354 5- We may cite an earlier result that every convergent sequence is eventually constant, hence coverges to the integer at which it is constant (Z+ contains all its limi points). Or we may show that its complemement is a union of open sets, hence open: (-infty, 1) U (!,2) U (2,3) U (3,4) U ... 16- A point in A is in every set whic contains A, hence in the intersection of any such collection of sets. If x is not in A-, there is a neighborhood around it outside A- since A- is closed (A-' is open). Therefore x is not a boundary point of A-, hence not in A-- If x is in (AUB)-, it is the limit of a sequence of points in A and B; hence of a subsequence in A or B, hence in A- or B-. If x is in A- (or B-), it is the limit of a sequence of points in A (or B), which sequence is in AUB. If x is in (A /int B)-, it is the limit of a sequence which is in both A and B, hence it is in both A- and B-; consider A=Q and B=(R\Q) The intersection (hence closure of the intersection) is the null set, but the closure of each is R. 17- Q 359 1- {(1+(1/n), 2+(1/n))} Note that what is happening at the left endpoint is what is important, I could use any number greater than 2 for the right endpoint. 4- Take any open cover of F, and add to it the open set F' (the complement of F). The open cover of F with F' provides an open cover of K, hence there is a finite subcover of K, and removing F' leaves a finite subcover of F. 9- Choose x_n in K_n for each n. Choose a convergent subsequence (such exists since K_1 is bounded, hence all K_i are bounded by the same bound). Let x be the limit of the convergent subsequence, it is in each K_n since the tail of the subsequence is in each K_n (perhaps a later tail for larger n) and each K_n is closed.

Sun Feb 15 1998