If two sequences are Cauchy, then d(sequence_1, sequence_2) is cauchy in R The question says this:
If $(X,d)$ is a metric space and $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences, prove that $\{d(x_n,y_n)\}$ is a Cauchy sequence in $R$.
I see that I would have to show that $d_1(d(x_n,y_n), d(x_m,y_m)) < \epsilon $ for all $m,n \ge N $ for some metric on R.  If I assume that $d_1$ is the Euclidean metric then the reverse triangle inequality leads to the sequence being Cauchy, but how would I show this for an arbitrary metric $d_1$ in R?
 A: You need to do this for the standard metric on the reals. If not, it should have been specified, as Cauchy-ness depends on the metric used, and it can fail for other choices of the metric on the reals.
So you need to show that for every $\varepsilon > 0$ there exists some $N$ such that for all $n,m \ge N$ we have 
$$\left| d(x_n, y_n) - d(x_m,y_m) \right| < \varepsilon\text{.}$$
So find $N_1$ for $(x_n)$ and $\frac{\varepsilon}{2}$, $N_2$ for $(y_n)$ and $\frac{\varepsilon}{2}$. Then 
$$d(x_n, y_n) \le d(x_n, x_m) + d(x_m,y_m) + d(y_m, x_m)$$
so for $n,m \ge \max(N_1,N_2)$ we have 
$$d(x_n, y_n) < \frac{\varepsilon}{2} + d(x_m,y_m) + \frac{\varepsilon}{2}$$
from which we have 
$$ d(x_n, y_n) - d(x_m,y_m) < \varepsilon$$
and then we interchange $n,m$ in the above argument to get the other order on the left, and take the max to get the absolute value.
A: The intended metric in $\mathbb{R}$ is probably the usual metric here, namely  the one defined by the absolute value of the difference.  The proof is particularly simple if you are allowed to use a metric completion $\mathbf{X}$ of $X$. Passing to the limits $x=\lim_{n\to\infty}x_n$ and $y=\lim_{n\to\infty}y_n$ you can easily show from the triangle inequality that $d(x_n,y_n)$ necessarily converges to $d(x,y)$.
The existence of metric completion is immediate by passing to the ultrapower and does not require delicate analysis of Cauchy sequences.
