# sol-07 - MATH 453 SOLUTIONS TO ASSIGNMENT 7 NOVEMBER 1 2004...

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M  453 S   A  7 N  1, 2004 Exercise 3 from Section 23, page 152 Let B α = A A α . Then B α is connected for each α since A and A α are connect and have a point in common. Now S B α = A S A α and the B α ’s have a point in common ( A is nonempty) so A S A α is connected. ± Exercise 10 from Section 24, page 158 Suppose x 0 U and let V be the set of points in U that can be joined to x 0 by a path in U . V is non-empty since it contains x 0 . If x V , then there is an open ball B ( x , ε ) around x contained in U as U is open. Each point of B ( x , ε ) can be connected to x via a straight line (path) in B ( x , ε ), and x can be connected to x 0 via a path in U . Hence each element of B ( x , ε ) can be joined to x 0 via a path in U . Thus B ( x , ε ) V and so V is open. Now let x be a limit point of V in U . Again we can ﬁnd an open ball B ( x , ε ) around x that is contained in
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