07. Completeness

7.1 Basic Definitions and Examples

Bounded Sequence

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence in some metric space $X$

Sequence is bounded if

$$\text{set }\{x_n:n\ge 1\}\text{ is bounded}$$

In other words there exists $R>0$ and $a$ such that

$$x_n\in B(a,R)\text{ for all }n\ge 1$$

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Cauchy Sequence

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence in some metric space $X$

Sequence is Cauchy if for all $ϵ>0$ there exists $N$ such that for all $n,m\ge N$ then

$$d(x_n,x_m)<ϵ$$

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Convergent Sequence

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence in some metric space $X$

Sequence is convergent if there exists $a\in X$ such that

$$\underset{n\to \infty }{\lim }{x}_n=a$$

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Properties of Subsequence for Bounded, Cauchy, Convergent Sequences

If sequence $\sigma$ have any one of Bounded, Cauchy or Convergent then

$$\text{subsequence }\sigma \text{ also has the property}$$

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Completeness

Metric Space is complete if

$$\text{every Cauchy Sequence converges }$$

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7.2 First Properties of Complete Metric Spaces

Relation between Completeness and Closure

Let $X$ be a complete metric space and $Y\subseteq X$

$$Y\text{ is Complete}⟺Y\text{ is Closed}$$

Proof

Let $X$ be a complete metric space and let $Y\subseteq X$ then

Suppose $Y$ is closed then
Let $(y_n{)}_{n=1}^{\infty }$ be a Cauchy Sequence in $Y$ then it is a Cauchy Sequence in $X$
Since $X$ is complete, it converges so suppose ${\lim }_{n\to \infty }{y}_n=a$
By Interior points and convergent sequences then $a\in Y$ hence $Y$ is complete

Suppose $Y$ is complete then
Let $(y_n{)}_{n=1}^{\infty }$ be a sequence of elements of $Y$ with ${\lim }_{n\to \infty }{y}_n=a$ for some $a\in X$ then
$(y_n{)}_{n=1}^{\infty }$ is a Cauchy Sequence in $Y$ so by completeness it converges to an element in $Y$
By uniqueness of limits then $a\in Y$ thus $Y$ contains limits of such sequence
So by Interior points and convergent sequences then $Y$ is closed

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Diameter

Let $X$ be a metric space and $Y\subseteq X$ a non-empty subset then

Diameter of $Y$ is defined as

$$\mathrm{diam}(T):=\sup \{d(x,y):x,y\in Y\}$$

when the set is bounded otherwise it is infinity

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Cantor's Intersection Theorem lemma

Let $X$ be a complete metric space and suppose that $S_1\supseteq S_2\supseteq \cdots$
such that it forms a nested sequence of non-empty closed sets in $X$ with property

$$\mathrm{diam}(S_n)\to 0\text{ as }n\to \infty$$

Then

$$\bigcap _{n=1}^{\infty }{S}_n\text{ contains a unique point a}$$

Proof

For each $n\in N$, pick $x_n\in S_n$
Let $ϵ>0$ so there exists large $N>0$ such that

$$diam(S_N)<ϵ$$

If $n,m\ge N$ then since $S_i$ are nested then

$$x_n,x_m\in S_n$$

By definition of diameter then

$$d(x_n,x_m)\le \mathrm{diam}(S_N)<ϵ$$

Hence $(x_n{)}_{n=1}^{\infty }$ is Cauchy

As $X$ is complete then ${\lim }_{n\to \infty }{x}_n=a$ for some $a\in X$
For each $i$, by the nesting property of sets $S_i$ then

$$x_n\in S_i\text{ for all }n\ge i$$

Since $S_i$ is closed then by Interior points and convergent sequences then

$$a\in S_i$$

As this holds true for all $i$ then $a\in {\bigcap }_{i=1}^{\infty }{S}_i$

Suppose there exists $b\in {\bigcap }_{i=1}^{\infty }{S}_i$ then

$$d(a,b)\le \mathrm{diam}(S_i)\text{ for all }i$$

Since $diam(S_i)\to 0$ then $d(a,b)=0$ so $a=b$

Hence $a$ is unique

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7.3 Completeness of Function Spaces

08 - Completeness of Normed Vector Spaces for Bounded Functions

Completeness of Normed Vector Spaces for Bounded Functions

Let $X$ be a set then

$$B(x)\text{ is complete}$$

where $B(X)$ denotes the normed vector space of bounded functions $f:X\to \mathbb{R}$ with norm

$$||f|{|}_{\infty }=\underset{x\in X}{\sup }|f(x)|$$

Proof

Let $(f_n{)}_{n=1}^{\infty }$ be a Cauchy Sequence in $B(X)$ then
For each $x$, sequence $(f_n(x){)}_{n=1}^{\infty }$ is a Cauchy Sequence of Real Numbers

As $\mathbb{R}$ is complete then each sequence has a limit written as $f(x)$ so that

$$\underset{n\to \infty }{\lim }{f}_n(x)=f(x)$$

Showing that $f$ is bounded
Take $ϵ=1$ in Cauchy so there exists $N$ such that for $n,m\ge N$ then

$$\underset{x\in X}{\sup }|f_n(x)-f_m(x)|\le 1$$

Taking $m=N$ then

$$|f_N(x)-f_n(x)|\le 1\text{ for all }n\ge N\text{ and }x\in X$$

As $n\to \infty$ then we get

$$|f_N(x)-f(x)|\le 1\text{ for all }x\in X$$

As $f_N$ is a bounded function then so is $f$

Showing that $f_n\to f$ in the norm $||\cdot |{|}_{\infty }$ (have pointwise convergence currently)
Let $ϵ>0$ so there exists $N$ such that if $n,m\ge N$ then

$$|f_n(x)-f_m(x)|\le ϵ\text{ for all }x\in X$$

For fixed $n\ge N$ and $x\in X$ as $m\to \infty$ hence

$$|f_n(x)-f(x)|\le ϵ$$

Hence for al $n\ge N$ we have

$$||f_n-f|{|}_{\infty }\le ϵ$$

Hence $f_n\to f$ in the $||\cdot |{|}_{\infty }-$norm

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09 - Completeness of the Space of Bounded Continuous Functions

Completeness of the Space of Bounded Continuous Functions

Let $X$ be a metric space then

$$C_b(X)\text{ is complete}$$

where $C_b(X)$ denotes normed vector space of bounded continuous functions $f:X\to \mathbb{R}$ with

$$||f|{|}_{\infty }=\underset{x\in X}{\sup }|f(x)|$$

Proof

By Completeness of Normed Vector Spaces for Bounded Functions, $B(X)$ is complete
So by Relation between completeness and closure then need to show

$$C_b(X)\text{ is c alosed subset of }B(X)$$

By Interior points and convergent sequences then just need to show that
If $(f_n{)}_{n=1}^{\infty }$ is a sequence of elements of $C_b(X)$ then $f\in C_b(X)$ ($f$ is continuous)

Let $a\in X$ and let $ϵ>0$
Since $f_n\to f$ in the $||\cdot |{|}_{\infty }$-norm then there exists $n$ such that

$$||f_n-f||\le \frac{ϵ}{3}$$

As $f_n$ is continuous then there is $\delta >0$ such that

$$|f_n(x)-f_n(a)|<\frac{ϵ}{3}\text{ for all }x\in B(a,\delta )$$

But for $x\in B(a,\delta )$ then

$$\begin{array}{rl}|f(x)-f(a)& |\le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|\\ & <\frac{ϵ}{3}+\frac{ϵ}{3}+\frac{ϵ}{3}=ϵ\end{array}$$

Hence $f$ is continuous at $a$ and as $a$ was arbitrary then $f$ is a continuous function on $X$

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7.4 The Contraction Mapping Theorem

Lipschitz Map / Continuous

Let $(X,d_X)$ and $(Y,d_Y)$ be metric space
Suppose $f:X\to Y$

$f$ is Lipschitz Map or Lipschitz Continuous if there exists constant $K\ge 0$ such that

$$d_Y(f(x),f(y))\le K{d}_X(x,y)$$

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Contraction Mapping

Let $(X,d_X)$ be a metric space then

$f$ is a Contraction Mapping if $f:X\to X$ is a Lipschitz Map with constant $K\in [0,1)$

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10 - Contraction Mapping Theorem

Contraction Mapping Theorem

Let $X$ be a non-empty complete metric space
Suppose that $f:X\to X$ is a contraction then

$f$ has a unique fixed point so there exists unique $x\in X$ such that

$$f(x)=x$$

Proof

Need to show that it is unique

Suppose we have $f(x_1)=x_1$ and $f(x_2)=x_2$ then

$$d(x_1,x_2)=d(f(x_1),f(x_2))\le Kd(x_1,x_2)$$

Since $d(x_1,x_2)\ge 0$ and $0\le K<1$ then

$$d(x_1,x_2)=0$$

Hence

$$x_1=x_2$$

Showing there is a fixed point
Proof is constructive and may be used in practical situations to find fixed point numerically
Pick an arbitrary $x_0\in X$and form iterates

$$\begin{array}{rl}{x}_1& :=f(x_0)\\ x_2& :=f(x_1)\\ & \text{etc...}\end{array}$$

Claim that $(x_n{)}_{n=1}^{\infty }$ converges to some limit $x$ and $f(x)=x$
Using contraction property then

$$d(x_n,x_{n-1})\le Kd(x_{n-1},x_{n-2})\le K^{2}d(x_{n-2},x_{n-3})\le \cdots \le K^{n-1}d(x_0,x_1)$$

Hence for $n>m$ then

$$\begin{array}{rl}d(x_n,x_m)& \le d(x_n,,x_{n-1})+\cdots +d(x_{m+1},x_m)\\ & \le (K^{n-1}+K^{n-2}+\cdots +K^m)d(x_0,x_1)\\ & =K^m(1+K+K^2+\cdots K^{n-1-m})d(x_0,x_1)\\ & \le K^m(1+K+K^2+\cdots )d(x_0,x_1)\\ & =K^m\cdot \frac{d(x_0,x_1)}{1-K}=C{K}^m\end{array}$$

where $C=\frac{d(x_0,x_1)}{1-K}$
Since $K<1$ for any $ϵ>0$ there exists $N$ such that $C{K}^N<ϵ$
Thus $(x_n{)}_{n=1}^{\infty }$ is a Cauchy Sequence

As $X$ is complete then $x_n\to x$ for some $x\in X$
Since $f$ is continuous then

$$f(x)=\underset{n\to \infty }{\lim }f(x_n)=\underset{n\to \infty }{\lim }{x}_{n+1}=x$$

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