350 Math - Assignment 3 - Solutions
1. Let A be a set of real numbers. Show that if every open interval that contains L also contains a point of the set A which is different from L, then for every > 0, (L - , L + ) contains infinitely many points of A.
Solution: Assume that there is an such that the interval (L - , L + ) only contains finitely many points of A. Let S = {x1 , . . . xN } be the set of these points which are diferet from L. Let m = min{|xj - L| : xj S}. Then m > 0 and the interval (L - m, L + m) does not contain any point Of A that is different from L.
2. Construct a sequence {an } such that lim inf an = +. Solution: an = n. 3. Let {an } be a sequence. And let A be the set of all subsequential limit points of this sequence. Prove that sup A is a subsequential limit point. Solution: Let A = sup{S}, where S is a set of limit points of {an }, call them . For each > 0, , limit points of {} such that A- < . We can 1 then take = n , and for each n consider bn such that A - bn < . Now we have a sequence {bn } limit of points converging to A. For every k, consider 1 1 1 = k and some element that we name ank such that ank (bk - k , bk + k ) and nj < nk , if j < k. We show now that ank A, which implies that A is a subsequenctial limit of {an }. We need to show that given > 0, N such that if n > N , then |ank - A| < . But we know that for a given , k 1 such that k < 2 and |ank -A| = |ank -bk +bk -A| |ank -bk |+|bk -A| < 1 k + 2 < 2 + 2 = . 4. Give an example of infinitely many closed sets whose union is not closed. Solution: Let Fn = [1/n, 1]. Then
nN
fn = (0, 1].
Solution: Let x A B. Since A is o...