[proofplan] The proof is a direct comparison between the Cauchy condition and the two possible values of the [discrete metric](/page/Discrete%20Metric). In the forward direction, applying the Cauchy condition at radius less than $1$ forces all sufficiently late terms to have distance $0$, hence to be equal. In the reverse direction, an eventually constant sequence has zero pairwise distance after some index, which immediately satisfies the Cauchy condition for every positive radius. [/proofplan] [step:Use the Cauchy condition at radius less than one to force equality of late terms] Assume that the sequence $x:\mathbb{N}\to X$ is Cauchy in $(X,d)$. Define the positive real number $\varepsilon_0$ by \begin{align*} \varepsilon_0=\frac{1}{2}. \end{align*} By the Cauchy property applied to $\varepsilon_0$, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m\ge N$ and $n\ge N$, \begin{align*} d(x_m,x_n)<\varepsilon_0. \end{align*} Since $d$ is the discrete metric, if $x_m\ne x_n$ then $d(x_m,x_n)=1$, which contradicts \begin{align*} d(x_m,x_n)<\frac{1}{2}. \end{align*} Therefore $x_m=x_n$ for all $m,n\ge N$. Taking $a:=x_N\in X$, we obtain $x_n=a$ for every $n\ge N$, so the sequence is eventually constant. [guided] Assume that $x:\mathbb{N}\to X$ is Cauchy in the [metric space](/page/Metric%20Space) $(X,d)$. The Cauchy condition says that for every real number $\varepsilon>0$, there is an index after which all pairwise distances are smaller than $\varepsilon$. To exploit the discrete metric, we choose a positive radius strictly smaller than the nonzero distance between two distinct points. Define \begin{align*} \varepsilon_0=\frac{1}{2}. \end{align*} Since $x$ is Cauchy, there exists $N\in\mathbb{N}$ such that whenever $m,n\in\mathbb{N}$ satisfy $m\ge N$ and $n\ge N$, one has \begin{align*} d(x_m,x_n)<\varepsilon_0. \end{align*} Now use the special form of the discrete metric. For any two points $u,v\in X$, the value $d(u,v)$ is either $0$ if $u=v$, or $1$ if $u\ne v$. Thus for indices $m,n\ge N$, the inequality \begin{align*} d(x_m,x_n)<\frac{1}{2} \end{align*} rules out the possibility $x_m\ne x_n$, because that possibility would give $d(x_m,x_n)=1$. Hence $x_m=x_n$ for all $m,n\ge N$. To translate this pairwise equality into eventual constancy, fix the late term \begin{align*} a:=x_N. \end{align*} Putting $m=N$ in the equality just proved gives $x_n=x_N=a$ for every $n\ge N$. Therefore there exist $N\in\mathbb{N}$ and $a\in X$ such that $x_n=a$ for all $n\ge N$, which is exactly eventual constancy. [/guided] [/step] [step:Show that eventual constancy gives the Cauchy condition] Conversely, assume that there exist $N\in\mathbb{N}$ and $a\in X$ such that $x_n=a$ for every $n\ge N$. Let $\varepsilon>0$ be arbitrary. If $m,n\in\mathbb{N}$ satisfy $m\ge N$ and $n\ge N$, then $x_m=a$ and $x_n=a$, so \begin{align*} d(x_m,x_n)=d(a,a)=0. \end{align*} Since $0<\varepsilon$, this gives $d(x_m,x_n)<\varepsilon$. Hence for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that all terms after index $N$ are pairwise within $\varepsilon$, so $x$ is Cauchy in $(X,d)$. [/step] [step:Conclude the equivalence] The first step proves that every [Cauchy sequence](/page/Cauchy%20Sequence) in the discrete metric is eventually constant. The second step proves that every eventually constant sequence is Cauchy. Therefore a sequence $x:\mathbb{N}\to X$ is Cauchy in $(X,d)$ if and only if it is eventually constant. [/step]