Page 103 - DMTH401_REAL ANALYSIS

Unit 8: Completeness and Compactness

(f ) Cauchy in C(C, M) so converge to cts f : C ® M. f(C) dense in M. Also compact, so closed, hence Notes
k
f(C) = M.
Corollary: continuous surjective map f : [0, 1] ® [0, 1].

2
Proof: Extend surjective cts f : (C ® [0, 1] linearly to each interval removed during construction
of C.
Self Assessment

Fill in the blanks:

1. A complete .................... is not meagre in itself.
2. The Cantor set C is .........................
3. Let f: [1, ¥) ®  be cts s.t. for some a    ....................... x with f(x) < a. Then " k   : S =
 ¥ n k {x [1, ): f(nx) a} ¥  is nowhere x with dense.
=
4. Subspace S of complete metric M totally ....................... compact.

8.7 Summary

 S closed.
n
 S nowhere dense as has dense complement and closed.
n
 If f'(x) exists for some x then f  S for some n.
n
 Let f  S , f ® f. Find x  [0, 1] s.t. " y [0, 1],
k n k k
|f (y) – f (x )|  n|y – x |
k k k k
x has convergent subsequence so assume x ® x. By uniform convergence
k k
|f(y) – f(x)|  n|y – x|
Therefore f  S , so S closed.
n n
 Let g  C[0, 1],  > 0. g uniformly cts so  > 0 s.t.

|x – y||g(x) – g(y)|< . . . (1)
4
i 
Let x = (x) = kmin |x – x |. Then 0   show suffices to show f =  + g  S .
i 0 i  k i n
k 2
Suppose f  S and find x "responsible for it".
n
x + x
-
Choose 1  j  k s.t. x  [x , x]. Let y = j 1 j
j–1 j
2

= |(y) – (x )|
2 i
 |f(y) – f(x)| + |g(y) – g(x)|
j j

 |f(y) – f(x)| + |f(x) – f(x)| +
(1) j 4

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