Euclidean space is the space in which everyone is most familiar. In Euclidean k-space, the distance between any two points is
where k is the dimension of the Euclidean space. Since the Euclidean k-space as a metric on it, it is also a topological space.
Definition: A sequence of real numbers is said to converge to the real number s provided for each there exists a number such that implies .
Definition: A sequence of real numbers is called a Cauchy sequence if for each there exists a number such that .
Convergent sequences are Cauchy Sequences.
Suppose that .
Let . Then such that
Hence, is a Cauchy sequence.
Convergent sequences are bounded.
Proof: Let be a convergent sequence and let . From the definition of convergence and letting , we can find N such that
From the triangle inequality;
for all . Thus is a bounded sequence.
In a complete space, a sequence is a convergent sequence if and only if it is a Cauchy sequence.
Convergent sequences are Cauchy sequences. See Lemma 1.
Consider a Cauchy sequence . Since Cauchy sequences are bounded, the only thing to show is:
Let . Since is a Cauchy sequence, such that
So, for all . This shows that is and upper bound for and hence for all . Also is a lower bound for . Therefore . Now:
Since this holds for all , . The opposite inequality always holds and now we have established the theorem.
Note: The preceding proof assumes that the image space is . Without this assumption, we will need more machinery to prove this.
It is possible to have more than one metric prescribed to an image space. It is often better to talk about metric spaces in a more general sense. A sequence in a metric space converges to s in S if . A sequence is called Cauchy if for each there exists an such that:
The metric space is called complete if every Cauchy sequence in converges to some element in .
Let be a complete metric and be a subspace of . Then is a complete metric space if and only if is a closed subset of .
Suppose is a closed subset of . Let be a Cauchy sequence in .
Then is also a Cauchy sequence in . Since is complete, converges to a point in . However, is a closed subset of so is also complete.
Left as an exercise.
1. Let . Let where and
a.) and are metrics for .
b.) and form a complete metric space.
2. Show that every open set in is the disjoint union of a finite or infinite sequence of open intervals.
3. Complete the proof for theorem 3.
4.Consider: Let and be metric spaces.
a.) A mapping is said to be a Lipschitz mapping provided that there is some non-negative number c called a Lipschitz constant for the mapping such that:
b.) A Lipschitz mapping that has a Lipschitz constant less than 1 is called a contraction.
Suppose that and are both Lipschitz. Is the product of
these functions Lipschitz?