# Subsequence of a Cauchy Sequence is Cauchy

• Sep 26th 2010, 08:23 PM
magus
Subsequence of a Cauchy Sequence is Cauchy
Show that a subsequence of a Cauchy sequence is a Cauchy sequence. The catch is that you can't use the fact that every Cauchy subsequence is convergent.

I really don't know where to begin without assuming what the problem says I can't assume.

Any direction would be appreciated.
• Sep 26th 2010, 09:30 PM
tonio
Quote:

Originally Posted by magus
Show that a subsequence of a Cauchy sequence is a Cauchy sequence. The catch is that you can't use the fact that every Cauchy subsequence is convergent.

I really don't know where to begin without assuming what the problem says I can't assume.

Any direction would be appreciated.

Let $\{x_n\}\,,\,\,\{x_{n_k}\}$ be the Cauchy seq. and one of its subsequences, and let $\epsilon>0$

1) There exists $M_\epsilon\in\mathbb{N}\,\,s.t.\,\,n,m>M_\epsilon\ Longrightarrow |x_n-x_m|<\epsilon$

2) By the very definition of "infinite subsequence", there exist $K\in\mathbb{N}\,\,s.t.\,\,\forall k>K\,,\,n_k>M_\epsilon$

Now end the proof.

Tonio
• Sep 27th 2010, 10:04 AM
magus
Thanks. Doesn't that nearly end the proof? Not that I'm mad at you or anything for taking me that far :P

I can't think of anything after that but then going right to $|x_{n_k}-x_m|\leq \epsilon$ which completes the proof.

Is this a valid step or do I have more work to do?
• Sep 27th 2010, 10:33 AM
Plato
The ‘trick’ comes from the nature of subsequences and the definition of Cauchy Sequences.

If $\left(x_{n_k}\right)$ is a subsequence of $\left(x_n\right)$ then $\left(n_k\right)\ge k$.

So if $k~\&~j~\ge N$ then $n_k~\&~n_j~\ge N$.
• Sep 27th 2010, 01:42 PM
magus
Two questions though. Doesn't that statement near complete the proof and also can $k~\&~j~\ge N$ be the same $N$ as in $n_k~\&~n_j~\ge N$?

It would seem to me that the subsub scripts would need to be greater then some other number. Maybe I'm being too nitpicky but just to be sure I'm getting what you two are saying I'm going to write it all out.

We start with the known that ${s_n}$ is convergent and therefore a Cauchy sequence. Then we know from that that there exists some $m,n \ge N$ such that $|s_n-s_m|\le \epsilon$.

We then choose some $j,k\le M$such that the subsequences of ${s_n}$ $s_{n_j}$ and $s_{m_k}$ such that $n_j, n_k\ge N$

We're then want to prove that $s_{n_j}-s_{m_k} \le \epsilon$

since ${s_n}$ is convergent it is bounded and therefore so is $s_{n_j}$ and $s_{m_k}$.

If we choose a large enough $N$ the sequences will be bounded into smaller and smaller intervals because they are bound by the sequences bounds. The process of tightening this window is by choosing larger values of N and therefore forcing larger values of M. This is exactly the Cauchy criterion therefore the subsequence of a Cauchy sequence is a Cauchy sequence.

Does that fly?
• Sep 27th 2010, 01:58 PM
Plato
Quote:

Originally Posted by magus
Two questions though. Doesn't that statement near complete the proof and also can $k~\&~j~\ge N$ be the same $N$ as in $n_k~\&~n_j~\ge N$? It would seem to me that the subsub scripts would need to be greater then some other number.

The point is: that for Cauchy Sequences it is true that for each $\varepsilon > 0$ there is one $N$ that works for $k\ge N~\&~j\ge N$ then $\left| {x_k - x_j } \right| < \varepsilon$.

I am puzzled by this question.

If the sequence $\left( {x_n } \right)$ converges then each subsequence $\left( {x_{n_k} } \right)$ converges to the same limit.

Moreover, any convergent sequence is a Cauchy Sequence.
So any subsequence of a Cauchy Sequence must be a Cauchy Sequence.
• Sep 27th 2010, 02:24 PM
magus
Quote:

Originally Posted by Plato
Moreover, any convergent sequence is a Cauchy Sequence.
So any subsequence of a Cauchy Sequence must be a Cauchy Sequence.

Right but for some reason they don't want me to use that information even though it seems the simplist way to prove it.
• Sep 27th 2010, 02:51 PM
Plato
Nevertheless, do you understand why for each $\varepsilon > 0$ one $N$ works?
• Sep 27th 2010, 04:50 PM
magus
Yes I do now. Thank you all for all your help.
• Sep 30th 2010, 01:29 AM
raimis
Quote:

Originally Posted by magus
Right but for some reason they don't want me to use that information even though it seems the simplist way to prove it.

Maybe because a Cauchy sequence is not necessarily convergent unless it is defined in a complete metric space.