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Tutorial-3

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

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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 = { 0 , 1 } ; ∂B = { 1 / 2 , 2 } ; ( A B ) = { 0 , 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 [email protected] All tutorial solutions in one pdf can be found at http://www.cs.mcgill.ca/ ~ atombe2/AnalysisTutorial.pdf . 1

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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 φ ( t ) : [ a, b ] R , defined by φ ( t ) = F ( t ) - | f ( t ) | . Note that f S means that t [ a, b ], φ ( t ) > 0. Also, φ is continuous as a sum and composition of continuous functions. Thus, by a theorem from Analysis I, φ attains its minimum on [ a, b ], and thus, δ = min t [ a,b ] φ ( t ) = φ ( t min ) > 0. Take = δ/ 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 ) | ) φ ( t ) - > δ - δ/ 2 > 0 . 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 Φ( t ) = | f ( t ) | - F ( t ). Since f C c , Φ( t ) > 0 for some t o [ a, b ], so that δ = Φ( t o ) > 0. Then D ( f, δ/ 2) C c because k g - f k < δ/ 2 = | g ( t o ) | - F ( t o ) ≥ | f ( t o ) | - | g ( t o ) - f ( t o ) | - F ( t ) > δ - δ/ 2 > 0 . Since cl S is the smallest closed set containing S , we certainly have that cl S C . Conversely, let us show that C cl S , i.e. ( > 0 , f C ) ( D ( f, ) S 6 = ) .
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