Cauchy sequences in metric spaces Let $(X,d)$ be a metric space and let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $X$, i.e. $d(x_n,x_m)$ goes to $0$ when $n,m\rightarrow\infty$. The sequence does not necessarily have a limit in $X$, however.
I'm wondering if for fixed $k$, the sequence $d(x_k,x_l)$ has a limit in $\mathbb{R}$ when $l\rightarrow\infty$? I know that $\lim\sup_{l\to\infty}d(x_k,x_l)$ exists (this is always true for Cauchy sequences), but what about the limit?
Thank you very much in advance :)
 A: Given any point $w\in X$ the function "distance from $w$"
$$
d(w,\cdot):X\longrightarrow\Bbb R
$$
is continuous and transforms Cauchy sequences (in $X$) into Cauchy sequences (in $\Bbb R$). But $\Bbb R$ is complete!
A: Yes, the limit in question always exists.  
One conceptual way to see it is to observe this first for convergent sequences, and then apply that case to the metric completion $\overline{X}$ of $X$.  
On the other hand, this can certainly be done by a straightforward $\epsilon$-$n$ argument: the basic idea is that for any fixed $\epsilon > 0$ there is $L$ such that for $l_1,l_2 \geq L$, $d(x_{l_1},x_{l_2}) < \epsilon$.  Combining this with the triangle inequality shows that $|d(x_k,x_{l_1}) - d(x_k,x_{l_2})| < \epsilon$.  
Added: The sketch in the last paragraph above is showing that, for fixed $k$, the sequence $\{d(x_k,x_l)\}_{l=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}$.  So it is closely related to Andrea Mori's answer.  Note also that the completeness of $\mathbb{R}$ is needed here: e.g. even if for all $k,l \in \mathbb{Z}^+$ we have $d(x_k,x_l) \in \mathbb{Q}$, $\lim_{l \rightarrow \infty} d(x_k,x_l)$ need not be in $\mathbb{Q}$.
A: First, let us fix some $l\in\mathbb N$.
Let us denote $S:=\limsup\limits_{k\to\infty} d(x_k,x_l)$.
Suppose we are given some $\varepsilon>0$.
Since the sequence is Cauchy, there exists $k_0$ such that
$$p,q\ge k_0 \Rightarrow d(x_p,x_q) \le\frac\varepsilon2.$$
From the definition of limit superior we get that there exists $q\ge k_0$ such that
$$d(x_l,x_q) \ge S-\frac\varepsilon2.$$
Using triangle inequality we get
$$d(x_l,x_p) \ge d(x_l,x_q)-d(x_q,x_p) \ge S-\varepsilon$$
for every $p\ge k_0$.
This shows that $\liminf\limits_{k\to\infty} d(x_k,x_l) \ge S-\varepsilon$. 
Since $\varepsilon>0$ can be chosen arbitrary, we get
$$\liminf\limits_{k\to\infty} d(x_k,x_l) \ge S.$$
Hence both limit inferior and limit superior are equal to the same value $S$.
