2.4. SEPARABILITY AND COMPLETE SEPARABILITY 43

Corollary 2.3.6 Let {zk} be a Cauchy sequence in C. Then it converges.

Proof: Say zk = xk + iyk. Then from the way we define distance in C, {zk} is Cauchyif and only if {xk} and {yk} are both Cauchy. Indeed, |zk− zm|2 ≡ |xk− xm|2 + |yk− ym|2 .Therefore, there exists x,y such that xk→ x,yk→ y. It follows zk→ x+ iy. ■

Now consider the case that the metric space is Fp.

Theorem 2.3.7 A sequence{

xk}

converges in Fp if and only if it is a Cauchy se-quence.

Proof:⇐Let xk =(xk

1, · · · ,xkp). Then since this is a Cauchy sequence, the components

are Cauchy, and so, from what was just shown, limk→∞ xki = xi which, from the way we

define the norm implies xk→ x≡(

x1 · · · xp)

⇒ If limk→∞ xk = x, then for k large enough,∣∣xk−x

∣∣ < ε/2. Hence if k,m are largeenough, ∣∣∣xk−xm

∣∣∣≤ ∣∣∣xk−x∣∣∣+ |x−xm|< ε

2+

ε

2= ε ■

2.4 Separability and Complete SeparabilityDefinition 2.4.1 A metric space is called separable if there exists a countable densesubset D. This means two things. First, D is countable, and second, that if x is any pointand r > 0, then B(x,r)∩D ̸= /0. A metric space is called completely separable if thereexists a countable collection of nonempty open sets B such that every open set is the unionof some subset of B. This collection of open sets is called a countable basis.

For those who like to fuss about empty sets, the empty set is open and it is indeed theunion of a subset of B namely the empty subset.

Theorem 2.4.2 A metric space is separable if and only if it is completely separable.In fact a separable metric space has a countable basis of balls. Also Fp is separable.

Proof: ⇐= Let B be the special countable collection of open sets and for each B ∈B,let pB be a point of B. Then let P ≡ {pB : B ∈B}. To be specific, let pB be the center ofB. If B(x,r) is any ball, then it is the union of sets of B and so there is a point of P in it.Since B is countable, so is P .

=⇒ Let D be the countable dense set and let

B ≡{B(d,r) : d ∈ D,r ∈Q∩ [0,∞)}

Then B is countable because the Cartesian product of countable sets is countable. Itsuffices to show that every ball is the union of these sets. Let B(x,R) be a ball. Lety ∈ B(y,δ ) ⊆ B(x,R) . Then there exists d ∈ B

(y, δ

10

). Let ε ∈ Q and δ

10 < ε < δ

5 . Theny ∈ B(d,ε) ∈B. Is B(d,ε) ⊆ B(x,R)? If so, then the desired result follows because thiswould show that every y∈ B(x,R) is contained in one of these sets of B which is containedin B(x,R) showing that B(x,R) is the union of sets of B. Let z∈ B(d,ε)⊆ B

(d, δ

5

). Then

d (y,z) ≤ d (y,d) + d (d,z) < δ

10 + ε < δ

10 + δ

5 < δ . Hence B(d,ε) ⊆ B(y,δ ) ⊆ B(x,r).Therefore, every ball is the union of sets of B and, since every open set is the union ofballs, it follows that every open set is the union of sets of B.