Let $\{a_n\}$ and $\{b_n\}$ be Cauchy sequences. Let $s_n = d(a_n,b_n)$. Show that $\{s_n\}$converges

Question

I have been trying to solve this question in my book but I am having a hard time with so many variables like the $m$'s for $|s_n - s_m|$ where $s_n =d(a_n,b_n)$ and $s_m = d(a_m,b_m)$ and for $d(a_n,a_m)$ and $d(b_n,b_m)$. I belive I can remove the confusion by replacing it with a variable $K$, but ultimately I am not sure.

I would really appreciate any insight\advice you can offer.

Question

Let $\{a_n\}$ and $\{b_n\}$ be Cauchy sequences in a metric space $(X,d)$. For each $n \in \mathbb{N}$ let $s_n = d(a_n,b_n)$. Show that $\{s_n\}$converges.

Hint: Use $d(a_n,b_n) \leq d(a_n,a_m) + d(a_m,b_m) + d(b_m,b_n)$ to show that $|s_n - s_m|$ is small for large $m$ and $n$.

My proof attempt

Let $\epsilon > 0$, since $\frac{\epsilon}{4} > 0\ $there exists $N_1\in \mathbb{N}$ s.t for $m,n \geq N_1, \ d(a_n,a_m) < \frac{\epsilon}{4}$

Similarly, there exists $N_2$ s.t for $m,n \geq N_2 \in \mathbb{N}, \ d(b_n,b_m) < \frac{\epsilon}{4}$

Let $K= max\{N_1,N_2\}$

Now let $s_n =d(a_n,b_n) \leq d(a_n,a_K) + d(a_K,b_K) + d(b_K,b_n)$

Thus for $m,n \geq K$,

$|s_n - s_m| = |d(a_n,b_n) - d(a_m,b_m)|$

$ \ \ =|d(a_n,a_K) + d(a_K,b_K) + d(b_K,b_n)- ( d(a_m,a_K) + d(a_K,b_K) + d(b_K,b_m))|$

$ \ \ =|d(a_n,a_K) + d(b_K,b_n)- ( d(a_m,a_K) + d(b_K,b_m))|$

$ \ \ \leq d(a_n,a_K) + d(b_K,b_n) + d(a_m,a_K) + d(b_K,b_m)$

$ \ \ < \frac{\epsilon}{4} + \frac{\epsilon}{4}+ \frac{\epsilon}{4}+\frac{\epsilon}{4}$

$ \ \ = \epsilon$

This concludes my proof attempt

Answer 1

No. You have $|s_n - s_m| = |d(a_n,b_n) - d(a_m,b_m)|=$

$ \ \ =|d(a_n,a_K) + d(a_K,b_K) + d(b_K,b_n)- ( d(a_m,a_K) + d(a_K,b_K) + d(b_K,b_m))|$

but this assumes $d(a_n,b_n) =d(a_n,a_K) + d(a_K,b_K) + d(b_K,b_n)$ and $d(a_m,b_m)= d(a_m,a_K) + d(a_K,b_K) + d(b_K,b_m)$ but these are only "$\le$", not "$=$".

For example $4\le 3+2+1$ and $2\le 1+2+2$ does not imply $|4-2|\le |(3+2+1)-(1+2+2)|.$

The correct way to use the hint is $$d(a_n,b_n) \leq d(a_n,a_m) + d(a_m,b_m) + d(b_m,b_n)$$ $$d(a_m,b_m) \leq d(a_m,a_n) + d(a_n,b_n) + d(b_n,b_m)$$ The first of these implies $$X \leq d(a_n,a_m) + d(b_m,b_n)$$ and the second implies $$(-X) \leq d(a_m,a_n)+d(b_n,b_m)$$ where $X= d(a_n,b_n) - d(a_m,b_m). $

Therefore $$|X|\le d(a_n,a_m) + d(b_n,b_m).$$

Answer 2

$\begin{align} d(a_m, b_m) &\le d(a_m, a_n) +d(a_n, b_n) +d(b_n, b_m)\end{align}$

$\begin{align} d(a_m, b_m) -d(a_n, b_n) &\le d(a_m, a_n) +d(b_n, b_m) \end{align}$

Similarly, $\begin{align} d(a_n, b_n) -d(a_m, b_m) &\le d(a_n, a_m) +d(b_m, b_n) \end{align}$

Hence, $\begin{align} |d(a_m, b_m) -d(a_n, b_n) | &\le d(a_m, a_n) +d(b_n, b_m) \end{align}$

Since, $(a_n),( b_n) $ are cauchy, $\begin{align} |d(a_n, b_n) -d(a_m, b_m) |&\le d(a_n, a_m) +d(b_n, b_m) \to 0 \text{ as $m, n\to \infty$}\end{align} $

Hence, $\{d(a_n,b_n)\}$ is cauchy sequence in $\Bbb{R}$

Since, $(\Bbb{R}, d_{euclidean}) $ is complete, there is no difference between cauchy sequence and convergent sequence. i.e $\{d(a_n,b_n)\}$ is convergent.