Naturals

$\mathbb{N}=\{1,2,3,4,5,\dots \}$

Integers

$\mathbb{Z}=\{\dots ,-3,-2,-1,0,1,2,3,\dots \}$

Rational number

any number that can be expressed in the form $\frac{p}{q}$ where $p,q\in \mathbb{Z}$ and $q\ne 0$

1.1.1 pf $\sqrt{2}$ is not rational

$\begin{array}{rl}& {\left(\frac{p}{q}\right)}^2=2\text{where gcd(p,q)=1}\\ & p^2=2q^2⟹p\text{ is even by Euclid}\\ & p=2r⟹2r^2=q^2\\ & \text{ so }q\text{ is even which is a contradiction}\end{array}$

Field

$\mathbb{Q}$ makes up a field
any set where addition and multiplication are well-defined operators:

Commutative, associative, distributive, additive identity, additive inverse, multiplicative inverse, multiplicative identity, and closure.

Ordered
$x\le y$ or $y\le x$
if $x\le y$ and $y\le x$ then $x=y$
if $x\le y$ and $y\le z$ then $x\le z$
if $y\le z$ then $x+y\le x+z$
if $0\le x$ and $0\le y$ then $0\le xy$

We assume that $\mathbb{R}$ is an ordered field that contains $\mathbb{Q}$.

Functions

Given two sets $A$ and $B$ a function from $A$ to $B$ is a mapping that takes each element $x\in A$ with a single element of $B$. So $f:A\to B$

Range: $\{y\in B:y=f(x)\text{ for some }x\in A\}$

Drichlet's unruly function

$g(x)=\{\begin{array}{l}1\text{ if }x\in \mathbb{Q}\\ 0\text{ if }x\notin \mathbb{Q}\end{array}$

Absolute Value Function

$|x|=\{\begin{array}{l}x\text{ if }x\ge 0\\ -x\text{ if }x<0\end{array}$

Properties

$\begin{array}{rl}& |ab|=|a||b|\\ & |a+b|\le |a|+|b|\mathbf{\text{ triangle inequality}}\end{array}$
by subbing $a=(a+c-c)$ into the triangle inequality we get:
$|a-b|\le |a-c|+|c-b|$

Axiom of Completeness

Every nonempty set of real numbers that is bounded above has a least upper bound

Bounded Above

$A\subseteq \mathbb{R}$ is bounded above if there exists a number $b\in \mathbb{R}$ such that $a\le b$ for a $a\in A$

Least Upper bound (supremum)

$s\in \mathbb{R}$ is the least upper bound for $A\subseteq \mathbb{R}$ if it meets both criteria:

1 $s$ is an upper bound for $A$
2 if $b$ is any upper bound for $A$, then $s\le b$

Greatest lower bound is the infimum.

Maximum/Minimum

$a_0$ is a maximum of the set $A$ if it is an element of $A$ and $a_0\ge a$ for all $a\in A$.

$a_1$ is a minimum if $a_1\in A$ and $a_1\le a$ for every $a\in A$.

When a maximum exists it is a supremum and when a minimum exists it is a infimum.

Example 1.3.7

$A\subseteq \mathbb{R}$ is nonempty and bounded above, let $c\in \mathbb{R}$. Now we have the set:
$c+A=\{c+a:a\in A\}$
Claim: sup($c+A$) = $c$ + sup$A$

$s=\text{sup}A⟹a\le s$ for all $a\in A$
$a+c\le s+c$
so $s+c$ is an upper bound.

Let $b$ be another upper bound for $c+A$.

$\begin{array}{c}c+a\le b\\ a\le b-c\end{array}$
so $b-c$ is an upper bound for $a$ so $s<b-c$ since it is sup$A$. So $s+c<b$. Therefore $s+c$ is sup$(c+A)$.

Lemma 1.3.8 Supremum

Assuming $s\in \mathbb{R}$ is an upper bound for $A\subseteq \mathbb{R}$,

then $s=supA$ iff for every choice of $\epsilon >0$, there exists an element $a\in A$ such that $s-\epsilon <a$.

Proof

$\Rightarrow$
Assume $s=supA$. $s-\epsilon <s$ so $s-\epsilon$ is not an upper bound for $A$. Then there must be $a\in A$ so that $s-\epsilon <a$.

$\Leftarrow$
$s$ is an upper bound. For any $\epsilon >0$, $\mathrm{\exists }a$ where $s-\epsilon <a$ so $s-\epsilon$ is no longer a upper bound. Let $b<s$ so $s-\epsilon$ for some $\epsilon >0$. Therefore $b$ is not an upper bound, therefore $s$ is the lowest upper bound. $s=supA$.

Consequences of Completeness

Thm 1.4.1 Nested Interval Property

For each $n\in \mathbb{N}$ assume that there is some closed interval $I_n=[a_n,b_n]=\{x\in \mathbb{R}:a_n\le x\le b_n\}$
If $I_1\supseteq I_2\supseteq I_3\supseteq \dots$ then $\bigcap _{n=1}^{\mathrm{\infty }}{I}_n\ne \mathrm{\varnothing }$

Proof

Let $A=\{a_n:n\in \mathbb{N}\}$ which is the lower bound of each interval. Since they are nested each $b_n$ serves as an upper-bound to $A$. Let $x=supA$ which exists by the axiom of completeness.

So for a particular $I_n$ we have $a_n\le x$ and $x\le b_n$ which means $x\in I_n$ for every choice of $n\in \mathbb{N}$. Therefore the intersection is non-empty.

Thm 1.4.2 Archimedean Property

Statement 1: Given any number $x\in \mathbb{R}$ $\mathrm{\exists }$ an $n\in \mathbb{N}$ where $n>x$

Statement 2: Given any $y>0$ there exists an $n\in \mathbb{N}$ satisfying $\frac{1}{n}<y$

Proof

Proof (statement 1)
States that $\mathbb{N}$ is not bounded above. For contradiction assume that $\mathbb{N}$ is bounded above. $\mathbb{N}\subset \mathbb{R}$ so by the axiom of completeness there exists an $\alpha =sup\mathbb{N}$. So, $\alpha -1$ is no longer an upper bound so there is an $n\in \mathbb{N}$ so that $\alpha -1<n$ therefore $\alpha <(n+1)\in \mathbb{N}$ which is a contradiction.

($\mathbb{N}$ is closed under addition)

(statement 2) From our previous proof let $x=\frac{1}{y}⟹\frac{1}{y}<n$ for some $n\in \mathbb{N}$.

Thm 1.4.3 Density of Q in R

For every two real numbers $a$ and $b$ with $a<b$ $\mathrm{\exists }$ $r\in \mathbb{Q}$ such that $a<r<b$.

Proof

Proof want a $m\in \mathbb{Z}$ and $n\in \mathbb{Z}$ so that
$a<\frac{m}{n}<b$
by Archimedean Property $\mathrm{\exists }n\in \mathbb{N}$ where
$\frac{1}{n}<b-a$

Inequality (1) is equivalent to $na<m<nb$

Pick an $m\in \mathbb{Z}$ so that
$m-1\stackrel{(3)}{\le }na\stackrel{(4)}{<}m$
Inequality (4) gives us $a<\frac{m}{n}$

Inequality (2) is equivalent to $a<b-\frac{1}{n}$.

$\begin{array}{rl}\text{by (3)}& m& \le na+1& <n(b-\frac{1}{n})+1\\ & =nb\end{array}$
so $n<nb⟹\frac{m}{n}<b$ and we are done.

Corollary 1.4.4 Irrational Dense in R

Given any two real numbers $a<b$ there exists a $t\in \mathbb{R}\setminus \mathbb{Q}$ satisfying $a<t<b$

Proof

$r\in \mathbb{Q}$ if $x$ is irrational then so is $r+x$ otherwise if $r+x\in \mathbb{Q}$ then $r+x-r$ since $\mathbb{Q}$ is a field which contradicts $r$ being irrational.

So let for any $a,b\in \mathbb{R}$ we have $a-\sqrt{2}<a$ and $b-\sqrt{2}<b$.

by density in $\mathbb{R}$ $\mathrm{\exists }r\in \mathbb{Q}$ so that
$\begin{array}{rl}a-\sqrt{2}& <r<b-\sqrt{2}\\ a& <r+\sqrt{2}<b\end{array}$
and $r+\sqrt{2}$ is irrational.

Thm 1.4.5 Existence of Square Roots

There exists an $\alpha \in \mathbb{R}$ such that ${\alpha }^2=2$
Proof
$T=\{t\in \mathbb{R}:t^2<2\}$

Assume $\alpha =supT$
Assume for contradiction that
${\alpha }^2<2$

$\begin{array}{rl}{(a+\frac{1}{n})}^2& ={\alpha }^2+\frac{2\alpha }{n}+\frac{1}{n^2}\\ & <\alpha +\frac{2\alpha }{n}+\frac{1}{n}\\ & ={\alpha }^2+\frac{2\alpha +1}{n}\end{array}$
$\begin{array}{r}{\alpha }^2+\frac{2\alpha +1}{n}<2+\frac{2\alpha +1}{n}\end{array}$
By Archimedes there is a $\frac{1}{n_0}<\frac{2-a^2}{2\alpha +1}$ so

$\begin{array}{rl}\frac{2\alpha +1}{n_0}& <2-{\alpha }^2\\ {(\alpha +\frac{1}{n_0})}^2& <{\alpha }^2+(2-{\alpha }^2)\end{array}$

so $(\alpha +\frac{1}{n})\in T$ contradicting that $supT=\alpha$

Assume now that ${\alpha }^2>2$

$\begin{array}{rl}{(\alpha -\frac{1}{n})}^2& ={\alpha }^2-\frac{2\alpha }{n}+\frac{1}{n^2}\\ & >{\alpha }^2-\frac{2\alpha }{n}\end{array}$

by Archimedes there is a $n_0$ such that

$\begin{array}{rl}\frac{1}{n_0}& <\frac{{\alpha }^2+2}{2\alpha }\text{ so}\\ -\frac{2\alpha }{n_0}& >-(a^2+2)\end{array}$

so therefore
$\begin{array}{r}{(\alpha -\frac{1}{n_0})}^2>a^2-(a^2+2)\end{array}$
$\alpha -\frac{1}{n_0}\notin T$ and $\alpha -\frac{1}{n_0}<\alpha$ so $a-\frac{1}{n_0}$ is a lower bound.

A sequence is a function whose domain is $\mathbb{N}$
$f:\mathbb{N}\to \mathbb{R},f(n)$

Convergence of a Sequence

A sequence ($a_n$) converges to $a\in \mathbb{R}$ if for every $ϵ>0$ there exists a $N\in \mathbb{N}$ such that if $n>\mathbb{N}$ then $|a_n-a|<ϵ$

$ϵ$-Neighbourhood of a

$V_{ϵ}=\{x\in \mathbb{R}:|x-a|<ϵ\}$

So an equivalent definition is:
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n=a}$ iff $\mathrm{\forall }\epsilon >0,B_{\epsilon }=\{m:a_n\notin V_{\epsilon }(a)\}$ is finite.

Thm 2.2.7 Uniqueness of Limits

If $(a_n)\to a$ and $(a_n)\to b$ then $a=b$

Proof

There exists and $N_a$ and $N_b$ where
for all $\frac{ϵ}{2}>0$ $|a_n-a|<\frac{ϵ}{2}$ and $|a_n-b|<\frac{ϵ}{2}$

$|a_n-a-a_n+b|\le |a_n-a|+|a_n-b|<ϵ$
$|b-a|<ϵ$

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Algebraic and Order Limit Theorems

Algebraic Theorems

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_n=b$ then

$\begin{array}{}\text{(i)}& & \underset{n\to \mathrm{\infty }}{lim}(c\cdot a_n)=c\cdot a,\text{ for all }c\in \mathbb{R}\text{(ii)}& & \underset{n\to \mathrm{\infty }}{lim}(a_n+b_n)=a+b\text{(iii)}& & \underset{n\to \mathrm{\infty }}{lim}(a_n\cdot b_n)=a\cdot b\text{(iv)}& & \underset{n\to \mathrm{\infty }}{lim}\left(\frac{a_n}{b_n}\right)=\frac{a}{b},\text{if }b\ne 0\end{array}$

Proofs

Proof i
There is an $N$ so that $n>N$
$|a_n-a|<\frac{ϵ}{|c|}$
$\begin{array}{r}|c{a}_n-ca|=|c||a_n-a|<|c|\frac{ϵ}{|c|}=ϵ\end{array}$

Proof ii

We know for some $N_a$ that $|a_n-a|<\frac{\epsilon }{2}$ and some $N_b$ that $|b_n-b|\le \frac{ϵ}{2}$. Take $max\{N_a,N_b\}$ and the rest is trivial

Proof iii

$\begin{array}{rcl}|a_n{b}_n-ab|& =& |a_n{b}_n-a{b}_n+a{b}_n-ab|\\ & \le & |a_n{b}_n-a{b}_n|+|a{b}_n-ab|\\ & =& |b_n||a_n-a|+|a||b_n-b|.\end{array}$

We know there is a $N_b$ so that
$n\ge N_b⟹|b_n-b|<\frac{1}{|a|}\frac{ϵ}{2}$
Also $b_n$ converges so it is bounded $|b_n|\le M$
$n\ge N_a⟹|a_n-a|<\frac{1}{M}\frac{ϵ}{2}$
$n\ge max\{N_a,N_b\}$
$\begin{array}{rl}|a_n{b}_n-ab|& \le |b_n||a_n-a|+|a||b_n-b|\\ & \le M|a_n-a|+|b_n-b|\\ & \le M\left(\frac{ϵ}{2M}\right)+|a|\left(\frac{ϵ}{|a|2}\right)=ϵ\end{array}$

Proof iv
$(b_n)\to b\stackrel{(iii)}{⟹}\left(\frac{1}{b_n}\right)\to \frac{1}{b}$ if $b\ne 0$

Since $(b_n)\to b$ there is a $N_1$ so that $|b_n-b|<|b|/2$ for any $n>N_1$ so $b_n<|b|/2$

there is an $N_2$ where $|b_n-b|<\frac{ϵ|b{|}^2}{2}$ so take $max\{N_1,N_2\}$
$|\frac{1}{b_n}-\frac{1}{b}|=|b-b_n|\frac{1}{|b||b_n|}<\frac{ϵ|b{|}^2}{2}\frac{1}{|b{|}^{\frac{|b|}{2}}}=ϵ.$

Order Theorems

Assume $\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_n=b$

$\begin{array}{r}\text{(i)}& \mathrm{\forall }n,a_n\ge 0⟹a\ge 0\text{(ii)}& \mathrm{\forall }n,a_n\le b_n⟹a\le b\text{(iii)}& \mathrm{\forall }n,c\le b_n⟹c\le b{a}_n\le c⟹a\le c\end{array}$

Proofs

Proof i
For contradiction assume that $a<0$
$ϵ=|a|$ so for some $N$, $n\ge N$, $|a_n-a|<|a|$ so $a_N<0$ which is a contradiction

Proof ii
by the algebraic limit theorems $(b_n-a_n)\to b-a$. Since $b_n-a_n\ge 0$ using i we know $b-a\ge 0$

Proof iii
Take $a_n=c$ for all $n\in \mathbb{N}$ and apply ii

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Bounded Sequence

A sequence ($x_n$) is bounded if there exists a number $M>0$ so that $|x_n|\le M$ for all $n\in \mathbb{N}$

Thm 2.3.2 Convergent then bounded

Proof

Assume $(x_n)\to l$. So $\mathrm{\forall }ϵ>0\mathrm{\exists }N$ so that when $n>N⟹|x_n-l|<ϵ$

So let $ϵ=1$, for some $N$, all $n>N$ we have
$|x_n-l|<1⟹|x_n|<|l|+1$
Now the set of $x_i$ not in $V_{\epsilon =1}$ is finite. Let
$M=max\{|x_1|,|x_2|,\dots ,|x_{N-1}|,|l|+1\}$
therefore $|x_n|\le M$ for all $n\in \mathbb{N}$

Monotone Sequence

A sequence ($a_n$) is increasing if $a_n\le a_{n+1}$ for all $n\in \mathbb{N}$ and decreasing if $a_n\ge a_{n+1}$ and monotonic if either.

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Thm Squeeze

$y_m\le x_n\le z_n$ for all $n\in \mathbb{N}$
Assume $(y_n)\to L$ and $(z_n)\to L$. Then $(x_n)\to L$

Proof

$\mathrm{\exists }ϵ>0$ since $(y_n)\to L$ there exists an $N_y\in \mathbb{N}$ so that for all $n>N_y$, $L-ϵ<y_n<L_{ϵ}$

Since $(z_n)-L$ there exists an $N_z\in \mathbb{N}$ so that for all $n>N_y$ we have $L-ϵ<z_n<L+ϵ$

let $N=max\{N_y,N_z\}$
$n\ge N$
$L-ϵ<y_n\le x_n\le z_n<L+ϵ$

Thm Ratio Test

Let $x_n>0$ for all $n\in \mathbb{N}$ and $lim\frac{x_{n+1}}{x_n}=L$ exists. If $L<1$ then $(x_n)$ converges and $(x_n)\to 0$

Proof

Let $r\in \mathbb{R}$ so that $L<r<1$. Set $ϵ=r-L$, there exists $k\in \mathbb{N}$ so that $\frac{x_{n+1}}{x_n}<L+ϵ=r$ for all $n>k$

So, $0<x_{k+1}<r{x}_k$ so

$\begin{array}{rl}{x}_{k+1}& <r{x}_k\\ x_{k+2}& <r{x}_{k+1}<r^2{x}_k\\ x_{k+m}& <r^m{x}_k\\ \\ \text{(0<r<1)}& r^m& \to 0\end{array}$

So $(x_{k+m})\to 0$
$(x_n)\to 0$

Thm 2.4.2 Monotone Convergence

If $(a_n)$ is monotonic bounded then it converges if and only if it is bounded

Proof

Assume $a_n$ is increasing and bounded and set $s=sup\{a_n:n\in \mathbb{N}\}$,
$s-ϵ<a_N\le a_n\le s<s+ϵ$
by lemma 1.3.8

so $|a_n-s|<ϵ$

by thm 2.3.2 it is bounded.

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Subsequences

Let ($a_n$) be a sequence and let $n_1<n_2<\cdots <n_5$ be a sequence of increasing natural numbers. The then sequence
$(a_{n_1},a_{n_2},\dots )$
is a subsequence of ($a_n$) and is denoted by $a_{n_k}$ where $k\in \mathbb{N}$ indexes the subsequence.

Thm Convergence and subsequences (divergence)

The subsequences of a convergent sequence converges to the same limit as the original sequence.

Proof

Assume $(a_n)\to a$ and let$(a_{n_k})$ be a subsequence.

For any $\epsilon >0$ there exist an $N\in \mathbb{N}$ so that $|a_n-a|<\epsilon$ for any $n>N$. $n_k\ge k$ for all $k$, then $|a_{n_k}-a|<\epsilon$ for any $k\ge N$

Thm Bolzano Weierstrass

Every bounded Sequence contains a convergent sequence

Proofs

Using Nested Interval Property

Let ($a_n$) be bounded so that there exists an $M>0$ satisfying $|a_n|\le M$ for all $n\in \mathbb{N}$.
Take $[-M,0]$ and $[0,-M]$. At least one interval must have an infinite number of terms outside the epsilon neighbourhood.

Let $I_1$ be the the interval with infinite elements. Let $a_{n_1}$ be a term in the sequence $(a_n)$ such that $a_{n_1}\in I_1$.

Bisect $I_1$ and let $I_2$ have the infinite number of terms.

Prick $a_{n_2}$ from the original sequence with $n_2>n_1$ and $a_{n_2}\in I_2$ In general, construct the closed interval $I_k$ taking a half of $I_{k-1}$ containing an infinite number of terms of $(a_n)$ and then select $n_k>n_{k-1}>···>n_2>n_1$ so that $a_{n_k}\in I_k$
$I_1\supseteq I_2\supseteq I_3\supseteq \dots$
By the the nested interval property there exists at least one $x\in \mathbb{R}$ for every $I_k$. WTS that $(a_{n_k})\to x$

lets $\epsilon >0$ the length of $I_k$ is $M{\left(\frac{1}{2}\right)}^{k-1}$ which converges to 0. Choose $N$ so that $k\ge N$ the length of $I_k<\epsilon$. Since $x,a_{n_k}\in I_k$ it follows that $|a_{n_k}-x|<\epsilon$

Using Monotone Convergence Theorem

Lemma:
Every Sequence has a monotone subsequence
Let element $a_n$ of $(a_n)$ be called a castle if $a_n\ge a_k$ for all $k\ge n$

Case 1: There are infinitely many castles
$a_{n1},a_{n_2},\dots$ then $a_{n_1}\ge a_{n_2}\ge a_{n_3}\ge \dots$
and so $(a_{n_i})$ is a decreasing sequence

Case 2: Only finitely many castles in ($a_n$)
There $\mathrm{\exists }$ $a_{n_1}>a_k$ for every castle $a_k$ where any element past $a_{n_1}$ is not a castle.
so $\mathrm{\exists }$ $a_{n_2}>a_{n_1}$ and $a_{n_2}$ is not a castle. $a_{n_3}>a_{n_2}$ and so on, so strictly increasing.

Monotone + bounded = convergent

Limit Superior

Let $(x_n)$ be bounded
is the infimum of the set $V$ of $v\in \mathbb{R}$ such that $v<x_n$ for at most finitely many $n\in \mathbb{N}$
denoted $lim sup{a}_n$ or $\overline{lim}{x}_n$

Alternatively $lim sup{x}_n=infsup{x}_m$ for $n=1,2,3,\dots m=n,n+1,\dots$

Limit Inferior

Let $(x_n)$ be bounded
is the supremum of the $W$ of $w\in \mathbb{R}$ such that $x>a_n$ for at most finitely many $n\in \mathbb{N}$

Alternatively $lim inf{x}_n=supinf{x}_m$ for $n=1,2,3,\dots m=n,n+1,\dots$

Example:

$(a_n)=2-\frac{1}{2},6+\frac{1}{3},2-\frac{1}{4},6+\frac{1}{5},\dots$

inf$a_n=2-\frac{1}{2}$

throw away 2 terms
inf $a_n=2-\frac{1}{4}$

throw away 2$n$ terms
inf $a_n=2-\frac{1}{2n}$
take the collection and lim inf = 2

Now:
$\begin{array}{r}\{a_k:k\ge 2\}\\ sup:6+\frac{1}{3}\\ \{a_k:k\ge 4\}\\ sup:6+\frac{1}{5}\\ \\ \{a_k:k\ge 2n\}\\ sup:6+\frac{1}{2n+1}\\ \underset{n\in \mathbb{N}}{inf}\underset{m=n,n+1,\dots }{sup}=6\end{array}$

Thm Limit with limsup and liminf

Let $a_n$ be bounded then $lim(a_n)=a$ iff $lim sup(a_n)=liminf(a_n)=a$

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Cauchy:

if the terms of $(a_n)$ become "closer" as $n\to \mathrm{\infty }$ then $a_n$ should converge

A sequence is a Cauchy Sequence if $\mathrm{\forall }\epsilon >0$ there exists $k\in \mathbb{N}$ so that if n,$m\ge k$ then $|a_n-a_m|<\epsilon$

note m and n are independent can assume $m>n$

Thm Convergent iff Cauchy

Proof

$⟹$

Converges then Cauchy

If $(a_n)\to l$ then $a_n$ is Cauchy
Let $\epsilon >0$ and $K$ be such that $|a_{n-L}|<\frac{\epsilon }{2}$ for $n\ge K$

If $m,n\ge k$

$\begin{array}{rl}|a_n-a_m|=|a_n-L+L-a_m|& \le |a_n-L|+|a_n-L|\\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$

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Lemma Cauchy then Bounded

If $(a_n)$ is Cauchy then it is Bounded

From the definition of a Cauchy Sequence then there exists $k\in \mathbb{N}$ so that for $n\ge k$ that $|a_n-a_k|<\epsilon \stackrel{set}{=}1$

By triangle

$|a_n|\le |a_n-a_k|+|a_k|$
$|a_n|\le |a_k|+1$ for all $n\ge k$
Set $M=max\{|a_1|,|a_2|,\dots ,|a_k|,|a_k+1|\}$

Then $|a_n|\le M$ for all $n\in \mathbb{N}$

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Since $(a_n)$ is Cauchy it is bounded so by BW there exists and convergent subsequence $(a_{n_k})\to L$

Let $\epsilon >0$ since $a_n$ is Cauchy, there exists $M\in \mathbb{N}$ so that if $m,n\ge M$, then $|a_n-a_m|<\epsilon /2$

Also since $(a_{n_k})\to L$ there exists $n_l\ge M$ so that $|a_{n_l}-L|\le \frac{\epsilon }{2}$

So if

$\begin{array}{r}n\ge M\\ \\ |a_n-L|& =|a_n-a_{n_l}+a_{n_l}-L|\\ & \le |a_n-a_{n_l}|+|a_{n_l}-L|\\ & \le \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$

Example

$(1+(-1{)}^n)$

There exists some value ${\epsilon }_0>0$ so that for every $k\in \mathbb{N}$ there exists at least one value $n\ge k$ and $m\ge k$ so that
$|a_n-a_m|>{\epsilon }_0$
Have $a_{2n}=2$ and $a_{2n+1}=0$. Take ${\epsilon }_0=1$ then for any $k\in \mathbb{N}$ choose $n$ even and $m=n+1$ then

$|a_n-a_{n+1}|=2>1$

Thm Archimedean and Cauchy to Completeness

Suppose every Cauchy Sequence in $\mathbb{R}$ converges to an element in $\mathbb{R}$ and that the Archimedean property holds. Then $\mathbb{R}$ satisfies the completeness property.

completeness -> MCT -> BW -> Cauchy

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