sol-10 - MATH 453 SOLUTIONS TO ASSIGNMENT 10 NOVEMBER 23,...

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M  453 S   A  10 N  23, 2004 Exercise 5 from Section 30, page 194 (a) Let D be a countable dense subset of the metrizable space X . I claim that B = { B ( x , 1 / n ) | x D and n Z + } is a countable basis. First, B is countable, since both D and Z + are. To show that it is a basis for the topology on X , let U be any open set, and let y U . Then there is an open ball, B ( y , ε ), around y that is contained in U . Choose n to be larger that 2 , and pick an x D such that x B ( y , 1 / n ). This gives y B ( x , 1 / n ) B ( y , ε ) U . Hence B is a basis. (b) Let X be a metrizable Lindel¨of space. For each n Z + , the collection of open sets { B ( x , 1 / n ) | x X } covers X . Pick a countable subcover and call it B n . Then the collection B = S n = 1 B n is a countable basis. Countability is immediate, and the proof that it is a basis is analogous to that in part (a). ± Exercise 6 from Section 30, page 194 We know from Example 3 (p. 192) that R ` is not second countable, but has a countable dense subset. Thus part (a) above implies that R ` cannot be metrizable. Suppose I 2 o were metrizable. Since I 2 o is also compact (and hence Lindel¨of), part (b) above implies that it is second countable. This in turn shows that the subspace A = I × (0 , 1) is second countable, and hence Lindel¨of. However, Example 5 (p. 193) shows that A is not Lindel¨of, so I 2 o cannot be metrizable. ± Exercise 12 from Section 30, page 194 I’ll show the case for second countability; the only di erence for first countability is to start with a countable basis at a point instead. Suppose B is a countable basis for X . Since f is an open map, B 0 = { f ( B ) | B B } is a collection of open sets in f ( X ). To show that it is a basis for
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sol-10 - MATH 453 SOLUTIONS TO ASSIGNMENT 10 NOVEMBER 23,...

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