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### Cauchy Sequences

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Date: 09/11/97 at 02:49:34
From: Suzanne Mooney
Subject: Cauchy Sequences

Hi,

I'm trying to prove that every Cauchy sequence has a sequential limit
point. I don't have a real proof, but an idea, but don't know if it's
right, and if it is, how to formulate it.

My idea is that, for some N a positive integer, then for every m,n>N,
a Cauchy sequence has |psubn - psubm| < epsilon for some epsilon >0.
Then - epsilon < psubn - psubm < epsilon, so

psubm - epsilon < psubn < psubm + epsilon

which means that psubm - epsilon and psubm + epsilon are,
respectively, lower and upper bounds on {psubn|n>N}.

So (psubm - epsilon, psubm + epsilon) would be a segment containing
all psubn for every n>N.

Then, no matter how small this segment becomes, i.e. for bigger and
bigger values of N, it still will contain its midpoint, psubm and all
psubn for n>N, so wouldn't that make psubm a sequential limit point?
And then I'm thinking psubm is just an arbitrary member of the set
{psubn|n>N}, so would that mean that every member of that set would
then be a sequential limit point of the Cauchy sequence?  Does that
make sense?  It seems to me so now, but it's very late and I've been
messing with this thing too long, so forgive me if my reasoning (or
lack thereof) is somewhat fuzzy.

Suzanne
```

```
Date: 09/11/97 at 09:07:42
From: Doctor Jerry
Subject: Re: Cauchy Sequences

COMMENT: "Wouldn't that make p_m a sequential limit point?"
For which m?

I don't know the tools you have available. For example, do you know or
can you assume that every bounded infinite set has a limit point? Or
do you know or can you assume that each subset of R that is bounded
above has a least upper bound?

It's possible to show that the set {p_1,p_2,...}, where {p_m} is a
Cauchy sequence, is bounded. For example, take epsilon = 1; then all
p_m are within 1 of p_N. The entries p_1,...,p_{N-1} can't be too
far away since they are only finite in number. So the entries of a
Cauchy sequence form a bounded set. So, this set has a limit point,
call it x. You can show that x is the sequential limit of the Cauchy
sequence. Since it is a limit point, then, given any eps>0, there is
at least one entry nearby. But, being a Cauchy sequence, the entries
tend to hang around each other for large n. Etc.  Rough, but the gist
is here, I think.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 09/11/97 at 17:50:07
From: Suzanne Mooney
Subject: Re: Cauchy Sequences

Thanks again, as always. Yes, the gist is there so that even I see it
(yay!). And I think may be I can even 'unrough' enough. Thanks,
thanks, thanks.

Suzanne :-)
```

```
Date: 09/21/97 at 16:36:32
From: Suzanne Mooney
Subject: Re: Cauchy Sequences

Okay. Let's see if I came up with something that works. We show that x
is the limit point of the sequence, along the lines you said. Then we
pick an arbitrary segment (a,b) containing x, so it has some p_m in
it. Then we choose a subsegment of (a,b) that contains x but not p_m,
so this subsegment must also have some p_n in it.

Now there exists some N_1 s.t. m,n > N_1 ==> |p_m - p_n| < epsilon_1.
Now this epsilon_1 must be less than |b-a|, since both are contained
in (a,b). Then (a,b) contains all p_m s.t. m>N_1, and since (a,b) is
an arbitrarily selected segment containing x, x is the sequential
limit point of the Cauchy sequence.

Is this an okay proof or does it have some gaping hole in it that I
haven't seen? Be kind, it took me a long time to come up with this!
Thanks as always, Suzanne  :-)
P.S.  If it does have holes, just point them out to me, but please
don't tell me how to fix them... I'm trying to get to where I can do
this on my own!

There's just not enough thanks to give you for all the help you've
given me!  Maybe thanks raised to the millionth power?  (I have to
save some for the next time you help me!)
```

```
Date: 09/22/97 at 09:34:48
From: Doctor Jerry
Subject: Re: Cauchy Sequences

Hi Suzanne,

Since I'm retired, I can vacation anytime.  I was in northern
Minnesota last week listening to the loons and watching the leaves
turn and so didn't see your reply until today. I'll respond on the
chance that you are still interested in these questions.

I made a mistake in my last reply. It's possible that the set of
entries of the Cauchy sequence is finite (1,2,3,3,3,3,3,3,...). If
this is so, let x be a p_j that is repeated for infinitely many values
of j. If the set of entries is infinite, then let x be the limit
comments in capital letters, not telling all.

>We show that x is the limit point of the sequence, along the lines
>you said. Then we pick an arbitrary segment (a,b) containing x so it
>has some p_m in it. Then we choose a subsegment of (a,b) that
>contains x but not p_m.

IT COULD HAPPEN THAT X IS THE SAME AS THE p_m.  I DON'T THINK THAT
THIS IS A BIG PROBLEM.

>So this subsegment must also have some p_n in it. Now there exists
>some N_1 s.t. m,n > N_1 ==> |p_m - p_n| < epsilon_1. Now this
>epsilon_1 must be less than |b-a|, since both are contained in (a,b).

IT'S FUZZY HERE.  GIVEN epsilon_1, there is N_1 and so on; however,
you haven't said how big epsilon_1 is at this point and so can't say
that epsilon_1 <|b-a|.  Also, m was prescribed and we don't know how
it relates to N_1.

DON'T READ WHAT I'm going to say next if you want to work entirely on
your own (this is often a good idea). I'll give a small hint. Choose
epsilon so that the interval (x-epsilon,x+epsilon) is inside (a,b).
Next, choose N so that m,n>N implies |p_m-p_m|<epsilon/2. Now,
convince yourself that you can find a p_r within (x-epsilon/2,x+
epsilon/2) and r>N.

>Be kind, it took me a long time to come up with this!

MANY PEOPLE TAKE A LONG TIME TO GET THEIR MATHEMATICS STRAIGHT. WHAT
COUNTS IN THE LONG RUN IS WHETHER THE PROOF IS CORRECT OR NOT.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
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