→ ℝ^{p} is called a sequence. We usually write it in the form
{aj}
where
it is understood that a_{j}≡ f
(j)
. In the same way as for sequences of real numbers, one can define what it
means for convergence to take place.
Definition 8.10.5A sequence,
{ak}
is said toconverge to a if for every ε > 0 there exists n_{ε}such that if n > n_{ε}, then
|a − an|
< ε. The usual notation for this is lim_{n→∞}a_{n} = a although it isoften written as a_{n}→ a.
One can also define a subsequence in the same way as in the case of real valued sequences.
Definition 8.10.6
{an }
k
is a subsequenceof
{an }
if n_{1}< n_{2}<
⋅⋅⋅
.
The following theorem says the limit, if it exists, is unique.
Theorem 8.10.7If 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.
|a− b| < |a − an |+|an − b | < ε + ε= ε.
2 2
Since ε is arbitrary, this proves the theorem. ■
The following is the definition of a Cauchy sequence in ℝ^{p}.
Definition 8.10.8
{an}
is a Cauchy sequence if for all ε > 0, there exists n_{ε}such that whenevern,m ≥ n_{ε},
|an− am| < ε.
A sequence is Cauchy, means the terms are “bunching up to each other” as m,n get large.
Theorem 8.10.9The set of terms in a Cauchy sequence in ℝ^{p}is bounded in the sense that for alln,
|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},
|an| < 1+ |an1|.
Therefore, for all n,
n
∑ 1
|an | ≤ 1+ |an1|+ |ak|. ■
k=1
Theorem 8.10.10If a sequence
{an}
in ℝ^{p}converges, then the sequence is a Cauchy sequence.Also, if some subsequence of a Cauchy sequence converges, then the original 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
|an− a| < ε
2
Therefore, if m,n ≥ n_{ε} + 1, it follows that
|a − a | ≤ |a − a|+ |a − a | < ε+ ε = ε
n m n m 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|
<
ε2
. 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},
|ank − a|
<
ε2
. Then letting n > max
(K, K1)
, pick k > max
(K, K1)
.
Then
ε ε
|a − an| ≤ |a − ank|+|ank − an| < 2 + 2 = ε.
Therefore, the sequence converges. ■
Definition 8.10.11A set K in ℝ^{p}is said to be sequentially compactif everysequence in Khas a subsequence which converges to a point of K.
Theorem 8.10.12If 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
[ a+b ]
ai, i2-i
or
[ci,di]
=
[a+b ]
-i2-i,bi
. 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,
− 1
|xi − yi| ≤ 2 |bi − ai|.
diam
(I0)
=
(∑p 2)
i=1|bi − ai|
^{1∕2},
|x − y|
=
( )
∑p 2
|xi − yi|
i=1
^{1∕2}
≤ 2^{−1}
( p )
∑ |b − a|2
i=1 i i
^{1∕2}≡ 2^{−1}diam
(I)
0
.
In particular, since d ≡
(d1,⋅⋅⋅,dp)
and c ≡
(c1,⋅⋅⋅,cp)
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 8.10.13Let 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 8.10.12, there exists a subsequence
{xnk}
such
that x_{nk}→ x ∈∏_{k=1}^{p}
[ak,bk]
. However, K is closed and each of the points of the sequence is in K so
x ∈ K. If not, then since K^{C} is open, it would follow that eventually x_{nk}∈ K^{C} which is impossible.
■
Theorem 8.10.14Every Cauchy sequence in ℝ^{p}converges.
Proof:Let
{ak}
be a Cauchy sequence. By Theorem 8.10.9, there is some box ∏_{i=1}^{p}
[ai,bi]
containing
all the terms of
{ak}
. Therefore, by Theorem 8.10.12, a subsequence converges to a point of ∏_{i=1}^{p}
[ai,bi]
.
By Theorem 8.10.10, the original sequence converges. ■