[proofplan] We prove completeness directly from the definition. Given a [Cauchy sequence](/page/Cauchy%20Sequence) in the [discrete metric](/page/Discrete%20Metric), we apply the Cauchy condition with the radius \begin{align*} \varepsilon = \frac{1}{2} \end{align*} Since distinct points in the discrete metric have distance exactly $1$, distance less than this displayed tolerance forces equality, so the sequence is eventually constant. An eventually constant sequence converges to its eventual value, and therefore every Cauchy sequence in $(X,d)$ converges. [/proofplan] [step:Use the Cauchy condition to force eventual equality] Let $\mathbb{N}:=\{1,2,3,\dots\}$, and let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $(X,d)$. We use the tolerance \begin{align*} \varepsilon = \frac{1}{2}. \end{align*} By the Cauchy property with this value of $\varepsilon$, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m,n\geq N$, \begin{align*} d(x_m,x_n)<\frac{1}{2}. \end{align*} Fix $n\geq N$. Since $d$ is the discrete metric, either $x_n=x_N$ and $d(x_n,x_N)=0$, or $x_n\ne x_N$ and $d(x_n,x_N)=1$. The inequality \begin{align*} d(x_n,x_N)<\frac{1}{2} \end{align*} excludes the second case. Hence $x_n=x_N$ for every $n\geq N$. [guided] Let $\mathbb{N}:=\{1,2,3,\dots\}$, and let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $(X,d)$. To prove completeness, we must show that this arbitrary Cauchy sequence converges to some point of $X$. The useful feature of the discrete metric is that there is a gap between equal and unequal points: equal points have distance $0$, while distinct points have distance $1$. We therefore choose a Cauchy tolerance strictly between these two values: \begin{align*} \varepsilon = \frac{1}{2}. \end{align*} By the definition of a Cauchy sequence, there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m,n\geq N$, \begin{align*} d(x_m,x_n)<\frac{1}{2}. \end{align*} Now fix any $n\geq N$ and compare $x_n$ with the single reference term $x_N$. Since $n\geq N$ and $N\geq N$, the Cauchy estimate gives \begin{align*} d(x_n,x_N)<\frac{1}{2}. \end{align*} Because $d$ is the discrete metric, there are only two possibilities: if $x_n=x_N$, then $d(x_n,x_N)=0$; if $x_n\ne x_N$, then $d(x_n,x_N)=1$. The inequality above rules out $d(x_n,x_N)=1$, so the only possible case is $x_n=x_N$. Thus every term after index $N$ is equal to $x_N$. The sequence is therefore eventually constant. [/guided] [/step] [step:Conclude convergence to the eventual value] Define $x_\ast:=x_N\in X$. We prove that $(x_n)_{n\in\mathbb{N}}$ converges to $x_\ast$ in the metric $d$. Let $\varepsilon>0$ be arbitrary. For every $n\geq N$, the eventual equality from the previous step gives $x_n=x_\ast$, hence \begin{align*} d(x_n,x_\ast)=0<\varepsilon. \end{align*} Therefore $x_n\to x_\ast$ in $(X,d)$. [/step] [step:Apply the definition of completeness] We started with an arbitrary Cauchy sequence $(x_n)_{n\in\mathbb{N}}$ in $(X,d)$ and proved that it converges to a point $x_\ast\in X$. Hence every Cauchy sequence in $(X,d)$ converges in $X$. By the definition of a [complete metric space](/page/Complete%20Metric%20Space), $(X,d)$ is complete. [/step]