Proof Suppose is convergent with . Then for any , there exists such that for all . Therefore, for all , we have Thus is a Cauchy sequence.
Proposition
Cauchy sequences in metric spaces are bounded.
Proof Let be a Cauchy sequence. Then there exists such that for all . Let Then for all n. Therefore for all , i.e. is bounded.
Proposition
Let be a metric space and let be a Cauchy sequence in . If has a convergent subsequence, then is convergent.
Proof Let be a Cauchy sequence and let be a convergent subsequence of with . We have for sufficiently large . By taking and using , we obtain Since is Cauchy, the right hand side goes to as . Therefore as , that is .
Theorem
Every Cauchy sequence in is convergent.
Proof Let be a Cauchy sequence in . Then is bounded. Thus, it follows from Bolzano-Weierstrass theorem that has a convergent subsequence. Consequently, by invoking previous theorem, is convergent.
Completeness
Complete Metric Space
A metric space is called complete if every Cauchy sequence in is convergent.
e.g.
with the standard Euclidean metric is complete.
with the induced Euclidean metric is not complete.
is complete under the metric
Proposition
Let be a complete metric space and . Then is complete iff is closed in .
Proof Let be a sequence in with . Then is a Cauchy sequence in and hence is a Cauchy sequence in . Since is complete, for some . By uniqueness of limit, we have and thus . This shows that is closed by corollary. Conversely, let be a Cauchy sequence in . Then is also a Cauchy sequence in . Since is complete, for some . Since is closed, . Therefore is complete.
Universal Property of Completion
Every metric space has a unique completion, i.e. there exists a complete metric space and an injection such that and .
Contraction Mapping Theorem
Fixed Point
Let be a metric space and a map. A point is called a fixed point of if .
Contraction
Let be a metric space and a map. is called a contraction if there exists such that where is called the contraction rate of .
Contraction Mapping Theorem
Let be a complete metric space. If is a contraction, then it has a unique fixed point.
Quotient of Complete Metric Spaces
Proposition
Suppose is a complete metric space and is a compact subspace. Then the topological quotient is metrizable, and complete under the metric
Proof We first show that is a well-defined metric on . For and , we have and . Therefore, , which shows that the metric is well-defined. The other properties of a metric (non-negativity, symmetry, and triangle inequality) can be verified similarly.
Now we show that the topology induced by this metric coincides with the quotient topology by showing that for all .