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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- Fall '10
- Topology, Empty set, Metric space, Open set, Finite set