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Unformatted text preview: Solutions homework 12. Page 78 Problem 5. (i) Let B = C . Then obvious that B A , as each interval ( r,s ) A . Conversely let x A . Then there exists > 0 such that ( x ,x + ) A (as A is open). Now take r Q such that x < r < x and s Q such that x < s < x + . Then ( r,s ) ( x ,x + ) A . Hence x ( r,s ) B . It follows that A B . (ii) The proof is identical to the first part as [ r,s ] ( x ,x + ) A . (iii) Obvious. (iv) If A had a finite subcover of D , then A would be a finite union of closed intervals and thus closed. On the other hand it is open, then it would be a finite or countable union of disjoint open intervals. Now none of the endpoints of these open intervals can be in A , as A is open, but they are in A when A is closed. Contradiction. (v) If A is a finite union of open intervals with rational endpoints, then that collection of intervals constitute a finite subcover of A from C . Cnversely if A has a finite subcover from...
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This note was uploaded on 02/05/2012 for the course MATH 554 taught by Professor Girardi during the Fall '10 term at South Carolina.
 Fall '10
 Girardi

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