Page 205 - DMTH401_REAL ANALYSIS

Unit 15: Uniform Convergence of Functions

I
2. Infinite collection indexed by  : { , I 2 , ... }. In this case A = .  Notes
1
n
I = ( - 1/3, n + 1/3).
Example: n
3. Infinite collection indexed by  In this case A = . 
.
Example: {( -x 1, + 1) } x Î  . This set contains all open intervals of length 2.
x
Definition 3: A collection of open intervals { }I a a Î A is a cover of a set S   if S  È aÎA I a .

Definition 4: Given a set S and its cover { }I a aÎA , a subcover of { }
I
a aÎA is a subcollection of
I
{ } , which itself is a cover for S.
a aÎA
Example: Let I 1 = (0, 2), I 2 = (4,5). A collection containing these two intervals covers
(0, 1) and {1} È (4, 4.5), but does not cover [0, 1).

Example: Let I n = ( - 1/3, n + 1/3), n Î  . Then { }I n n Î  covers  , but does not cover
n
+
}
 or {1/2}. Let =S {1} È (3 1/4, 3 1/4). Then { }I n n Î  is a cover for S. Moreover, { ,I I 3 is a
-
1
finite subcover. Consider another case where { }I n n Î  is a cover of . Here { }I n n Î  has no finite
subcover.
n
-
+
-
n
Example: Let I n = ( 1 1/ , 1 1/ ), n Î  \{1}. We see that the collection {(–3/4, 3/4)}
-
is a finite subcover for set =S [ 17/24, 17/24]. Now, consider set = ( 1,1). Is { }I n n ³ 2 a cover
S
-
for S? The answer is positive and { }I n n ³ 2 has no finite subcover.
Example: We can observe that given a set S and its cover { }I a a ÎA sometimes it’s possible
to extract a finite subcover and sometimes isn’t.
Theorem 3: (Heine-Borel). Every cover of closed interval [ , ]a b has a finite subcover.
b
a
a
x
Proof: Definite a set B = { Î [ , ] : [ , ] has a finite subcover. We see that ÎB since [ , ] is
a
,
a
x
a
a single point. We need to prove that Îb . B Define c = sub B. Now, we have to prove two claims:
Claim 1: =c b (Figure 13.6, left).
)
c
c
0
Assume that <c . b Then there is an interval I s.t. Îc I b . Pick e > s.t. ( - e , + e  I and
b
b
a
c
c
c
x
)
( - e , + e  [ , ]. Since c = sup B there is x ÎB s.t. x Î ( - e , ]. Since x ÎB , [ , ] can be
a
b
c
y
c
covered by finitely many intervals {I a 1 ,...,I am }. Pick y Î ( , c + e ) and see that [ , ] is covered
a
by finite subcover {I a 1 , ..., I a m , I b }. Hence y BÎ . But y > c what implies that c ¹ sup B? What is
contradiction? Therefore c = . b
B
Claim 2: b Î (Figure 15.6, right)
b
b
Pick I covering b. Take e > s.t. (b - e , ]  I b . Since b = supB there exists x Î s.t. x Î (b - e , ].
0
B
b
a
x
b
a
Since x Î B , [ , ] can be covered by finitely many intervals {I a 1 , ..., I a m }. Then [ , ] is covered
by finitely many intervals {I a 1 , ..., I a m , I b }.
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