Chapter 7: Compactness

Question 1

Lemma 7.3
subset of compact space

Proof
***

Answer 1

Proof

Let (aₙ) be a sequence in A. Then (aₙ) is a sequence in X and as X is compact, there is a subsequence (aₙ_ₖ) converging to a limit a∈ X. But each aₙ_ₖ ∈ A and A is closed so a ∈ A. Therefore A is compact

Question 2

Prop 7.4
compact , complete ,closed?
(2b)
Proof***

Answer 2

Proof
Let (aₙ) be a Cauchy sequence in A. Since A is compact,
there is a convergent subsequence (aₙ_ₖ), with limit a ∈ A. By Theorem 5.7, aₙ → a. Thus A is complete and, by Proposition 5.11, it is closed in X.

Question 3

DEF 7.1: compact

Answer 3

Let A ⊆ X be a subset of a metric space. We say that A is compact if every sequence in A has a subsequence that converges to a point of A.

Question 4

LEMMA 7.3: subset of compact space

Answer 4

Let A ⊆ X be a closed subset of a compact space X. Then A is compact.

(way of finding compact sets)

Question 5

PROP 7.4:for a compact set : completeness and closed?

Answer 5

Let A be a compact set in a metric space. Then A is complete. In particular, A is closed.

Question 6

DEF 7.5 bounded subset

Answer 6

A subset A of a metric space (X, d) is bounded if there is a D > 0 such that d(a, b) ≤ D for all a, b ∈ A.

eg in R, [a,b] is bounded with D=b-a

Question 7

LEMMA: bounded subset

Answer 7

A subset A of R k is bounded in the sense of Definition 7.5 if and only if it is bounded in the usual sense

that there exists M > 0 such that d(a, 0) ≤ M for all a ∈ A.

(eg D=2M in 7.5 )

Question 8

Example: compact?

[a,b], R, [a,∞) or (−∞, b],

Answer 8

• all closed intervals of the form [a,b] are compact by the Bolzano-Weierstrass thm 5.8
  *R is not compact because the sequence 0,1,2,3,4,.. has no convergent subseq
  *unbounded intervals [a,∞) or (−∞, b] are not compact because of sequences
  a,a+q,a+2,.. and b,b-1, b-2,…

Question 9

complete
closed
compact?

Answer 9

complete sets are closed, so that the property of being complete is stronger than the property
of being closed. The property of being compact is stronger still.

compact implies completeness implies closed

Question 10

PROP 7.7

compact and bounded?

Answer 10

Let A be a compact subset of a metric space (X,d). Then A is bounded.

Question 11

THM 7.8 ( Heine-Borel)

Answer 11

A subset K of Rᵏ with the Euclidean metric is compact if and only if it is closed and bounded.

(compact subsets of a metric space X are closed and bounded, X=Rᵏ… false for other spaces)

Question 12

Example:

Let A = {f ∈ C[0, 1]; f([0, 1]) ⊆ [0, 1]} and use supremum metric

Answer 12

For f, g ∈ A, d∞(f, g) ≤ 1. Thus A is bounded. Also A is closed because if gₙ → g where each gₙ ∈ A then, for x ∈ [0, 1], 0 ≤ gₙ(x) ≤ 1 from which it follows that 0 ≤ g(x) ≤ and g ∈ A.

We consider a sequence of functions fₙ ∈ A
eg defined by

fₙ(t) =
{ 2ⁿt if 0 ≤ t ≤ 1/2ⁿ
{1 if 1/2ⁿ ≤ t ≤ 1

If m is bigger than n then fₘ ( 1/ ( 2ⁿ ⁺¹) =1, as m ≥ n+1 so
1/(2ⁿ ⁺¹ ) ≥ 1/2ᵐ, but

fₙ(1/(2ⁿ ⁺¹ )) = 1/2, as 1/(2ⁿ ⁺¹ ) ≤ 1/2ⁿ. So

d∞(fₙ, fₘ) ≥ | fₙ (1/(2ⁿ ⁺¹ )) - fₘ(1/(2ⁿ ⁺¹ ))| = 1/2

whenever m≠ n

No subsequence of fn can be Cauchy, because any two terms are at
least 1/2 apart, so no subsequence can be convergent. Hence A is not
compact although it is closed and bounded.

Question 13

THM 7.10: continuity and compactness

Answer 13

Let f : X → Y be a continuous map between metric

spaces, and let K ⊆ X be compact. Then f(K) is compact.

Question 14

COROLLARY 7.11

Answer 14

A function f which is real-valued and continuous

on a compact set K is bounded on K and attains its bounds.

Question 15

UNIFORM CONTINUITY DEF 7.12

Answer 15

the same δ works for all x ∈ X.
Definition 7.12. A function f : X → Y between metric spaces
(X, dX) and (Y, dY ) is uniformly continuous if for all …… 0, there
exists δ > 0 such that f(B(x, δ)) ⊆ B(f(x), ) for all x ∈ X.

funct uniform continuity implies continuous
but not every continuous funct is uniformly cont

Question 16

UNIFORM CONTINUITY

Answer 16

Uniform Continuity Recall the first definition of continuity. A
function f : X → Y between two metric spaces (X, dX) and (Y, dY )
is continuous if for all x ∈ X and for all > 0, there exists δ > 0
such that f(B(x, δ)) ⊆ B(f(x), ). We could demand more, namely
that the same δ works for all x ∈ X.

Question 17

Example 7.13: Consider f:(0,1] → [1,∞), f(x) = 1/x

Answer 17

then

|f( 1/(2ⁿ ⁺¹ )) - f(1/(2ⁿ )) |
= 2ⁿ ⁺¹ - 2ⁿ = 2ⁿ

Set ε= 1, then for any choice of δ bigger thn 0, choose n large enough st
1/(2ⁿ ⁺¹ ) less than δ

and let x = 1/ 2ⁿ.
Then |(1/(2ⁿ ⁺¹ ) - x| = 1/(2ⁿ ⁺¹ )  less than δ.

but
|f( 1/(2ⁿ ⁺¹ )) - f(x)| = 2ⁿ ≥ ε. Thus f(B(x,δ)) NOT⊆NOT B(f(x), ε) and f is not uniformly continuous.

Question 18

THM 7.14

Answer 18

Let f : X → Y be a continuous function and suppose

that X is compact. Then f is uniformly continuous.

Question 19

DEF 7.15

cover of E

Answer 19

Let X be a metric space. A collection {Uᵢ: i ∈ I}
of subsets of X is a cover of E ⊂ X, or covers E ⊂ X, if

E ⊆UNION FOR[i∈I] OF Uᵢ

Question 20

FINITE COVER

Answer 20

If the indexing set I is a finite set then {Uᵢ: i ∈ I}

Question 21

OPEN COVER

Answer 21

If each of the Uᵢ

is an open set then the collection is an open cover

Question 22

FINITE SUBCOVER

Answer 22

If {Uᵢ: i ∈ I} is a cover for E, then a finite collection Uᵢ₁, . . . , Uᵢ_ₙ
with i₁, . . . , iₙ ∈ I is called a finite subcover of E if

it is itself a finite cover,

i.e. if E ⊆ Uᵢ₁ ∪ · · · ∪ Uᵢₙ

Question 23

EXAMPLE closed interval

[0.1] cover

Answer 23

eg break into two intervals length 1/2, 3 intervals length 1/3 etc

no way of writing R as a finite union of intervals of length one

Question 24

DEF 7.16

totally bounded

Answer 24

A metric space (X, d) is totally bounded if for each

ε > 0 there is a finite collection of open balls of radius ε which cover X

Question 25

PROP 7.17 compact metric space is…

2a

Answer 25

A compact metric space is totally bounded

Question 26

DEF 7.18: Heine-borel PROPERTY

Answer 26

A metric space X is said to have the Heine–Borel
property if every open cover of X has a finite subcover. That is, X
has the Heine-Borel property if for any open cover {U_i: i ∈ I} there
are finitely many indices i1, . . . , in such that Ui_1 ∪ · · · ∪ Ui_n = X.

Question 27

PROP 7.19: (3)

when can we say metric space has the heine-borel property

Answer 27

Let X be a totally bounded complete metric space. Then X has the Heine-Borel property.

Question 28

THM 7.2: metric spaces equivalent statements from (1) (2) (3)

Answer 28

Let (X, d) be a metric space. The following are
equivalent:

a. X is compact.
b. X is totally bounded and complete.
c. X has the Heine–Borel property.