# Thread: Cauchy sequence convergence in a general metric space

1. ## Cauchy sequence convergence in a general metric space

Hello!

My question is one from Baby Rudin... Chapter 3, Question 23.

The question is, given two Cauchy sequences $\displaystyle \{p_n\}$ and $\displaystyle \{q_n\}$, in a metric space X (with a metric "d"), we want to show that
$\displaystyle \{d(p_n, q_n)\}$ converges.

What I've got so far:

We are given $\displaystyle \{p_n\}$ and $\displaystyle \{q_n\}$ are Cauchy, so, by definition, given $\displaystyle \epsilon > 0$, $\displaystyle \exists N$ such that $\displaystyle d(p_n, p_m) < \epsilon$ and $\displaystyle d(q_n, q_m) < \epsilon$ whenever $\displaystyle n, m > N$.

So, by the triangle inequality we have

$\displaystyle d(p_n, q_n) \leq d(p_n, p_m) + d(p_m, q_m) + d(q_m, q_n)$

then it follows thats:

$\displaystyle |d(p_n, q_n) - d(p_m, q_m)| \leq d(p_n, p_m) + d(q_m, q_n)$

but, $\displaystyle d(p_n, p_m) < \epsilon$
and $\displaystyle d(q_m, q_n) < \epsilon$

so

$\displaystyle d(p_n, p_m) + d(q_m, q_n) < 2\epsilon$

so $\displaystyle |d(p_n, q_n) - d(p_m, q_m)| < 2\epsilon$ when $\displaystyle n > N$

that fulfils the definition of Cauchy, so
$\displaystyle \{d(p_n, q_n)\}$ is Cauchy.

but... we don't neccecarily know it converges, right?

because X is a general metric space... we don't know if Cauchy sequences converge in a general metric space... they only converge in complete metric spaces.

Any help much appriciated!! Thank you.

2. Originally Posted by matt.qmar

because X is a general metric space... we don't know if Cauchy sequences converge in a general metric space... they only converge in complete metric spaces.

Any help much appriciated!! Thank you.
Great job on seeing 'the trick' to show it's Cauchy! One thing...what is the codomain of $\displaystyle d$? Once you've answered this your question will be answered.

3. The codomain of d? well, all metrics are functions which map to the non-negative reals...

Yikes! Seems like you're giving me a hint.... but I still can't see it!

4. Originally Posted by matt.qmar
The codomain of d? well, all metrics are functions which map to the non-negative reals...

Yikes! Seems like you're giving me a hint.... but I still can't see it!
So, in fact, $\displaystyle \left\{d(x_n,y_n)\right\}_{n\in\mathbb{N}}$ is a sequence of real numbers, right?

5. aha!

the medium is the message... or some equivalent metaphorical saying.

the sequence is made up of reals!

thanks!!