* 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 be a sequence in the metric space . Then we say is Cauchy if , such that .

We also write in shorthand that as .

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 converge to . Given , then choose such that

. So then

Since was arbitrary, it follows 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 be a Cauchy sequence in . So for , using the fact

that is Cauchy, we can find an such that Now to show is bounded, we need to show that for some , such that for all . Now let . So now we have that for all with , by our choice of . Moreover, all with will have distance by our choice of . So taking and , we satisfy , as required .

Cauchy sequences in have a nice property; they always converge!

**Theorem:** Sequences in converge iff they are Cauchy.

*Proof*

We've already shown that convergent Cauchy, so now we will show the

reverse. So assume is Cauchy. We will restrict attention to

just for now, without loss of generality. Now, define

Necessarily, , since is the infimum of only a

subset corresponding to . Now is bounded since is Cauchy,

and also bounded. Moreover, is monotone. So is a bounded, monotone

sequence in , and must converge, to say . We will show that also.

So suppose . Since is Cauchy, such

that

We claim that

To see this, first

note that since for all by construction, it follows that for

all that ( is a lower bound for

anything above , so obviously so is ).

Now for the second side of that inequality, we focus on . Now by linearity of the infimum, . Hence is greater than any other lower

bound for that set, and it follows . Combining

inequalities, fixing and letting ,

but since was arbitrary, it follows by

comparison that , as . This result generalises to , but we'll omit the details here.

# Completeness

We say is complete if every Cauchy sequence in converges to a limit

in . We'll go through an example: with the is complete.

**Proof**

Recall that

The proof essentially follows form the fact that is complete, as we

will see.

So consider some sequence, of vectors, which is Cauchy. We wish to

show that there is some vector such that

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

, from satisfies

To see this,

consider that for all , it will clearly be the case that

Hence it follows

Now taking the limit as since is Cauchy,

and we confirm that the components are

Cauchy.

Now, since is complete, we have , there exists such that . Now

define by precisely

so each component is the

limit of the component sequences. Then

now taking the limit inside, which we

can do by limit laws (this is a finite sum),

since we defined . Hence we get

as , for an appropriately defined .

Since was arbitrary, it follows that is complete, as

desired.

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