Page 233 - DMTH503_TOPOLOGY

Unit 28: Compactness in Metric Spaces

28.1.1 Sequentially Compact Notes

A metric space (X, d) is said to be sequentially compact if every sequence in X has a convergent
sub-sequence.

Example 1: The set of all real numbers in (0, 1) is not sequentially compact.

1 1 1
For the sequence , , ,... in (0, 1) converges to 0  (0, 1), on the other hand [0, 1] is sequentially
2 3 4
compact.

28.1.2 Lebesgue Number

Let {G : i  } be an open cover for a metric space (X, d). A real number  > 0 is called a Lebesgue
i
number for the cover if any A  X s.t. d(A) <   A  G for at least one index i  .
0 i 0

Lebesgue Covering Lemma

Every open covering of a sequentially compact space has a lebesgue number.

28.1.3 Totally Bounded Set

Let (X, d) be a metric space. Let  > 0 be any given real number. A set A  X is called an - net
if
(i) A is finite set

(ii) X = U{S : a  A}
(a)
The metric space (X, d) is said to be topology bounded if it contains an – net for every  > 0.
Here (ii)  given any point p  X,  at least one point a  A s.t. d(p, a) < .

28.1.4 Compactness in Metric Spaces

If (X, d) be a metric space and A  X, then the statement that A is compact, A is countably compact
and A is sequentially compact are equivalent.

28.2 Theorems and Solved Examples

Theorem 1: A metric space is sequentially compact iff it has the Bolzano Weierstrass Property.
Proof: Let X be a metric space.

Let us suppose that it is sequentially compact.
Let A be an infinite subset of X.
Since A is infinite so let x  be any sequence of distinct points of A. Since X is sequentially
n
compact, so there exists a convergent subsequence x  of x . Let x be its limit and B be its
n k n
range.
Since x  is a sequence of distinct points, B is infinite.
n
We know that if the range of a convergent sequence is infinite then its limit point is the limit
point of the range.

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