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Real Analysis
Notes 4.7 Sequential Compactness
Theorem: Metric M is compact if f every sequence in M has convergent subsequence.
Lemma: A sequence of subsets of metric M. Then " x ¥ j 1 A x A s.t. x x.
k = j k k k
Proof: Take x A B (x, 1 ) .
k k k
Corollary: x M and ¥ j 1 {x , x ...}. then x have convergent
+
k = j j 1 k
Proof: Let x ¥ j 1 {x , x ...}. As k j s.t. x x. k ¥ so can choose subsequence k s.t. k >
=
j
j 1
+
ji+1
ji
j
j
kj
k (as ks not necessarily in order). Then x subsequence converging to x.
ji j kji
Proof of () of theorem 3.16. Let x M, F = {x , x ...}. F form decreasing sequence of non-
+
k j j j 1 j
empty closed subsets of M.
By corollary 3.6 ¥ j 1 F so x have convergent subsequence by corollary 3.18.
j
=
k
Notation
U open cover of M. " x M
r(x) = sup {r 1 : U U s.t. B(x, r) C U}
æ r(x)ö
Lemma: If y x K s.t. y B y , for k K.
k k+1 ç k ÷
è 2 ø
r(x)
B
Proof: Let U U be s.t. ( x, r(x) ) U. Take K s.t. d(y , x) < for k K. Then k K
2 k 16
æ r(x) ö æ r(x)ö r(x) r(x)
B y , 2 - d(x, y ) B x, 2 ø ÷ U, so r(y ) 2 – d(x, y ) 4 , so
ç
k ÷
ç
k
k
k
ø
è
è
r(x) r(y )
d(y , y ) d(y , x) + d(y , x) < k
k+1 k k+1 k 8 2
s 1 æ r(x )
M : = M, s : = sup r(x). Find x M s.t. r(x ) > , choose U U s.t. B x , 1 ö U .
1 1 xM1 1 1 1 2 1 ç è 1 2 ø ÷ 1
If x , ...., x have been defined,
1 j
j ö
j ö
æ r(x ) j æ r(x )
M : = M\ B x , = M \ B x ,
j+1 ç è j 2 ø ÷ i 1 è ç j 2 ø ÷
=
æ
i ö
j
If M = then M j i 1 B x , r(x ) i 1 U has finite subcover.
j+1 = ç i ÷ = i
è 2 ø
s æ r(x j 1 ö
)
+
If M let s = sup {r(x)}, find x s.t. r(x ) > j 1 , choose U U s.t. B x , + U .
+
+
j+1 j+1 xMJ+1 j+1 j+1 2 j+1 ç è j 1 2 ÷ ø j 1
If procedure stops we have finite subcover. If it runs forever we have infinite sequence x s.t. x
j i
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