# 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, $s_{n}<s_{m}+\epsilon$ 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?