= Exercise 1.5.8. Let (X, dд) be the metric space from Exercise 1.1.15. For each natural number n, let e(n) (e(n)) ao be the sequence in X such that e(" = 1 when n = j and e(n) := 0 when n ‡ j. Show that the set {e(n) : n = N} is a closed and bounded subset of X, but is not compact. (This is despite the fact that (X, d₁₁) is even a complete metric space - a fact which we will not prove here. The problem is that not that X is incomplete, but rather that it is "infinite-dimensional", in a sense that we will not discuss here.) Exercise 1.5.9. Show that a metric space (X, d) is compact if and only if every sequence in X has at least one limit point. Exercise 1.5.10. A metric space (X,d) is called totally bounded if for ev- ery > 0, there exists a positive integer n and a finite number of balls B(x(¹), ɛ),..., B(x(n), e) which cover X (i.e., X = U₁ B(x(i), ɛ). (a) Show that every totally bounded space is bounded. (b) Show the following stronger version of Proposition 1.5.5: if (X,d) is compact, then complete and totally bounded. (Hint: if X is not totally bounded, then there is some ε > 0 such that X cannot be covered by finitely many e-balls. Then use Exercise 8.5.20 to find an infinite se- quence of balls B(x(), /2) which are disjoint from each other. Use this to then construct a sequence which has no convergent subsequence.)