# Topology/Euclidean Spaces

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

editDefinition: 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

editConvergent sequences are Cauchy Sequences.

Proof : Suppose that .

Then,

Let . Then such that

Also:

so

Hence, is a Cauchy sequence.

## Theorem 1

editConvergent 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

editIn 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

editIt 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

editLet 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

edit1. 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?