# 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

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

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

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

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

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

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 $v_N - \epsilon$is 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

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

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

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?