Remark 1.4. Proof. 1. It suffices to take a point in each (nonempty) member of the basis and this is a dense subset, because each open set is the union of members of the basis, so it intersects this countable subset. The converse may fail for a non-Euclidean space; e.g. A metric space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X. R, C are separable. 3. Proof. Now let A = S∞ n=1 An. A metric space is separable space if it has a countable dense subset. 1. The analogy of the theories of separable and compact sets has an end when the countability of the set of radii D comes into play. A metric space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X. 10. A metric space is sequentially compact if and only if every infinite subset has an accumulation point. Theorem 1.5. ℓ∞ is not separable. Each compact metric space is complete, but the converse is false; the simplest example is an infinite discrete space with the trivial metric. If M is dense, then every ball in X contains a point of M. 2. (X, d) is second-countable, separable and Lindelöf – these three conditions are equivalent for metric spaces. A metric space is complete if every Cauchy sequence con-verges. De-note by U the collection of all compact ultrametric spaces. It suffices to take a point in each (nonempty) member of the basis and this is a dense subset, because each open set is the union of members of the basis, so it intersects this countable subset. (0,1] is not sequentially compact (using the Heine-Borel theorem) and not compact. Now every metric space having a countable basis is separable. Lemma 2 If a metric subspace M0of a space M is compact and we are given any collection of balls in Mthat cover M0, then there is a nite subcover. For a metric space the following two statements are equivalent.

on Xis tight. (0,1] is not sequentially compact (using the Heine-Borel theorem) and not compact. Theorem 2.6. X is separable if there is a countable subset which is dense in X. The converse is not true; e.g., a countable discrete space satisfies these three conditions, but is not compact. Problem 3 (WR Ch 2 #22). In general metric spaces, the boundedness is replaced by so-called total boundedness. A metric space X is compact if every open cover of X has a finite subcover. Proof.

comparing compact metric spaces. In general metric spaces, the boundedness is replaced by so-called total boundedness. X is closed and bounded (as a subset of any metric space whose restricted metric is d). A discrete metric space is separable if and only if it is countable. Theorem: A metric space is compact if and only if its space of bounded, continuous, real-valued functions is separable in the uniform topology. If (X;d) is a complete separable metric space, then every nite Borel measure on Xis tight. Typical examples are the real numbers or any Euclidean space. Proposition 2.3 Every totally bounded metric space (and in particular every compact met-ric space) is separable.