THEOREM 5.7. (1) If a set A is relatively compact, it is also totally bounded, in the metric space (X, d). (2) If (X, d) is complete and A is totally bounded, then A is relatively compact. ® How does one of the spheres have infinite no of xn Ⓒ How do diagonal elements yield couchy seq 7. © Request explain this step (1) Suppose A is relatively compact. We can apply the preceding theorem to assert that, from any sequence in A, a convergent subsequence can be extracted. We need only apply Theorem 5.3 now to conclude that A is totally bounded. (2) Let (X, d) be a complete metric space and let A be a totally bounded subset of X. By Theorem 5.6 we must show that it is possible to extract a convergent subsequence from every sequence of points in A. To this end, let {x} be a sequence of points from A. Since A is assumed to be totally bounded, for each &= 1/k, where k is a positive integer, there must be a finite e-net: {y(), y(), . Considering the collection of spheres of radius 1 about each point in the k = 1 net, we see that this collection must cover all of A. In particular, the x, (n = 1, 2, ...) must be contained in a finite number of these spheres. Further, one of these spheres must contain an infinite number of the x,. Hence we have some sphere, S, say, containing the subsequence of {x}: x₁(1), x₂(1), xy(1), ..., where x₁(¹) € {x₁, x₂,...} (for i=1, 2, ...). Reasoning along the same lines for the collection of neighborhoods of radius about the points in the k= 2 net, we can get some sphere of radius from this collection containing infinitely many of the x¹). Thus, calling this sphere S₂ and denoting the points it contains by x, (2), x₂(2), ..., we have S₂: x₁(2), x₂(²), x₂(2), ..., where x₁ (²) € (X₁, X₂, ...) Proceeding in this fashion, we arrive at Fig. 5.2. (i=1,2,...). S₁: x₁,x₂,... S₂: x₂ FIG. 5.2. It is now clear that selecting the “diagonal" terms (see arrows in Fig. 5.2, the sequence {x}) yields a Cauchy sequence which, since the space is complete, must have a limit in X. Since each of the sequences {x} (n is the index here, while i is fixed) is a subsequence of {x-1)} it is clear that {x} is a subsequence (3 of the original sequence {x}. Hence we need only apply the preceding theorem to complete the proof.