hmwk6sol_b

hmwk6sol_b - Math 202A Homework 6 Roman Vaisberg Problem...

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Unformatted text preview: Math 202A Homework 6 Roman Vaisberg Problem 1. (a) Let X be a topological space, { C n } be a collection of connected subsets such that any two of them have a point in common. Prove that G = [ C n is connected. Suppose G is not connected, and let U and V be open sets that separate G . Let x 1 ∈ G ∩ U , and x 2 ∈ G ∩ V . Without loss of generality we may suppose that C 1 3 x 1 and C 2 3 x 2 . Claim: C 1 ⊂ U . Suppose C 1 has a point in V . Then we have two open sets U and V in X that meet C 1 . We know C 1 ⊂ G ⊂ U ∪ V and C 1 ∩ U ∩ V = ∅ . ⇒ C 1 is not connected, which is a contradiction. Thus C 1 does not have a point in V and the claim is proven. A similar argument shows that C 2 ⊂ V . But this would imply that C 1 ∩ C 2 = ∅ , which contradicts the statement of the problem. So G must be connected. (b) Let X be a topological space and Y ⊂ X . Prove that if Y is connected, then ¯ Y is connected. Suppose ¯ Y is not connected and let U , V be open sets that separate ¯ Y . It is clear that Y ⊂ U ∪ V and Y ∩ U ∩ V = ∅ . But since U and V both contain a point of ¯ Y , they must each contain a point of Y . Thus Y ∩ U 6 = ∅ and Y ∩ V 6 = ∅ . Thus U and V are a separation of Y , which is a contradiction since Y is connected. Thus ¯ Y is connected. (c) Prove that the components of a locally connected topological space are open. Let B be a base of connected sets for X and let C be a component of X . Let x ∈ C and let B ∈ B be such that x ∈ B . Since B and C and connected we have that B ∪ C is connected. However, since C is maximally connected we must have that B ⊂ C . We conclude that C is open. (d) Prove that the topologist’s sine curve is not path connected. Let γ : [0 , 1] → X be the path that joins ( 1 π , 0) and (0 , 0) i.e. γ (0) = ( 1 π , 0), γ (1) = (0 , 0). Let π x : [0 , 1] × [0 , 1] → [0 , 1] be defined by π x ( x,y ) = x . It is clear that π x is continuous and thus so is π x ◦ γ . We observe that if x 1 ,x 2 ∈ (0 , 1 π ] with x 2 < x 1 and t 1 is such that π x ◦ γ ( t 1 ) = x 1 , then by the intermediate value theorem applied to π x ◦ γ , ∃ t 2 > t 1 such that π x ◦ γ ( t 2 ) = x 2 . This observation allows us to construct a sequence ( t n ) such that t i < t i +1 and π x ◦ γ ( t i ) = 1 (2 n + 1 2 ) π . Claim: ( t n ) → 1. Clearly the limit exists since ( t n ) is increasing and bounded above. Let L = lim( t n ). By continuity of π x ◦ γ we have: 0 = lim π x ◦ γ ( t n ) = π x ◦ γ ( L ), which implies that L must be 1. Claim proven. 1 But then by continuity of γ we have (0 , 1) = lim γ ( t n ) = γ (1) = (0 , 0). (The first equality follows from the fact that γ must lie on the topologist’s sine curve which takes on values of 1 at each t n .) Contradiction....
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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.

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hmwk6sol_b - Math 202A Homework 6 Roman Vaisberg Problem...

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