The following says that subsequences converge to the same thing that a convergent sequence
converges to. We leave the proof to you.
Theorem B.2.18Let
{xn}
be a vector valued sequence with lim_{n→∞}x_{n} = xand let
{xnk}
be a subsequence. Then lim_{k→∞}x_{nk} = x.
The following is the definition of a Cauchy sequence in ℝ^{p}.
Definition B.2.19
{an}
is a Cauchy sequence if for all ε > 0, there exists n_{ε}suchthat whenever n,m ≥ n_{ε},
|an− am| < ε.
One has the same set of Cauchy sequences if
|⋅|
is replaced with
∥⋅∥
thanks to equivalence
of norms.
A sequence is Cauchy, means the terms are “bunching up to each other” as m,n get
large.
Theorem B.2.20The set of terms in a Cauchy sequence in ℝ^{p}is bounded inthe sense that for all n,
|an|
< M for some M < ∞.
Proof: Let ε = 1 in the definition of a Cauchy sequence and let n > n_{1}. Then from the
definition,
|an− an1| < 1.
It follows that for all n > n_{1},
|a | < 1+ |a |.
n n1
Therefore, for all n,
∑n1
|an| ≤ 1+ |an1|+ |ak|.■
k=1
Theorem B.2.21If a sequence
{an}
in ℝ^{p}converges, then the sequence is aCauchy sequence. Also, if some subsequence of a Cauchy sequence converges, then theoriginal sequence converges.
Proof: Let ε > 0 be given and suppose a_{n}→ a. Then from the definition of convergence,
there exists n_{ε} such that if n > n_{ε}, it follows that
ε
|an− a| < 2
Therefore, if m,n ≥ n_{ε} + 1, it follows that
|an− am | ≤ |an− a|+ |a − am| <-ε+ ε = ε
2 2
showing that, since ε > 0 is arbitrary,
{an}
is a Cauchy sequence. It remains to that the last
claim.
Suppose then that
{an}
is a Cauchy sequence and a = lim_{k→∞}a_{nk} where
{ank}
_{k=1}^{∞} is a
subsequence. Let ε > 0 be given. Then there exists K such that if k,l ≥ K, then
|ak − al|
<
ε
2
. Then if k > K, it follows n_{k}> K because n_{1},n_{2},n_{3},
⋅⋅⋅
is strictly increasing as
the subscript increases. Also, there exists K_{1} such that if k > K_{1},
Theorem B.2.22Every Cauchy sequence in ℝ^{p}converges.
Proof:Let
{ak}
be a Cauchy sequence. Since it is Cauchy, it must be bounded (Why?).
Thus there is some box ∏_{i=1}^{p}
[ai,bi]
containing all the terms of
{ak}
. Therefore, by Theorem
B.2.6, a subsequence converges to a point of ∏_{i=1}^{p}
[ai,bi]
. By Theorem B.2.21, the original
sequence converges. ■
As mentioned above, sequential compactness and compactness are equivalent in ℝ^{p}. The
following is a very important property pertaining to compact sets. It is a surprising
result.
Proposition B.2.23Suppose ℱ is a nonempty collection of nonempty compactsets with the finite intersection property.This means that the intersection of any finitesubsetof ℱ is nonempty. Then ∩ℱ≠∅.
Proof:First I show each compact set is closed. Let K be a nonempty compact set and
suppose p
∕∈
K. Then for each x ∈ K, there are open sets U_{x},V_{x} such that x ∈ V_{x} and p ∈ U_{x}
and U_{x}∩ V_{x} = ∅. Just let these be balls centered at p,x respectively having radius
equal to one third the distance from x to p. Then since V is compact, there are
finitely many V_{x} which cover K say V_{x1},
⋅⋅⋅
,V_{xn}. Then let U = ∩_{i=1}^{n}U_{xi}. It follows
p ∈ U and U has empty intersection with K. In fact U has empty intersection with
∪_{i=1}^{n}V_{xi}. Since U is an open set and p ∈ K^{C} is arbitrary, it follows K^{C} is an open
set.
Consider now the claim about the intersection. If this were not so,
∪ {FC : F ∈ ℱ} = X
and so, in particular, picking some F_{0}∈ℱ,
{ }
FC : F ∈ ℱ
would be an open cover of F_{0}. A point in F_{0} is not in F_{0}^{C} so it must be in one of the
above sets. Since F_{0} is compact, some finite subcover, F_{1}^{C},
⋅⋅⋅
,F_{m}^{C} exists. But
then
m C
F0 ⊆ ∪k=1Fk
which means ∩_{k=0}^{m}F_{k} = ∅, contrary to the finite intersection property. ■