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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

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  • $\begingroup$ 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 $d(x,y)=|x-y|.$ $\endgroup$ Apr 16, 2019 at 0:01

2 Answers 2

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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.

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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 $n\geq k$.

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  • $\begingroup$ and if $\varepsilon>1$??? $\endgroup$
    – user523857
    Apr 15, 2019 at 17:50
  • $\begingroup$ We should take $\varepsilon \leq 1$ for this purpose. $\endgroup$
    – little o
    Apr 15, 2019 at 17:52
  • $\begingroup$ 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. $\endgroup$
    – hello
    Apr 15, 2019 at 17:53
  • $\begingroup$ 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. $\endgroup$
    – hello
    Apr 15, 2019 at 18:13
  • $\begingroup$ then stationary sequences are not the only cauchy sequences in the discrete space? $\endgroup$
    – user523857
    Apr 15, 2019 at 18:18

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