# 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

$d(x,y)={\sqrt {\sum _{i=1}^{k}{(x_{i}-y_{i})^{2}}}}$ 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}  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?