A sequence $\{c_n\}$ in a metric space $(X,d)$ $(euclidean$ $distance)$ satisfies the following condition:

There exists a positive real number $R\in \mathbb {R_{+}}$ such that for all $n\in \mathbb {N}:$


Show that $\{c_n\}$ is Cauchy.

So my approach is using the contraposition:

If $\{c_n\}$ is not Cauchy $\Longrightarrow $ there is no such R which satisfies the condition above.

Since $\{c_n\}$ is not Cauchy $\Longrightarrow$ $\exists$ $\epsilon>0$ such that $\forall$ $N$$\in\mathbb {N}$ : $d(c_n,c_m)\geq\epsilon,$ $n,m\ge N$.

$\Longrightarrow$ $\forall n\in \mathbb {N}$$:d(c_n,c_{n+1})\geq\epsilon>0$.

Now let $\{s_n\}=\{\sum\limits_{i=1}^{n}d(c_i,c_{i+1})\}$. Since all $d(c_i,c_{i+1})$ are bounded below by this $\epsilon>0$, the partial sums of $\{s_n\}$ do not converge to $0$ and thus $\{s_n\}$ diverges. But since for all $i\in\mathbb {N}$ $d(c_i,c_{i+1})\geq\epsilon>0$ and $\{s_n\}$ diverges, it implies that $\{s_n\}$ is not bounded above.

Thus, $\lim_{n \to \infty}\{s_n\}=\infty$. Therefore, there exists no such positive real number $R\in \mathbb {R_{+}}$ such that for all $n\in \mathbb {N}:$


Thank you in advance.


Careful when negating the statement! The correct negation is:

$\exists \epsilon>0:\forall$ $N$$\in\mathbb {N} \;\exists m,n\geq N$ : $d(c_n,c_m)\geq\epsilon$

You can easily fix your argument though. Hint: consider a subsequence of $c_n$.

  • $\begingroup$ Since there are only some $m,n\geq N$ which satisfy the condition, I create a subsequence$\{c_{n_i}\}$ consisting of all those $m,n$ and then $\{s_{n_i}\}=\{\sum\limits_{k=1}^{i}d(c_{n_k},c_{n_{k+1}})\}$. Since this diverges,$\{s_{n}\}$ also diverges. Is this what you mean? $\endgroup$ – user114193 Dec 9 '13 at 23:37
  • $\begingroup$ Yes, that works. $\endgroup$ – Listing Dec 10 '13 at 8:29

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