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

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

Lemma 1

Convergent sequences are Cauchy Sequences.

Proof : Suppose that $lim\;s_{n}=s$$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$$\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$$\epsilon >0$. Then $\exists N$$\exists N$ such that

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

Also:

$m>N\Rightarrow \mid s_{m}-s\mid <{\frac {\epsilon }{2}}$$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$$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}\}$$\{s_{n}\}$ is a Cauchy sequence.

Theorem 1

Convergent sequences are bounded.

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

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

From the triangle inequality;

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

for all $n\in \mathbb {N}$$n\in \mathbb {N}$. Thus $\{s_{n}\}$$\{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$$\Rightarrow$ Convergent sequences are Cauchy sequences. See Lemma 1.

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

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

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

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

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

Since this holds for all $\epsilon >0$$\epsilon >0$, $\limsup s_{n}\leq \liminf s_{n}$$\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}$$\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}\}$$\{s_{n}\}$ in a metric space $(S,d)$$(S,d)$ converges to s in S if $\lim _{n\rightarrow \infty }d(s_{n},s)=0$$\lim _{n\rightarrow \infty }d(s_{n},s)=0$. A sequence is called Cauchy if for each $\epsilon >0$$\epsilon >0$ there exists an $N$$N$ such that:

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

.

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

Theorem 3

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

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

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

$\Rightarrow$$\Rightarrow$ Left as an exercise.

Exercises

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

a.) $d_{1}$$d_1$ and $d_{2}$$d_2$ are metrics for $\mathbb {R} ^{k}$$\mathbb {R} ^{k}$.

b.) $d_{1}$$d_1$ and $d_{2}$$d_2$ form a complete metric space.

2. Show that every open set in $\mathbb {R}$$\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$$X$ and $Y$$Y$ be metric spaces.

a.) A mapping $T:X\rightarrow Y$$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$$d[T(x),T(y)]\leq cd(x,y)\;\forall x,y\in X$

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

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