Linear Algebra, Theory And Applications, 2012a

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|a k − a l | < ε 2 .Thenifk>K, it follows n k >Kbecause n 1 ,n 2 ,n 3 , ··· is strictly increasing
as the subscript increases. Also, there exists K 1 such that if k>K 1 , |a nk − a| < ε 2 . Then
letting n>max (K, K 1 ), pick k>max (K, K 1 ). Then
Therefore, the sequence converges.
|a − a n |≤|a − a nk | + |a nk − a n | < ε 2 + ε 2 = ε.
Definition D.0.13 AsetK in R p is said to be sequentially compact if every sequence
in K has a subsequence which converges to a point of K.
Theorem D.0.14 If I 0 = ∏ p
i=1 [a i,b i ] where a i ≤ b i ,thenI 0 is sequentially compact.
Proof: Let {a k } ∞ k=1 ⊆ I 0 and consider all sets of the form ∏ p
i=1 [c i,d i ]where[c i ,d i ]
equals either [ ]
a i , ai+bi
2 or [ci ,d i ]= [ a i+b i
]
2
,b i . 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
∑p
of these sets, |x i − y i |≤2 −1 |b i − a i | where diam (I 0 )=(
i=1 |b i − a i | 2) 1/2
,
( p∑
) 1/2 ( p∑
) 1/2
|x − y| = |x i − y i | 2 ≤ 2 −1 |b i − a i | 2 ≡ 2 −1 diam (I 0 ) .
i=1
i=1
In particular, since d ≡ (d 1 , ··· ,d p )andc ≡ (c 1 , ··· ,c p )aretwosuchpoints,
( p∑
) 1/2
D 1 ≡ |d i − c i | 2 ≤ 2 −1 diam (I 0 )
i=1
Denote by {J 1 , ··· ,J 2 p} 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, {a i } 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 {a k } for infinitely many values
of k. Continue in this way obtaining a nested sequence {I k } such that I k ⊇ I k+1 ,andif
x, y ∈ I k ,then|x − y| ≤2 −k diam (I 0 ), and I n contains {a k } 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 {a k } for infinitely many k. Thus the distance between a nk and c is no larger than
2 −k diam (I 0 ), and so lim k→∞ a nk = c ∈ I 0 .
Corollary D.0.15 Let K be a closed and bounded set of points in R p .ThenK is sequentially
compact.
Proof: Since K is closed and bounded, there exists a closed rectangle, ∏ p
k=1 [a k,b k ]
which contains K. Now let {x k } be a sequence of points in K. By Theorem D.0.14, there
exists a subsequence {x nk } such that x nk → x ∈ ∏ p
k=1 [a k,b k ]. However, K is closed and
each x nk is in K so x ∈ K.
Theorem D.0.16 Every Cauchy sequence in R p converges.
Proof: Let {a k } be a Cauchy sequence. By Theorem D.0.11, there is some box ∏ p
i=1 [a i,b i ]
containing all the terms of {a k }. Therefore, by Theorem D.0.14, a subsequence converges
to a point of ∏ p
i=1 [a i,b i ]. By Theorem D.0.12, the original sequence converges.