# Cauchy sequence in a discrete space

how to prove that any Cauchy sequence in a discrete space is stationary

Let $$(x_n)$$ be a cauchy sequence then $$\forall \varepsilon>0, \exists n_0\in \mathbb{N},\forall p,q \geq n_0\Rightarrow \begin{cases} 1\leq \varepsilon, x_p\neq x_q\\ 0\leq \varepsilon,x_p=x_q\end{cases}$$

how to continue?

thank you

• Terminology: The term "discrete space" means every subset is open, but does not automatically mean that it has a metric $d$ satisfying $d(x,y)=1$ whenever $x\ne y.$ That $d$ is called "the discrete metric" but other metrics may generate the discrete topology. For example $\{1/n: n\in \Bbb N\}$ is a discrete subspace of $\Bbb R$ and its topology is generated by the usual metric $d(x,y)=|x-y|.$ – DanielWainfleet Apr 16 at 0:01

The function $$d:\mathbb N\times \mathbb N\rightarrow[0,+\infty)$$ defined as $$d(m,n)=\left|\displaystyle\frac{1}{m}-\frac{1}{n}\right|$$ is a metric, which induces discrete topology on $$\mathbb N$$. The sequence $$(n,n\in\mathbb N)$$ is a Cauchy sequence in the space $$(\mathbb N,d)$$, and this sequence is not stationary.
If $$(x_n)$$ is Cauchy, then, for any $$\varepsilon>0,\exists k\in\mathbb{N}$$ such that $$d(x_n,x_m)<\varepsilon$$ for all $$n,m\geq k$$. Let $$\varepsilon\leq1$$, then if $$d$$ is the discrete metric, $$d(x,y)<1$$ if and only if $$x=y$$ since $$d=\begin{cases} 0&\mathrm{if}\,x=y\\ 1&\mathrm{if}\,x\neq y \end{cases}$$ so if $$d(x,y)<1$$ then $$d(x,y)=0$$. $$\exists k\in\mathbb{N}$$ such that $$d(x_n,x_m)<1$$ for all $$n,m\geq k$$, so $$d(x_n,x_m)=0$$ for all $$n,m\geq k$$, so $$x_n=x_m$$ for all $$n,m\geq k$$, i.e $$(x_n)$$ is stationary for all $$n\geq k$$.
• and if $\varepsilon>1$??? – Poline Sandra Apr 15 at 17:50
• We should take $\varepsilon \leq 1$ for this purpose. – Dbchatto67 Apr 15 at 17:52
• You're given that there is such a $k$ for all $\varepsilon$. For terms sufficiently early in the sequence, $\varepsilon\geq 1$, so there can be terms early in a Cauchy sequence in a discrete space where it is not stationary, but for terms sufficiently far through the sequence, where $k$ is large enough, $\varepsilon<1$, so the sequence must be stationary after a certain point. – windingnumberone Apr 15 at 17:53
• When $\varepsilon\geq 1$ the sequence is not necessarily stationary but you know from the definition of a Cauchy sequence that $\varepsilon<1$ for all terms sufficiently far through the sequence. – windingnumberone Apr 15 at 18:13