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Unformatted text preview: 4130 HOMEWORK 6 Due Thursday April 1 (1) Let A ⊂ R . A point x ∈ A is called isolated if it is not a cluster point of A . (a) Can an open set have an isolated point? Can a closed set have one? An open set U cannot have an isolated point because if x ∈ U and δ > 0 then ( x δ,x + δ ) contains an interval and hence contains infinitely many points of U . On the other hand, for any x , { x } is a closed set which does have an isolated point, namely x itself. (b) Give an example of a countable set with no isolated points. The set Q is countable and has no isolated points because if q ∈ Q and δ > 0, then ( q δ,q + δ ) contains infinitely many rational numbers, and so q is a cluster point of Q . (2) Section 3.3.1 # 8. If A ⊂ R is compact, then A is bounded, so sup( A ) and inf( A ) exist. For each n ∈ N , there exists a point a n ∈ A with sup( A ) ≥ a n > sup( A ) 1 n . The sequence { a n } converges in R to sup( A ). Since A is compact, we have that lim n →∞ a n ∈ A and thus sup( A ) ∈ A . Similarly, inf( A ) ∈ A . For the counterexample, take [ 1 , 0) ∪ (0 , 1]. This is not closed, hence noncompact, but it contains its sup and inf. (3) Section 4.2.4 # 3. (Recall that an interval is, by definition, a subset I of R such that for all x,y ∈ I and all z ∈ R with x < z < y , we have z ∈ I .) We are asked to show that the continuous image of an interval is an interval. Let I be an interval. Let f : I → R be a continuous function. Let X = f ( I ). We need to show that X is an interval. Let x,y ∈ X . Let z ∈ R be a number such that x < z < y . We need to show that z ∈ X ....
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 '08
 PROTSAK
 Topology, Intermediate Value Theorem, Continuous function, Metric space, Brother Albert

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