1701354 Introduction to Topology – Nguyen 4-13-06
SEQUENTIALLY COMPACT, CLOSED, AND BOUNDED SUBSETS
Class Handout
Note : The theorems numbered below refer to those in your textbook.
I. Sequential Compact and Closed Subsets
Recall the following definition:
Definition : A subset A

X of a metric space ( , ) is sequentially compact if every sequence in A has a subsequence that converges to an element of A .
To relate sequential compactness with closure, we introduce an equivalent version of closure based on sequences:
Definition : A subset A

X of a metric space ( , ) sequence in A converges to an element of A x n
is sequentially closed if every convergent

A and x n
 c , then c

A . Warning :
Do not confuse this definition with that for sequential compactness.
Lemma : A subset A is closed if and only if it is sequentially closed.
Exercise 1: Prove the lemma above.
Theorem 9.5
: Every sequentially compact subset A of a metric space ( , ) is (sequentially) closed in X .
Example:
(a) Prove the theorem above. Hint: Use the fact that if a sequence x converges to a limit c , n then any subsequence of x must also converge to c (Theorem 3.6). n
(b) The converse to the above theorem is false. Find a counterexample. Hint: Consider any subset of that is closed but not bounded.
Note: The following restricted converse is true:
Theorem 9.2
: Every closed subset A of a sequentially compact metric space ( , ) is sequentially compact.
II. Completeness and Closed Subsets
Theorem : Every complete subset A of a metric space ( , ) is (sequentially) closed.
Example:
(a) Prove the theorem above. Hint: Use the fact that every convergent sequence is Cauchy.
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1701354 Introduction to Topology – Nguyen 4-13-06
(b) The converse to the above theorem is false. Find a counterexample. Hint: Choose X to be any incomplete metric space and A any subset that is closed (in X ) that contains a nonconvergent Cauchy sequence.
Note: The following restricted converse is true.
Theorem 7.7
: Every closed subset A of a complete metric space ( , ) is complete under the induced metric.
III. Bounded Versus Totally Bounded Subsets
The converse of Theorem 9.5 is clearly not true as demonstrated by the counterexample above.
However, this can be fixed if we require subsets of n
to be both closed and bounded:
Theorem 9.6
: A subset of ( n
, d
E
) is sequentially compact if and only if it is closed and bounded.
Example: Prove the implication (

) in Theorem 9.6 for the case n = 1. Hint: Use the Bolzano-
Weierstrass Theorem.
Unfortunately, for subsets of an arbitrary metric space, Theorem 9.6 fails to hold. This is because for degenerate metrics like the trivial metric d
0
(defined on pp. 87-88), having a bounded subset no longer matches with our intuition of how it should behave. Therefore, we need to replace boundedness by a stronger version that does agree with our intuition.
Definition : Let X be a set.
(a) A collection

of subsets of X is said to cover X (or be a cover of X ) if X
 
, i.e. every point of X lies in a subset that belong to
(b) A cover
(c) If

and


is said to be
' finite if the number of subsets that belong to
are two covers of X, then

'

.
is called a subset that belong to

' also belongs to

. subcover of


if
is finite.

'
 
, i.e. every
Definition : A metric space X is called totally bounded (or precompact) if for every 0 exists a finite cover

 of X such that diam( )
 
for all A

 
, there
, i.e. every subset that belongs to


has diameter less than or equal to

.
Lemma : Every totally bounded metric space is bounded.
Example: Prove the lemma above. Hint: Let
 
{ ,..., A
N
} be a finite cover whose subsets have diameter less than or equal to 1 and select a base point x i

A i
for i

1,..., N . Then for any ,

X , use the fact that x
 m
,

A n
(and therefore are close to the points x m
, x n
) and the triangle inequality to bound the distance ( , ) .
Now we are ready for the metric space version of Theorem 9.6.
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1701354 Introduction to Topology – Nguyen 4-13-06
Theorem : A subset of a complete metric space ( , ) is sequentially compact if and only if it is closed and totally bounded.
Example: Prove the implication (

) in the theorem above. Hint: Let S

{ } n
be a sequence and for p = 1 let

1
be a finite cover of X whose subsets have diameters less than or equal to 1. Then one of these subsets must contain infinite many points of S . Call this set of points S
1
and choose x

(1)

S
1
. For p = 2 let

2
be a finite cover with diameters less than or equal to 1/2, S
2 be the infinite set of points in S
1
that are contained in some subset for

2
, and choose x

(2)

S
2
.
Repeat this process for every positive integer p . Now use the fact that diam( S p
) 1/ p to establish that the subsequence { x

} is Cauchy and hence convergent.
Exercise 2: The theorem fails to hold when we replace ‘totally bounded’ by ‘bounded’. Find a counterexample. Hint: Consider a metric space X d
0
) where X is arbitrary infinite set and d is
0 the trivial metric defined on pp. 87-88. Then show that every subset of X is closed and bounded, but only the finite subsets are sequentially compact.
What about a global version of the theorem above? Since every metric space is automatically closed (as a subset of itself), we have
Theorem 9.10
: A metric space ( , ) is sequentially compact if and only if it is complete and totally bounded.
Exercise 3: 9.A (p. 153)
Exercise 4: 9.D (p. 154)
Exercise 5: 9.E (p. 154)
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