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2.15 A metric space is called separable if it contains a countable dense subset. Show that Rk is separable. Hint: Consider the set of points which have only rational coordinates. Rudin’s Ex. 22 Proof Let Qk = {(x1 , . . . , xk ) ∈ Rk : xi ∈ Q for 1 ≤ i ≤ k}. Let > 0 be given. For any x = (x1 , . . . , xk ) ∈ Rk , by Theorem 1.20, Q is dense in R, so there exists p1 , . . . , pk ∈ Q such that |xi − pi | < √1k . Put p = (p1 , . . . , pk ). It is clear that p ∈ Qk , and d(p, x) = k X !1/2 2 |xi − pi | i=1 < k X 1 2 k i=1 !1/2 = . Hence, Qk is dense in Rk . By Theorem 2.13, Qk is countable. Therefore, Rk is separable. 2.16 Prove that every open set in R is the union of an at most countable collection of disjoint segments. Hint: Use Exercise 22. Rudin’s Ex. 29 Proof Let G be an open set in R. For each x ∈ G, there are y and z, with z < x < y, such that (z, y) ⊂ G. Let b = sup{y : (x, y) ⊂ G} and a = inf{z : (z, x) ⊂ G}. Then −∞ ≤ a < x < b ≤ ∞. Put Ix = (a, b). It is clear that Ix is a segment. We claim that b ∈ / G. In fact, there is nothing to prove if b = ∞. If b is finite, and b ∈ G, then there is some δ > 0 such that (b − δ, b + δ) ⊂ G since G is open. This contradicts to the definition of b. Similarly, a ∈ / G. We shall prove that Ix ⊂ G. Let w ∈ Ix , say x < w < b. By the definition of b, there is y > x such that (x, y) ⊂ G. Hence w ∈ G. For each x ∈ G, the above construction yields a collection of segments {Ix }. We claim that G = ∪Ix . In fact, since each Ix ⊂ G, we have ∪Ix ⊂ G. On the other hand, for any x ∈ G, we know there is Ix such that x ∈ Ix . This implies x ∈ ∪Ix , so that G ⊂ ∪Ix . It remains to show that the collection of segments {Ix } is disjoint and countable. To show that {Ix } is disjoint, we let (a, b) and (c, d) be any two segments in the collection with both containing a common point x. Since a < x < b and c < x < d, we have c < b and a < d. Since c ∈ / G, it does not belong to (a, b), so that c ≤ a. The reversed inequality a ≤ c holds by the same argument. Hence a = c. Similarly, b = d. Thus, any two different segments in the collection {Ix } are disjoint. To show that the collection {Ix } is countable, we choose a rational number in each Ix as its representative. This can be done since Q is dense in R. Since we have a disjoint collection, each segment contains a different rational number. Hence the collection can be put in one-to-one correspondence with a subset of the rational numbers. Thus it is an at most countable collection. 1 We can similarly discuss the case of a < w < x.