Prove that the distance between 2 Cauchy sequences is convergent.

Here is the exact question:

Let $(S,d)$ be a metric space. Let $(p_n)$ and $(q_n)$ be two Cauchy sequences in $(S,d)$(note that these two sequences are not necessarily convergent since $(S,d)$ is not necessarily complete). Define a sequence $a_n = d(p_n, q_n)\in\mathbb{R}$. Prove that $(a_n)$ is a convergent sequence.

My attempt:

Since $p_n$ is Cauchy, there exists $N_1$ such that $m,n> N_1 \implies d(p_m, p_n)< \epsilon$.

In particular, $d(p_m, p_n)< (m-n)\epsilon$ for $m,n> N_1$

Since $q_n$ is Cauchy, $d(q_m, q_n) < (m-n)\epsilon$ for $m,n> N_2$

Thus for $N= \max(N_1, N_2)$, and $m,n> N, d(p_n, q_n) = d[(d(p_N, q_N) +...+ d(p_m, q_m)),0]$, which is less than $(m-n)\epsilon- (m-n)\epsilon= 0$, and thus $<\epsilon$

I don't think the final line of my argument is correct.

• I've typeset your question with MathJax. Please make sure I have not altered the meaning of your question in any way. Oct 11 '14 at 22:04

Let $\epsilon>0$, and let $N$ such that for every $m,n>N$, $d(p_m,p_n),d(q_m,q_n)<\epsilon/2$. It follows that $$|d(p_m,q_m)-d(p_n,q_n)|\leq|d(p_m,p_n)|+|d(q_m,q_n)|<\epsilon,$$ and the sequence is Cauchy. Since $\mathbb{R}$ is complete, the sequence converges.

• Oh, so since a_n is in R, if we show a_n is Cauchy, then it converges. Is that the right reasoning? Oct 11 '14 at 22:15
• yeah, this is it. Oct 11 '14 at 22:31
• Could you explain how you got the first inequality in the second line? Oct 11 '14 at 22:50
• @Zslice: The first inequality on the second line is the reverse triangle inequality. Recall that the triangle inequality states that $|x+y|\leq|x|+|y|$. You can obtain the reverse triangle inequality by letting $z=x+y$. Then $x=z-y$. Substituting this into the triangle inequality, we get $|z-y+y|\leq |z-y|+|y|\implies |z|\leq |z-y| +|y|\implies |z|-|y|\leq |z-y|$. Oct 11 '14 at 23:02
• @Zslice $|d(p_m,q_m)-d(p_n,q_n)|\leq|d(p_m,q_m)-d(p_m,q_n)|+|d(p_m,q_n)-d(p_n,q_n)|\leq d(q_m,q_n)+d(p_m,p_n)$ Oct 11 '14 at 23:15

The last line of the argument indeed looks very suspect to me - you do seem to have nested occurances of the metric, which should not happen.

A simple proof (that may be using more facts than you are comfortable with) would be the following: Move from $(S, d)$ to its completion. Now the two sequences are convergent, and the metric is a sequentially continuous function into $\mathbb{R}$. Thus, it preserves convergence, hence your sequence converges to the distance of the limit points of the Cauchy sequences in their completion.

If you want a more "calculatory" argument, you are indeed of to a right start. You want to show that the sequence $d(p_n,q_n)$ is Cauchy in $\mathbb{R}$. Start with $\varepsilon > 0$. As $(p_n)$ is Cauchy, there is some $N_1$ such that $d(p_n,p_m) < 0.5\varepsilon$ for $n, m > N_1$. Likewise there is an $N_2$ for $q_n$ and $0.5\varepsilon$.

Now whenever $n,m > \max \{N_1, N_1\}$, then by the triangle inequality, $|d(p_n, q_n) - d(p_m, q_m)| < \varepsilon$.

• In my opinion, the major problem in this answer is that in order to prove the existence of the completion space, one needs first to confront the posted question... Oct 11 '14 at 22:00
• Indeed I was adding a more direct proof as you were writing your comment.
– Arno
Oct 11 '14 at 22:05

Let $M=\inf_m{a_m}$. Than we will prove that $\lim_n{a_n}=M$:

Since $p_n$ and $q_n$ are Chaucy, so is $a_n=d(p_n,q_n)$. Than we have:

$$|a_n-M|=|a_n-\inf_m{a_m}|<|a_n-(\epsilon_1-a_m)|<\epsilon_1+|a_n-a_m|<\epsilon_1+\epsilon_2=\epsilon.$$