Page 85 - DMTH401_REAL ANALYSIS

Unit 6: Completeness

Proposition: Notes
1. Complete subspace S of metric M is closed.
2. Closed subset S of complete M is complete.
Proof:

1. Let x  S, x  x  M. (x ) Cauchy in S so cvgs in S to y  S. S  M so x  y in M. By
n n n n
uniqueness of limits x = y  S.
2. Let (x ) S Cauchy. Cauchy in M so cvgs to point of M which in S as S closed.
n
Proposition: " S B(S) of bdd functions S   with sup norm is complete.
Proof: Let (f ) Cauchy,  > 0. 3 N s.t. ||f – f || <  for n, m  N. Hence for fixed x (f (x)) Cauchy
n m n n
in , so cvgs to f(x) .
For n  N|f (x) – f (x)|  " m  N. Let m  then
m n
|f(x) – f (x)|  " x  S, n  N
n
Then f bdd and f  f.
n
6.2 Proving Cauchy

Proposition: A sequence (x )  M is Cauchy iff  sequence   0 s.t.  ¾¾¾ and d(x , x ) 
0
n n n n m n n
for m > n.
Proof: () Suppose (x ) Cauchy. Then let  =  d(x , x ) ¾¾¾ 0.
n n m>n m n n
() Given  > 0 find k s.t.  <  for n  k. Then d(x , x )   <  for m > n  k. Exchanging m, n gives
n n n n
d(x , x ) <  " n, m  k.
m n
Proposition: (x )  M sequence s.t.   0 with å  n 1  <  and d(x , x )  " n. Then (x ) is
=
n
n n n n+1 n n
Cauchy.
Proof: Follows from 6.4 with  = å  n 1  . Then
=
n
n
-
m 1 m 1
-
d(x , x )  å d(x ,x )  å   
+
m n k k 1 n n
D ineq k n= k n
=
Example: If K compact topological space then space C(K) with sup norm is complete.
Proof: Each f bdd, attains max. Suffices to show C(K) closed in B(K).
Suppose f C(K) cvg to f  B(K). Then "  > 0  N s.t.
n
-

sup|f(x) f (x)|<  " n N
n

x K
" a  {x : f(x) > a} =  {x : f (x) a +  }
>
N
 > 0
RHS are pre-images of open sets so open. Hence LHS is open. Similarly {x : f(x) < a} open. (–, a),
(a, ) from sub-basis for  so f cts.
Example: C[0, 1] with norm ||f|| = f |f(x)|dx is incomplete.
1
1 0

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