Lecture: Properties of Compact Sets in Metric Spaces
Introduction
The lecture focuses on the properties of compact sets in metric spaces.
Main theorem discussed: A compact set must be closed and bounded.
Second property: If ( K ) is closed and ( X ) is compact, then ( K ) is also compact.
Main Theorem
Statement
If a set ( K ) is compact, it must be both closed and bounded.
If ( K ) is closed and ( X ) is compact, then ( K ) is compact.
Proof of Theorem
Proving ( K ) is Closed
Goal: Show the complement of ( K ) is open.
Take a point ( P ) in the complement.
Show ( P ) has a neighborhood fully contained in the complement.
Idea: Distance ( d(P, Q) ) for ( P ) not in ( K ) is positive.
Dividing this distance by 4 maintains positivity.
Create disjoint balls around ( P ) and ( Q ).
Using compactness, cover ( K ) with a finite number of such balls.
Conclude that ( K ) is closed as its complement is open.
Proving ( K ) is Bounded
Use the concept of covering ( K ) with balls of a fixed radius.
Compactness implies a finite number of such balls suffices.
For each point ( x ) in ( K ), there exists a finite bound ( M ) such that:
Distance ( d(x, q_j) < 1 ) for some center ( q_j ).
Hence, ( K ) is included in a ball of finite radius ( 2M ).
Concludes that ( K ) is bounded.
Second Property
Proving Closed Subset of Compact Set is Compact
Given: ( X ) is compact, ( K ) is closed.
Complement of ( K ) is open; take an arbitrary open cover.
( X ) can be covered by a finite number of open sets and possibly ( K^c ).
Since ( X ) is compact, a finite subcover exists for ( X ).
Conclude that ( K ), being part of ( X ), can also be finitely covered, proving compactness of ( K ).
Conclusion
The lecture concludes with the proof that compact sets are closed and bounded in metric spaces, and any closed subset of a compact set is also compact.