Download 2.15 A metric space is called separable if it contains a countable

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.