Prove that the distance between 2 Cauchy sequences is convergent.

Question

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.

Answer 1

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.

Answer 2

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, $|d(p_n, q_n) - d(p_m, q_m)| < \varepsilon$.

Answer 3

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:

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