# 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

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

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

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

## Lemma 1

Convergent sequences are Cauchy Sequences.

Proof : Suppose that ${\displaystyle lim\;s_{n}=s}$.

Then,

${\displaystyle \mid s_{n}-s_{m}\mid =\mid s_{n}-s+s-s_{m}\mid \leq \mid s_{n}-s\mid +\mid s-s_{m}\mid }$

Let ${\displaystyle \epsilon >0}$. Then ${\displaystyle \exists N}$ such that

${\displaystyle n>N\Rightarrow \mid s_{n}-s\mid <{\frac {\epsilon }{2}}}$

Also:

${\displaystyle m>N\Rightarrow \mid s_{m}-s\mid <{\frac {\epsilon }{2}}}$

so

${\displaystyle m,n>N\Rightarrow \mid s_{n}-s_{m}\mid \leq \mid s_{n}-s\mid +\mid s-s_{m}\mid <{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon }$

Hence, ${\displaystyle \{s_{n}\}}$ is a Cauchy sequence.

## Theorem 1

Convergent sequences are bounded.

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

${\displaystyle n>N\Rightarrow \mid s_{n}-s\mid <1}$

From the triangle inequality;

${\displaystyle n>N\Rightarrow \mid s_{n}\mid <\mid s\mid +1}$

Let ${\displaystyle M=max\{\mid s_{n}\mid +1,\mid s_{1}\mid ,\mid s_{2}\mid ,...,\mid s_{N}\mid \}}$.

Then,

${\displaystyle \mid s_{n}\mid \leq M}$

for all ${\displaystyle n\in \mathbb {N} }$. Thus ${\displaystyle \{s_{n}\}}$ 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:

${\displaystyle \Rightarrow }$ Convergent sequences are Cauchy sequences. See Lemma 1.

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

${\displaystyle \liminf s_{n}=\limsup s_{n}}$

Let ${\displaystyle \epsilon >0}$. Since ${\displaystyle \{s_{n}\}}$ is a Cauchy sequence, ${\displaystyle \exists N}$ such that

${\displaystyle m,n>N\Rightarrow \mid s_{n}-s_{m}\mid <\epsilon }$

So, ${\displaystyle s_{n} for all ${\displaystyle m,n>N}$. This shows that ${\displaystyle s_{m}+\epsilon }$ is and upper bound for ${\displaystyle \{s_{n}:n>N\}}$ and hence ${\displaystyle v_{N}=\sup\{s_{n}:n>N\}\leq s_{m}+\epsilon }$ for all ${\displaystyle m>N}$. Also ${\displaystyle v_{N}-\epsilon }$is a lower bound for ${\displaystyle \{s_{m}:m>N\}}$. Therefore ${\displaystyle v_{n}-\epsilon \leq \inf\{s_{m}:m>N\}=u_{N}}$. Now:

${\displaystyle \limsup s_{n}\leq v_{N}\leq u_{N}+\epsilon \leq \liminf s_{n}+\epsilon }$

Since this holds for all ${\displaystyle \epsilon >0}$, ${\displaystyle \limsup s_{n}\leq \liminf s_{n}}$. The opposite inequality always holds and now we have established the theorem.

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

${\displaystyle m,n>N\Rightarrow d(s_{m},s_{n})<\epsilon }$

.

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

## Theorem 3

Let ${\displaystyle X}$ be a complete metric and ${\displaystyle Y}$ be a subspace of ${\displaystyle X}$. Then ${\displaystyle Y}$ is a complete metric space if and only if ${\displaystyle Y}$ is a closed subset of ${\displaystyle X}$.

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

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

${\displaystyle \Rightarrow }$ Left as an exercise.

## Exercises

1. Let ${\displaystyle x,y\in \mathbb {R} ^{k}}$. Let ${\displaystyle d_{1}(x,y)=max\mid x_{i}-y_{i}\mid }$ where ${\displaystyle \{i=1,2,...,k\}}$ and ${\displaystyle d_{2}(x,y)=\sum _{i}^{k}\mid x_{i}-y_{i}\mid }$. Show:

a.) ${\displaystyle d_{1}}$ and ${\displaystyle d_{2}}$ are metrics for ${\displaystyle \mathbb {R} ^{k}}$.

b.) ${\displaystyle d_{1}}$ and ${\displaystyle d_{2}}$ form a complete metric space.

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

3. Complete the proof for theorem 3.

4.Consider: Let ${\displaystyle X}$ and ${\displaystyle Y}$ be metric spaces.

a.) A mapping ${\displaystyle T:X\rightarrow Y}$ 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:

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

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

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