# If $K$ is an unbounded subset of a metric space $(X, d)$, then $K$ contains a non-cauchy sequence

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.

You should exhibit some $$\varepsilon>0$$ such that, given any $$N$$, there are $$m,n>N$$ with $$d(a_m,a_n)\ge\varepsilon$$.
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 $$2N+2>N$$.
• 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? Jul 17, 2022 at 22:00
• @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? Jul 17, 2022 at 22:16