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.
Best Answer
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.