21.4. CLOSED SUBSPACES 553

First suppose cp ≡(cp

1 , · · · ,cpn)

is not bounded in Fn. Then dp ≡ cp/ |cp|Fn is a unit vectorin Fn and so there exists a subsequence, still denoted by dp which converges to d where|d|= 1. Then

0= limp→∞

x∥cp∥ = lim

p→∞

n

∑k=1

dpk xk =

n

∑k=1

dkxk

where ∑k |dk|2 = 1 in contradiction to the linear independence of the {x1, · · · ,xn}. Hence itmust be the case that cp is bounded in Fn. Then taking a subsequence, still denoted as p, itcan be assumed cp→ c and then in 21.10 it follows x = ∑

nk=1 ckxk ∈ span(x1, · · · ,xn). ■

Proposition 21.4.2 Let E be a separable Banach space. Then there exists an increas-ing sequence of subspaces, {Fn} such that dim(Fn+1)−dim(Fn)≤ 1 and equals 1 for all nif the dimension of E is infinite. Also ∪∞

n=1Fn is dense in E. In the case where E is infinitedimensional, Fn = span(e1, · · · ,en) where for each n

dist(en+1,Fn)≥12

(21.11)

and defining,Gk ≡ span

({e j : j ̸= k

})dist(ek,Gk)≥

14

. (21.12)

Proof: Since E is separable, so is ∂B(0,1) , the boundary of the unit ball thanks toCorollary 3.4.3. Let {wk}∞

k=1 be a countable dense subset of ∂B(0,1).Let e1 = w1. Let F1 = Fe1. Suppose Fn has been obtained and equals the following:

span(e1, · · · ,en) where {e1, · · · ,en} is independent, ∥ek∥= 1, and

dist(en,span(e1, · · · ,en−1))≥12.

For each n, Fn is closed by Theorem 21.4.1.If Fn contains {wk}∞

k=1 , let Fm = Fn for all m > n. Otherwise, pick w ∈ {wk} to be thepoint of {wk}∞

k=1 having the smallest subscript which is not contained in Fn. Then w is ata positive distance λ from Fn because Fn is closed. Therefore, there exists y ∈ Fn such thatλ ≤ ∥y−w∥ ≤ 2λ . Let en+1 =

w−y∥w−y∥ . It follows

w = ∥w− y∥en+1 + y ∈ span(e1, · · · ,en+1)≡ Fn+1

Then if x ∈ span(e1, · · · ,en) ,

∥en+1− x∥ =

∥∥∥∥ w− y∥w− y∥

− x∥∥∥∥= ∥∥∥∥ w− y

∥w− y∥− ∥w− y∥x∥w− y∥

∥∥∥∥≥ 1

2λ∥w− y−∥w− y∥x∥ ≥ λ

2λ=

12.

This has shown the existence of an increasing sequence of subspaces, {Fn} as describedabove. It remains to show the union of these subspaces is dense. First note that the union ofthese subspaces must contain the {wk}∞

k=1 because if wm is missing, then it would contradict