→ ℝ^{p} is called a sequence. We usually write it in the form
{a}
j
where it is understood that a_{j}≡ f
(j)
.
Definition E.2.1A sequence,
{ak}
is said toconverge to a if for every ε > 0 there existsn_{ε}such that if n > n_{ε}, then
|a− an|
< ε. The usual notation for this is lim_{n→∞}a_{n} = aalthough it is often written as a_{n}→ a. A closed set K ⊆ ℝ^{n}is one which has the propertythat if
{kj}
_{j=1}^{∞}is a sequence of points of K which converges to x, then x ∈ K.
One can also define a subsequence.
Definition E.2.2
{a }
nk
is a subsequenceof
{a }
n
if n_{1}< n_{2}<
⋅⋅⋅
.
The following theorem says the limit, if it exists, is unique.
Theorem E.2.3If a sequence,
{an}
converges to a and to b thena = b.
Proof:There exists n_{ε} such that if n > n_{ε} then
|an − a|
<
ε
2
and if n > n_{ε}, then
|an − b|
<
ε
2
. Then pick such an n.
-ε ε
|a − b| < |a− an|+ |an − b| < 2 + 2 = ε.
Since ε is arbitrary, this proves the theorem. ■
The following is the definition of a Cauchy sequence in ℝ^{p}.
Definition E.2.4
{an}
is a Cauchy sequence if for all ε > 0, there exists n_{ε}such thatwhenever n,m ≥ n_{ε}, if follows that
|an− am |
< ε.
A sequence is Cauchy, means the terms are “bunching up to each other” as m,n get
large.
Theorem E.2.5The set of terms in a Cauchy sequence in ℝ^{p}is bounded in the sense thatfor 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 − an|
1
< 1.It follows that for all n > n_{1},
|an|
< 1 +
|an |
1
.Therefore, for all
n,
n∑1
|an| ≤ 1 +|an1|+ |ak|.■
k=1
Theorem E.2.6If a sequence
{an}
in ℝ^{p}converges, then the sequence is a Cauchysequence. Also, if some subsequence of a Cauchy sequence converges, then the originalsequence 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
|an− a|
<
ε2
. Therefore, if m,n ≥ n_{ε} + 1, it follows
that
ε ε
|an− am| ≤ |an− a|+|a− am | < 2 + 2 = ε
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
{an }
k
_{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},
|an − a|
k
<
ε
2
. Then letting n > max
(K, K1)
,
pick k > max
(K,K1 )
. Then
|a− an| ≤ |a− ank|+ |ank − an| < ε + ε = ε.
2 2
Therefore, the sequence converges. ■
Definition E.2.7A set K in ℝ^{p}is said to be sequentially compactif everysequence inK has a subsequence which converges to a point of K.
Theorem E.2.8If I_{0} = ∏_{i=1}^{p}
[ai,bi]
where a_{i}≤ b_{i}, then I_{0}is sequentially compact.
Proof:Let
{ak}
_{k=1}^{∞}⊆ I_{0} and consider all sets of the form ∏_{i=1}^{p}
[ci,di]
where
[ci,di]
equals either
[ai, ai+bi]
2
or
[ci,di]
=
[ai+bi,bi]
2
. Thus there are 2^{p} of these sets because there are
two choices for the i^{th} slot for i = 1,
⋅⋅⋅
,p. Also, if x and y are two points in one of these sets,
|xi − yi|
≤ 2^{−1}
|bi − ai|
where diam
(I0)
=
(∑p 2)
i=1 |bi − ai|
^{1∕2},
( p )1∕2 ( p )1 ∕2
|x − y | = ∑ |x − y|2 ≤ 2−1 ∑ |b − a |2 ≡ 2−1diam (I).
i=1 i i i=1 i i 0
In particular, since d ≡
(d ,⋅⋅⋅,d)
1 p
and c ≡
(c ,⋅⋅⋅,c )
1 p
are two such points,
( ∑p )1∕2
D1 ≡ |di − ci|2 ≤ 2−1diam (I0)
i=1
Denote by
{J1,⋅⋅⋅,J2p}
these sets determined above. Since the union of these sets
equals all of I_{0}≡ I, it follows that for some J_{k}, the sequence,
{ai}
is contained in
J_{k} for infinitely many k. Let that one be called I_{1}. Next do for I_{1} what was done for
I_{0} to get I_{2}⊆ I_{1} such that the diameter is half that of I_{1} and I_{2} contains
{ak}
for
infinitely many values of k. Continue in this way obtaining a nested sequence
{Ik}
such
that I_{k}⊇ I_{k+1}, and if x,y∈ I_{k}, then
|x − y|
≤ 2^{−k}diam
(I0)
, and I_{n} contains
{ak}
for infinitely many values of k for each n. Then by the nested interval lemma, there
exists c such that c is contained in each I_{k}. Pick a_{n1}∈ I_{1}. Next pick n_{2}> n_{1} such that
a_{n2}∈ I_{2}. If a_{n1},
⋅⋅⋅
,a_{nk} have been chosen, let a_{nk+1}∈ I_{k+1} and n_{k+1}> n_{k}. This can be
done because in the construction, I_{n} contains
{ak}
for infinitely many k. Thus the
distance between a_{nk} and c is no larger than 2^{−k}diam
(I0)
, and so lim_{k→∞}a_{nk} = c ∈ I_{0}.
■
Corollary E.2.9Let K be a closed and bounded set of points in ℝ^{p}. Then K is sequentiallycompact.
Proof:Since K is closed and bounded, there exists a closed rectangle, ∏_{k=1}^{p}
[ak,bk]
which
contains K. Now let
{xk}
be a sequence of points in K. By Theorem E.2.8, there exists a
subsequence
{xn }
k
such that x_{nk}→ x ∈∏_{k=1}^{p}
[ak,bk]
. However, K is closed and each x_{nk} is in
K so x ∈ K. ■
Theorem E.2.10Every Cauchy sequence in ℝ^{p}converges.
Proof:Let
{ak}
be a Cauchy sequence. By Theorem E.2.5, there is some box
∏_{i=1}^{p}
[ai,bi]
containing all the terms of
{ak}
. Therefore, by Theorem E.2.8, a subsequence
converges to a point of ∏_{i=1}^{p}
[ai,bi]
. By Theorem E.2.6, the original sequence converges.
■
Appendix F Some Topics Flavored With Linear Algebra