I am trying to prove that if $K$ is a (non-empty) unbounded subset of a metric space $(X, d)$, then $K$ contains a non-cauchy sequence. Is the following correct?

If $K$ is unbounded, then, for any $M>0$, there exists $x,y\in K$ s.t. $d(x,y)>M$. As such, we may construct the following sequence, in $K$. Let $a_{1}, a_{2}$ be chosen s.t. $d(a_{1}, a_{2})>1$. Generally, for each $n \in \{1,3,5,...\}$, let $a_{n}, a_{n+1}$ be chosen such that $d(a_{n}, a_{n+1})>n$. This completes the construction of $\{a_{n}\}$, which we now prove does not satisfy the Cauchy Criterion.

Consider some arbitrary $\epsilon>0$, and $N \in\mathbb{N}$. Choose $j\in \{1, 3, 5,...\}$ such that $j>\max\{N, \epsilon\}$. Then, set $n=j, m=j+1$. We have that $d(a_{n}, a_{m})=n>\epsilon$, where $m,n>N$. Hence, $\{a_{n}\}_{n\in\mathbb{N}}$ is not Cauchy.


1 Answer 1


You should exhibit some $\varepsilon>0$ such that, given any $N$, there are $m,n>N$ with $d(a_m,a_n)\ge\varepsilon$.

Your idea is good, but requires a bit of extra work.

You can take $\varepsilon=1/2$. Given $N$, by construction $d(a_{2N+1},a_{2N+2})>2N+1>1/2$, where $2N+1>N$ and $2N+2>N$.

  • $\begingroup$ Regarding the necessity of exhibiting a particular $\epsilon>0$: does not, what i've produced, imply the negation of the Cauchy Criterion—and so is not, what i've done, in itself sufficient to establish that $\{a_{n}\}$ fails to satisfy the Cauchy Criterion? $\endgroup$
    – Charles
    Jul 17, 2022 at 22:00
  • $\begingroup$ @Charles The Cauchy criterion is: for every $\varepsilon>0$ there exists $N$ such that, for all $m,n>N$, it holds $d(a_m,a_n)<\varepsilon$. Now work out the negation. Of course there may be other methods, but why complicating things? $\endgroup$
    – egreg
    Jul 17, 2022 at 22:16
  • $\begingroup$ I am trying to use my above proof to prove that the constructed sequence cannot have a convergent subsequence. Is what I have proven at all useful for this purpose? $\endgroup$
    – Charles
    Jul 18, 2022 at 2:17

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