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.
Sequences[edit]
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
.
Lemma 1[edit]
Convergent sequences are Cauchy Sequences.
Proof : Suppose that
.
Then,
Let
. Then
such that
Also:
so
Hence,
is a Cauchy sequence.
Theorem 1[edit]
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;
Let
.
Then,
for all
. Thus
is a bounded sequence.
Theorem 2[edit]
In a complete space, a sequence is a convergent sequence if and only if it is a Cauchy sequence.
Proof:
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.
Definition[edit]
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
.
Theorem 3[edit]
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
.
Proof:
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.
Exercises[edit]
1. Let
. Let
where
and
. Show:
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?