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

d(x,y) = \sqrt {\sum^k_{i=1} {(x_i - y_i)^2}}

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 \{s_n\} is said to converge to the real number s provided for each \epsilon > 0 there exists a number N such that n > N implies \mid s_n - s \mid  < \epsilon.

Definition: A sequence \{s_n\} of real numbers is called a Cauchy sequence if for each \epsilon > 0 there exists a number N such that m,n > N \Rightarrow \mid s_n -s_m \mid  < \epsilon.

Lemma 1[edit]

Convergent sequences are Cauchy Sequences.

Proof : Suppose that lim \; s_n = s.

Then,

\mid s_n - s_m \mid =  \mid s_n - s + s - s_m \mid \leq \mid s_n - s\mid + \mid s - s_m \mid

Let \epsilon > 0. Then \exists N such that

n > N \Rightarrow \mid s_n - s \mid < \frac {\epsilon}{2}

Also:

m > N \Rightarrow \mid s_m - s \mid < \frac {\epsilon}{2}

so

m,n > N \Rightarrow \mid s_n - s_m \mid \leq \mid s_n - s\mid + \mid s - s_m \mid < \frac {\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

Hence, \{s_n\} is a Cauchy sequence.

Theorem 1[edit]

Convergent sequences are bounded.

Proof: Let \{s_n\} be a convergent sequence and let lim \; s_n = s. From the definition of convergence and letting \epsilon = 1, we can find N \in \mathbb{N} such that

n > N \Rightarrow \mid s_n - s \mid < 1

From the triangle inequality;

n > N \Rightarrow \mid s_n \mid < \mid s \mid + 1

Let M = max \{\mid s_n \mid + 1, \mid s_1 \mid , \mid s_2 \mid , ... , \mid s_N \mid \}.

Then,

\mid s_n \mid \leq M

for all n \in \mathbb{N}. Thus \{s_n\} 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:

\Rightarrow Convergent sequences are Cauchy sequences. See Lemma 1.

\Leftarrow Consider a Cauchy sequence \{s_n\}. Since Cauchy sequences are bounded, the only thing to show is:

\liminf s_n = \limsup s_n

Let \epsilon > 0. Since \{s_n\} is a Cauchy sequence, \exists N such that

m,n > N \Rightarrow \mid s_n - s_m \mid  <  \epsilon

So, s_n < s_m + \epsilon for all m,n > N. This shows that s_m + \epsilon is and upper bound for \{s_n : n > N\} and hence v_N = \sup \{s_n : n > N\} \leq s_m + \epsilon for all  m > N. Also v_N - \epsilonis a lower bound for \{s_m : m > N\}. Therefore v_n - \epsilon \leq \inf \{s_m : m > N\} = u_N. Now:

\limsup s_n \leq v_N \leq u_N + \epsilon \leq \liminf s_n + \epsilon

Since this holds for all \epsilon > 0, \limsup s_n \leq \liminf s_n. The opposite inequality always holds and now we have established the theorem.

Note: The preceding proof assumes that the image space is \mathbb{R}. 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 \{s_n\} in a metric space (S,d) converges to s in S if \lim _{n \rightarrow\infty} d(s_n,s) = 0. A sequence is called Cauchy if for each \epsilon > 0 there exists an N such that:

m,n > N \Rightarrow d(s_m,s_n) < \epsilon

.

The metric space (S,d) is called complete if every Cauchy sequence in S converges to some element in S.

Theorem 3[edit]

Let X be a complete metric and Y be a subspace of X. Then Y is a complete metric space if and only if Y is a closed subset of X.

Proof: \Leftarrow Suppose Y is a closed subset of X. Let \{s_n\} be a Cauchy sequence in Y.

Then \{s_n\} is also a Cauchy sequence in X. Since X is complete, \{s_n\} converges to a point s in X. However, Y is a closed subset of X so Y is also complete.

\Rightarrow Left as an exercise.



Exercises[edit]

1. Let x,y \in \mathbb{R}^k. Let d_1(x,y) = max \mid x_i - y_i \mid where \{i = 1,2,...,k\} and d_2(x,y) = \sum_i^k \mid x_i - y_i \mid. Show:

a.) d_1 and d_2 are metrics for \mathbb{R}^k.

b.) d_1 and d_2 form a complete metric space.

2. Show that every open set in \mathbb{R} is the disjoint union of a finite or infinite sequence of open intervals.

3. Complete the proof for theorem 3.

4.Consider: Let X and Y be metric spaces.

a.) A mapping T : X \rightarrow Y 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:

d[T(x),T(y)] \leq cd(x,y) \; \forall x,y \in X

b.) A Lipschitz mapping T : X\rightarrow Y that has a Lipschitz constant less than 1 is called a contraction.

Suppose that f : \mathbb{R} \rightarrow\mathbb{R} and g :\mathbb{R} \rightarrow\mathbb{R} are both Lipschitz. Is the product of these functions Lipschitz?