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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 :)

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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!

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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 $\mathbb{Q}$.

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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 $S$.

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