250 CHAPTER 10. NORMED LINEAR SPACES
Example 10.1.23 Let n∈N. Cn with distance given by d (x,y)≡max j∈{1,··· ,n}{∣∣x j− y j
∣∣}is a complete space. Recall that |a+ jb| ≡
√a2 +b2. Then Cn is complete. Similarly Rn is
complete.
To see that this is complete, let{xk}∞
k=1 be a Cauchy sequence. Observe that for each
j,{
xkj
}∞
k=1. That is, each component is a Cauchy sequence in C. Next,∣∣∣Rexk
j−Rexk+pj
∣∣∣≤ ∣∣∣xkj− xk+p
j
∣∣∣Therefore,
{Rexk
j
}∞
k=1is a Cauchy sequence. Similarly
{Imxk
j
}∞
k=1is a Cauchy sequence.
It follows from completeness of R shown above, that these converge. Thus there existsa j,b j such that
limk→∞
Rexkj + i Imxk
j = a j + ib j ≡ x
and so xk→x showing thatCn is complete. The same argument shows thatRn is complete.It is easier because you don’t need to fuss with real and imaginary parts.
10.1.3 Closure of a SetNext is the topic of the closure of a set.
Definition 10.1.24 Let A be a nonempty subset of (X ,d) a metric space. Then A is definedto be the intersection of all closed sets which contain A. Note the whole space, X is onesuch closed set which contains A. The whole space X is closed because its complement isopen, its complement being /0. It is certainly true that every point of the empty set is aninterior point because there are no points of /0.
Lemma 10.1.25 Let A be a nonempty set in (X ,d) . Then A is a closed set and A = A∪A′
where A′ denotes the set of limit points of A.
Proof: First of all, denote by C the set of closed sets which contain A. Then A = ∩Cand this will be closed if its complement is open. However,
(A)C
= ∪{
HC : H ∈ C}.Each
HC is open and so the union of all these open sets must also be open. This is because if x isin this union, then it is in at least one of them. Hence it is an interior point of that one. Butthis implies it is an interior point of the union of them all which is an even larger set. ThusA is closed.
The interesting part is the next claim. First note that from the definition, A ⊆ A so ifx ∈ A, then x ∈ A. Now consider y ∈ A′ but y /∈ A. If y /∈ A, a closed set, then there existsB(y,r)⊆ AC
. Thus y cannot be a limit point of A, a contradiction. Therefore, A∪A′ ⊆ ANext suppose x ∈ A and suppose x /∈ A. Then if B(x,r) contains no points of A different
than x, since x itself is not in A, it would follow that B(x,r)∩A = /0 and so recalling thatopen balls are open, B(x,r)C is a closed set containing A so from the definition, it alsocontains A which is contrary to the assertion that x ∈ A. Hence if x /∈ A, then x ∈ A′ and soA∪A′ ⊇ A ■