[proofplan] The proof is to show that every singleton is open in the metric topology. For each $x\in X$, the open ball of radius $1/2$ around $x$ contains $x$ and no other point, because every distinct point has distance $1$ from $x$. Hence every subset of $X$ is a union of open singletons, so every subset is $\tau_d$-open. Since $\tau_d$ is a topology on $X$, it consists only of subsets of $X$, and therefore equals $\mathcal{P}(X)$. [/proofplan] [step:Compute the small open balls in the discrete metric] For $x\in X$ and $r>0$, let $B_d(x,r)$ denote the open ball in the [metric space](/page/Metric%20Space) $(X,d)$: \begin{align*} B_d(x,r)=\{y\in X:d(x,y)<r\}. \end{align*} Fix $x\in X$. We claim that \begin{align*} B_d(x,1/2)=\{x\}. \end{align*} Indeed, $d(x,x)=0<1/2$, so $x\in B_d(x,1/2)$. Conversely, if $y\in B_d(x,1/2)$ and $y\neq x$, then the definition of the [discrete metric](/page/Discrete%20Metric) gives $d(x,y)=1$, contradicting $d(x,y)<1/2$. Therefore $y=x$, and hence $B_d(x,1/2)=\{x\}$. [guided] We want to prove that the metric topology is discrete, so the key point is to isolate individual points by metric balls. Let $x\in X$ be arbitrary. For every real number $r>0$, the open ball centered at $x$ with radius $r$ is the subset of $X$ defined by \begin{align*} B_d(x,r)=\{y\in X:d(x,y)<r\}. \end{align*} We choose the radius $1/2$ because it lies strictly between the two possible values of the discrete metric: $0$ for equal points and $1$ for distinct points. First, $x$ belongs to this ball because \begin{align*} d(x,x)=0<1/2. \end{align*} Now let $y\in B_d(x,1/2)$. By definition of the ball, this means \begin{align*} d(x,y)<1/2. \end{align*} If $y\neq x$, then the defining property of the discrete metric gives \begin{align*} d(x,y)=1, \end{align*} which is incompatible with $d(x,y)<1/2$. Thus $y\neq x$ is impossible, so $y=x$. We have shown both inclusions, and therefore \begin{align*} B_d(x,1/2)=\{x\}. \end{align*} This is the whole mechanism of the proof: every point is separated from every other point by a uniform positive distance, so a small enough ball contains exactly one point. [/guided] [/step] [step:Show that every subset is open in the metric topology] Let $A\subset X$ be arbitrary. If $A=\varnothing$, then $A$ is open because every topology contains the empty set. Suppose otherwise that $A$ is any subset of $X$. Using the singleton decomposition of a set, we have \begin{align*} A=\bigcup_{x\in A}\{x\}. \end{align*} By the previous step, for each $x\in A$, \begin{align*} \{x\}=B_d(x,1/2). \end{align*} Each open ball belongs to the metric topology $\tau_d$ by definition of the metric topology. Hence each singleton $\{x\}$ with $x\in A$ is $\tau_d$-open. Since a topology is closed under arbitrary unions, the union $\bigcup_{x\in A}\{x\}$ is $\tau_d$-open. Therefore $A\in\tau_d$. Because $A\subset X$ was arbitrary, every subset of $X$ is $\tau_d$-open. Thus \begin{align*} \mathcal{P}(X)\subset \tau_d. \end{align*} [/step] [step:Identify the metric topology with the discrete topology] Since $\tau_d$ is a topology on the set $X$, every element of $\tau_d$ is a subset of $X$. Hence \begin{align*} \tau_d\subset \mathcal{P}(X). \end{align*} The previous step proved the reverse inclusion $\mathcal{P}(X)\subset\tau_d$. Therefore \begin{align*} \tau_d=\mathcal{P}(X). \end{align*} By definition, the discrete topology on $X$ is $\mathcal{P}(X)$. Hence the metric topology induced by the discrete metric is the discrete topology on $X$. [/step]