Page 97 - DMTH401_REAL ANALYSIS

Unit 8: Completeness and Compactness

Notes
Theorem: Subspace S of complete metric M totally bounded iff S compact.
Proof: () S totally bounded and so compact.

() S totally bounded so is S  S .

8.2 Cantor's Theorem

Definition: Diameter of 0  S  M defined by

diam (S) = supd(x, y)
x,y S

Theorem: Cantor
Let F decreasing sequence of non-empty closed subsets of metric M s.t. diam (F ) ¾¾¾® 0.
n n n®¥
Then  ¥ n 1 F  .
=
n
Proof: Pick x  F . Then " i  n, x  F  F .
n n i i n
Hence, for i, j n, d(x , x)  diam (F ). Hence (x ) Cauchy. Converges to some x as M complete.
i j n n
F closed so x  F . Hence x   ¥ n 1 F .
n n = n

8.3 Perfect Set

A set S is perfect if it is closed and every point of S is an accumulation point of S.

Example: Find a perfect set. Find a closed set that is not perfect. Find a compact set that
is not perfect. Find an unbounded closed set that is not perfect. Find a closed set that is neither
compact nor perfect.
Solution:

 A perfect set needs to be closed, such as the closed interval [a, b]. In fact, every point in that
interval [a, b] is an accumulation point, so that the set [a, b] is a perfect set.
 The simplest closed set is a singleton {b}. The element b in then set {b} is not an accumulation
point, so the set {b} is closed but not perfect.
 The set {b} from above is also compact, being closed an bounded. Hence, it is compact but
not perfect.
 The set {–1} [0, ¥) is closed, unbounded, but not perfect, because the element –1 is not an
accumulation point of the set.

 The set {–1}[0, ¥) from above is closed, not perfect, and also not compact, because it is
unbounded.

Example: Is the set {1, 1/2, 1/3, ...} perfect? How about the set {1, 1/2, 1/3, ...}{0}?
Solution: The first set is not closed. Hence it is not perfect.
The second set is closed, and {0} is an accumulation point. However, every point different from
0 is isolated, and can therefore not be an accumulation point. Therefore, this set is not perfect
either.

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