MAT 3749 1.3 Part III Handout
Closed Sets
A set K  is closed if its complement K c 
Example 1
(a) a is closed.
(b) For a, b 
with a  b ,  a, b is closed.
(c) The empty set  is closed.
(d)
is closed.
Last Edit: 2/6/2016 1:09 AM
\ K is open.
(e) Any finite set a1 , a2 ,
(f) The set S 
1 
  is not closed.
n 1  n 

  1 
(g) The set K     
 n 1  n  
(h)  and
, an  is closed.
0
is closed.
are the only two subsets of
that are both open and closed.
Theorem
1. The intersection of arbitrary collection of closed sets is closed.
2. The union of a finite number of closed sets is closed.
Proof
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Open Cover
An open cover of a set S is a collection U  of open sets such that S  U .

Finite Subcover
 
A finite subcover of U  is a finite collection Uk
S
n
k 1
of sets of U  such that
n
k 1
U k .
Compact Sets
A set K is compact if every open cover of K has a finite subcover.
Example 2
 0,1 is not compact.
Analysis
We need to construct an open cover without a finite subcover.
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Proof

 1

,1 for nN . Then  0,1  U n . (HW)
Let U n  
n 1
 n 1 
 
Suppose U n n 1 has a finite subcover U nk

If ni  n j , then ni  1  n j  1 
N
k 1
with n1  n2 
 nN .

 1
  1
1
1


,1  
,1  U ni  U n j .
ni  1 n j  1  ni  1   n j  1 
Thus, U nk  U nN for all k  1, 2,..., n .
 1

U nk  U nN  
,1 .
k 1
 nN  1 
N
1
1
 U nk .
  0,1 and
Now,
nN  1 k 1
nN  1
N
So,
This implies that  0,1 

N
k 1
 
U nk which contradicts the assumption that U nk
N
k 1
is a finite
subcover.
So, U n n 1 does not have a finite cover. Thus,  0,1 is not compact.

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Recall: The Completeness Axiom
Let E be a non-empty subset of .
1. If E is bounded above, then E has a least upper bound.
2. If E is bounded below, then E has a greatest lower bound.
Bounded Sets
A set K  R is bounded if it is bounded above and below.
Heine-Borel Theorem
A subset of R is compact iff it is closed and bounded.
Theorem 13
The continuous image of a compact set is compact.
Lemma
If S  R is closed and s  sup S exists, then s  S . (HW)
Extreme Value Theorem (Generalized)
A continuous function f on a (non-empty) compact set K attains its maximum value at
some point of K .
Analysis
This is a generalized version of the one we see in 2.3 Part III. As a closed interval
a, b are both closed and bounded. Thus it is compact.
We know …
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Proof
Since f is continuous and K is compact, by Theorem 13, f  K  is compact.
So, by the H-B Theorem, f  K  is closed and bounded.
Since K is non-empty, f  K  is nonempty.
Since f  K  is nonempty and bounded above, by the completeness axiom,
sup f  K  exists. Let s  sup f  K  .
Since f  K  is closed, by lemma, s  f  K  .
So, c  K such that f  c   s .
Since s  sup f  K  , f  c   f  x  , for all x  K
Therefore, f attains its maximum value at some point of K .
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