Topology/Euclidean Spaces

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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?