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Unformatted text preview: U.C. BERKELEY MATH 202A FINAL, FALL 2007 This is a 3 hour exam, 12:30–3:30 p.m. There are 8 problems; all will be scored and weighted equally . Please write your answers on separate paper (your own paper or the blank paper handed out). At the conclusion of the exam, you need only submit your solutions and may keep the exam question sheet. I will have a stapler at the front of the room. Solutions will be posted online immediately following the exam ( http://math.berkeley.edu/ ∼ holmer/ ). Problem 1 . Suppose X is a complete metric space that is not compact. Prove that there exists > 0 and a sequence x n in X such that for n 6 = m , d ( x n ,x m ) ≥ . Solution . Since X is complete but not compact, it is not totally bounded. Thus there exists > 0 such that X cannot be covered by a finite number of balls of radius . Pick x 1 ∈ X . Then B ( x 1 , ) does not cover X , and thus ∃ x 2 / ∈ B ( x 1 , ). Since B ( x 1 , ) ∪ B ( x 2 , ) does not cover X , there exists x 3 / ∈ B ( x 1 , ) ∪ B ( x 2 , ). Etc. Problem 2 . Suppose X is a topological space, { Y α } α ∈ I is a (possibly uncountable) collection of topological spaces, and for each α ∈ I , there is given a function f α : X → Y α . In terms of these functions f α , we can define the function F : X → Q α ∈ I Y α by F ( x ) = ( f α ( x )) α ∈ I . Prove that F : X → Q α ∈ I Y α is continuous if and only if for each α ∈ I , f : X → Y α is continuous. Solution . Suppose F is continuous. Fix α ∈ I . Let U α be an open set in Y α . Then f 1 α ( U α ) = F 1 Y α ∈ I U α ! where U α = U α if α = α and U α = Y α if α 6 = α . But the righthand side is open in X since F is continuous. Conversely, suppose that ∀ α ∈ I , f α : X → Y α is continuous. Let Q α ∈ I U α be a subbase element of the product topology, i.e. there exists α ∈ I such that for all α 6 = α , U α = Y α and U α is open in Y α . Then F 1 Y α ∈ I U α ! = f 1 α ( U α ) and the righthand side is open in X since f α is continuous. Since it suffices to check that the preimage of each subbase element is open, we have established that F is continuous. Problem 3 . (a) Let X be a topological space and Y be a Hausdorff topological space. Let A be a dense subset of X , and let f : X → Y and g : X → Y be two 1 2 U.C. BERKELEY MATH 202A FINAL, FALL 2007 given continuous functions. Show that if f and g agree on A (i.e. f ( x ) = g ( x ) for all x ∈ A ), then they agree on all of X (i.e. f and g are identical). (b) Now further suppose that X is T3 1 2 (completely regular and points are closed), and A is a subset of X that is not dense. Show that there exists two continuous functions f : X → R and g : X → R that agree on A but not on all of X ....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
 Spring '10
 tao/analysis
 Math

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