Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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!

share|improve this answer

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 $\mathbb{Q}$.

share|improve this answer

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 $S$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.