )- | f ( t ) | . Note that f S means that t [ a,b ], ( t ) > 0. Also, is continuous as a sum and composition of continuous functions. Thus, by a theorem from Analysis I, attains its minimum on [ a,b ], and thus, = min t [ a,b ] ( t ) = ( t min ) > 0. Take = / 2. Then k f- g k < means that for all t [ a,b ], | f ( t )- g ( t ) | < . Then t [ a,b ], F ( t )- | g ( t ) | F ( t )- ( | g ( t )- f ( t ) | + | f ( t ) | ) ( t )- > - / 2 > . Thus k f- g k < implies that g S , i.e. D ( f, ) S . Done. 2. Let C = { f X : | f ( t ) | F ( t ) for all t [ a,b ] } , we will prove that cl( S ) = C . First, C is closed, i.e. C c is open, where C c = { f X : | f ( t ) | > F ( t ) for some t [ a,b ] } . To see this, we use the same trick as in 1: pick an arbitrary f C c and define a continuous function ( t ) = | f ( t ) | - F ( t ). Since f C c , ( t ) > 0 for some t o [ a,b ], so that = ( t o ) > 0. Then D ( f,/ 2) C c because k g- f k < / 2 = | g ( t o ) |- F...
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Tutorial-3 - Analysis II Tutorial Solution sketches...

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