Given some set , a function is a metric if it satisfies the following properties.
if ; .
, for any .
With this presentation, is a metric space
Definitions
Let be a metric space and . Then we have the following definitions
An -neighborhood of is
A point is a limit point if every neighborhood of contains a point such that .
If a point is not a limit point, then it is an isolated point
is closed if every limit point of is a point of
A point is an interior point of if there is a neighborhood of such that .
The complement of is
is perfect if is closed and if every point of is a limit point of
is bounded if there is a real number and point such that for all
is dense in if every point is a limit point of , or a point of .
Complete Spaces
A subset of a metric space is complete if every Cauchy sequence in converges and its limit is in .
Compact Sets
An open cover of a set in a metric space is a collection of open subsets of such that . A subset of is compact if every open cover of contains a finite subcover.
Let be a metric space. is totally bounded if for every , the set can be covered by finitely many balls of radius . That is, there are such that .
Characterizations of Compactness
Info
If is a subset of a metric space , the following are equivalent
is complete and totally bounded
Every sequence in has a convergent subsequence.
Every open cover of has a finite subcover.
. Let . Since is totally bounded, it can be covered in finitely many balls of radius . Let be one of these balls that contain infinitely many points of . Recursively, note can be covered with finitely balls of radius . Pick to be one of these balls such that has infinitely points of . Pick such that , so this sequence is Cauchy and converges (by completeness).
. If is not complete, then there is some Cauchy sequence which does not converge in , violating (2). If is not totally bounded, then there is some where cannot be covered in finitely many -balls. Pick , and . Clearly for , thus no subsequence can converge. By contraposition we have proved the claim.
. Let be an open cover of . It suffices to show that there exists where every -ball intersecting is contained in some . Suppose this is not true. Then for every , there is a ball where and not contained in any . Pick , which has a subsequence converging to . There is some where for some . For large enough , and . Fixing , for any , . Thus , giving a contradiction.
. If has no convergent subsequence, then for every , there is a neighborhood of containing finitely many points. These form an infinite cover of , which clearly has no finite subcover.
Connected Sets
Two subsets and of a metric space are separated if and are empty. A set is connected if is not a union of two nonempty separated sets.