* The following were written up as revision during my mid-semester break, with some corners cut and proofs added where needed. I’ve skipped the first couple of chapters because they’re boring and largely irrelevant.*

# Cauchy Sequences

Our motivation for defining and investigating Cauchy sequences is to have a notion akin to convergence which *isn’t* dependent on the limit of the sequence.

**Definition:** Let $(x_n)_{n=1}^{\infty}$ be a sequence in the metric space $(X,d)$. Then we say $x_n$ is Cauchy if $\forall \epsilon > 0$, $\exists N \in \N$ such that $m,n \geq N \implies d(x_m,x_n)< \epsilon$.
We also write in shorthand that $d(x_m,x_n) \rightarrow 0$ as $m,n \rightarrow \infty$.
An important thing to notice is that this is a *stronger statement than having the distance between consecutive terms going to zero*.
**Theorem:** In a metric space, every convergent sequence is Cauchy.
The proof is rather straightforward. Simply use the fact that it converges in conjuction with the triangle inequality to get the Cauchy criterion directly.
*Proof*
Let $x_n$ converge to $x \in X$. Given $\epsilon > 0$, then choose $N$ such that

$n \geq 0 \implies d(x_n,x) < \epsilon$. So then $\forall m ,n \geq
N,$ $$d(x_n,x_m) \leq d(x_m,x)+d(x_n,x) \quad \text{by triangle ineq}$$ $$ \leq
2\epsilon$$ Since $\epsilon$ was arbitrary, it follows $x_n$ is Cauchy.
**Theorem:** Cauchy Sequences are bounded
This one is not as straightforward or obvious. We use a similar method as for showing convergent sequences are bounded; consider two halves of the sequences, one of which is finite. Use the Cauchy criterion to bound the sequence past the cut off point, and define a maximum between that Cauchy bound and the maximum distance between the cut off point and all the points before (of which there are finitely many.)
*Proof*
Let $x_n$ be a Cauchy sequence in $(X,d)$. So for $\epsilon = 1$, using the fact
that $x_n$ is Cauchy, we can find an $N \in \N$ such that $\forall m,n > N,

d(x_n,x_m < 1).$ Now to show $x_n$ is bounded, we need to show that for some $a
\in X$, $\exists r
\in \R$ such that $d(x_n,a) \leq r$ for all $x_n \in (x_n)$. Now let $k =
\max\{d(x_N,x),...,d(x_N,x_{N-1})\}$. So now we have that for all $x_n$ with $n
\geq N$, $d(x_n,x_N)\leq 1$ by our choice of $N$. Moreover, all $x_n$ with $n <
N$ will have distance $d(x_n,x_N)\leq k$ by our choice of $k$. So taking $a =
x_N$ and $r = \max{k,1}$, we satisfy $d(x_n,a)\leq r$, as required $\square $.
Cauchy sequences in $\R^n$ have a nice property; they always converge!
**Theorem:** Sequences in $\R^n$ converge iff they are Cauchy.
*Proof*
We've already shown that convergent $\implies$ Cauchy, so now we will show the
reverse. So assume $(x_n) \subset \R^n$ is Cauchy. We will restrict attention to
just $\R$ for now, without loss of generality. Now, define $$y_n =
\inf\{x_n,x_{n+1},...\} \quad \forall n \in \N$$
Necessarily, $y_{n+1} \geq y_n$, since $y_{n+1}$ is the infimum of only a
subset corresponding to $y_n$. Now $(y_n)$ is bounded since $(x_n)$ is Cauchy,
and also bounded. Moreover, $y_n$ is monotone. So $y_n$ is a bounded, monotone
sequence in $\R$, and must converge, to say $a$. We will show that $x_n
\rightarrow a$ also.
So suppose $\epsilon >0$. Since $(x_n)$ is Cauchy, $\exists N(\epsilon)$ such

that $$x_n – \epsilon \leq x_m \leq x_l +\epsilon , \quad \forall l,m,n \geq

N $$ We claim that $$y_n-\epsilon\leq x_m \leq y_n+\epsilon$$ To see this, first

note that since $y_n \leq x_n$ for all $n$ by construction, it follows that for

all $m,n \geq N$ that $y_n – \epsilon \leq x_m$ ($y_n$ is a lower bound for

anything above $x_m$, so obviously so is $y_n – \epsilon$).

Now for the second side of that inequality, we focus on $x_m \leq x_l

+\epsilon$. Now by linearity of the infimum, $y_n + \epsilon = \inf\{x_l +

\epsilon \ | \ l \geq n\}$. Hence $y_n+\epsilon$ is greater than any other lower

bound for that set, and it follows $x_m \leq y_n + \epsilon$. Combining

inequalities, fixing $m$ and letting $n \rightarrow \infty$, $$a-\epsilon \leq

x_m \leq a + \epsilon$$ but since $\epsilon >0$ was arbitrary, it follows by

comparison that $x_m \rightarrow a$, as $m \rightarrow \infty$. This result generalises to $\R^n$, but we’ll omit the details here.

# Completeness

We say $(X,d)$ is complete if every Cauchy sequence in $X$ converges to a limit

in $X$. We’ll go through an example: $\R^n$ with the $p-metric$ is complete.

**Proof**

Recall that $$d_p(x,y) = \Big(\sum_{i=1}^n |x_i-y_i|^p \Big)^{\frac{1}{p}}$$

The proof essentially follows form the fact that $(\R,d)$ is complete, as we

will see.

So consider some sequence, $(x^{(k)})$ of vectors, which is Cauchy. We wish to

show that there is some vector $x \in \R^n$ such that $$\lim_{k \rightarrow

\infty}d_p(x^{(k)},x)=0$$

We start by claiming that each of the components themselves are Cauchy. That is,

$\forall i = 1,..,n$, $(x_i^{(k)})$ from $(x^{(k)})$ satisfies $$lim_{l,m

\rightarrow \infty}\left| x_i^{(l)}-x_i^{(m)} \right| = 0$$ To see this,

consider that for all $i = 1,…,n$, it will clearly be the case that $$\left|

x_i^{(l)}-x_{i}^{(m)} \right|\leq \sum_{i=1}^n\left| x_i^{(l)}-x_i^{(m)}

\right|^p$$ Hence it follows $$0 \leq \left| x_i^{(l)}-x_i^{(m)} \right| \leq \Big(\sum_{i=1}^n|x_i^{(l)}-x_i^{(m)} |^p\Big)^{\frac{1}{p}}$$

Now taking the limit as $l,m \rightarrow \infty,$ since $x^{(k)}$ is Cauchy,

$$0 \leq \left| x^{(l)}-x^{(m)} \right| \leq \lim_{l,m \rightarrow

\infty}d_p(x^{(l)},x^{(m)}) =0$$ and we confirm that the components are

Cauchy.

Now, since $(\R,d)$ is complete, we have $\forall i =1,…,n$, there exists $x_i

\in \R$ such that $\lim_{k\ ra \infty} \left| x_i^{(k)}-x_i \right|=0$. Now

define $x \in \R^n$ by precisely $$x = (x_1,x_2,…)$$so each component is the

limit of the component sequences. Then $$\lim_{k \ra

\infty}d_p(x^{(k)},x)=\lim_{k \ra \infty}\Big(\sum_{i=1}^n\left|

x_i^{(k)}-x_i \right|^p)^{\frac{1}{p}}$$ now taking the limit inside, which we

can do by limit laws (this is a finite sum), $$= \Big(\sum_{i=1}^n \big(

\lim_{k \ra \infty} \left| x_i^{(k)}-x_i \right|\big)^p

\Big)^{\frac{1}{p}}=0$$ since we defined $x_i^{(k)}\ra x_i$. Hence we get

$x^{(k)}\ra x$ as $k \ra \infty$, for an appropriately defined $x \in \R^n$.

Since $x^{(k)}$ was arbitrary, it follows that $(\R^n,d_p)$ is complete, as

desired. $\square$

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