Page 102 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
Then by Cantor (5.15)  ¥ k 1 B(x ,  k ) 0 / . Let x be in this intersection. Then x  B(x ,  )  G " k

=
k
k
k
k
so x  S " k. Hence, there is a point x that is not in the union of all nowhere dense subsets of M,
k
so M cannot be meagre.
Proposition: The Cantor set C is uncountable.
Proof: " x  C there are points y  C, y  x arbitrarily close to x. In other words, C\{x} is dense in
C. Therefore {x} is nowhere dense in C as it is closed.
If C were countable would have C =  ¥ j {x } showing C meagre in itself. This contradicts Baire's
j
theorem.
Lemma: Let f: [1, ¥) ®  be cts s.t. for some a    arbitrarily large x with f(x) < a. Then " k 
 : S =  ¥ n k {x [1, ): f(nx) a} ¥  is nowhere x with dense.
=
Proof: f cts so S closed. Let 1  < < ¥. RTP (,)\S  0 /

n 1  ¥
+
For large n, < so (n + 1) < n. Then  n k (n , n )  contains some (r, ¥) and so a point y
=
n 
s.t. f(y) < a.
y
Find n s.t. y  (n, n). Then x =  (,) and f(nx) < a so x  S.
n
Proposition: Let f: [1, ¥) ®  be cts s.t. " x  1, limf(nx) exists. Then limf(x) exists.
n®¥ x®¥
Proof: If limf(x) not exist then  a, b; a < b s.t.  arbitrarily large x,y with f(x) < a, f(y) > b.
x®¥
Then by previous lemma:

¥ ¥ ¥ ¥
  {x [1, ) : f(nx) a}   {x [1, ) : f(nx) b}
¥


¥

=
=
=
k 1 n 1 k 1 n 1
=
is meagre. By Baire  x T.
x not in first union so " k  n  k s.t. f(nx) < a. x not in second union so " k  m  k s.t. f(mx) > b.
Hence f(nx) not converge.
Theorem:  f  C[0, 1] not differentiable at any point.
Proof: IDEA: C[0, 1] is complete. Functions with derivative at at least one point form a meagre
subset. Result by Baire.
Define S :
n
S = {f  C[0, 1] : ( x  [0, 1])( " y  [0, 1]) |f(y) – f(x)| n|y – x|}
n
8.6 Compactness and Cantor Set

Theorem: Every compact metric M is continuous image of Cantor set C.
–k
Proof: Let A  M be finite s.t. " x M d(A , x)  2 .
k k
k
By induction construct sequence of cts functions f : C ® M s.t. f (C) = A , d(f (x), f (x))  2 " x
k k k k k+1
C.

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