[proofplan] We prove both directions directly from the definitions. If $X$ has at most one point, then there are no two distinct points to separate, so the $T_0$ condition holds vacuously. Conversely, if $(X,\tau)$ is $T_0$ and $X$ contained two distinct points, the $T_0$ condition would require an [open set](/page/Open%20Set) containing exactly one of them. The [indiscrete topology](/page/Indiscrete%20Topology) has only $\varnothing$ and $X$ as open sets, and neither can contain exactly one of two distinct points. [/proofplan] [step:Verify the $T_0$ condition when there are no two distinct points] Assume that $X$ has at most one point. To prove that $(X,\tau)$ is $T_0$, let $x,y\in X$ be distinct points. No such pair exists, because $X$ has at most one point. Therefore the defining condition for a $T_0$ space is satisfied vacuously, and $(X,\tau)$ is $T_0$. [/step] [step:Use the $T_0$ condition to force an open set separating two chosen points] Assume that $(X,\tau)$ is $T_0$. Suppose, toward a contradiction, that $X$ has at least two points. Choose distinct points $x,y\in X$. By the definition of the $T_0$ separation axiom, since $x\neq y$, there exists an open set $U\in\tau$ such that either $x\in U$ and $y\notin U$, or $y\in U$ and $x\notin U$. [guided] Assume that $(X,\tau)$ is $T_0$, and suppose for contradiction that $X$ has at least two points. This means there exist points $x,y\in X$ with $x\neq y$. The $T_0$ axiom is an asymmetric separation condition: for every pair of distinct points, at least one of the two points has an open neighbourhood that excludes the other. Applying this definition to the distinct pair $x,y$, we obtain an open set $U\in\tau$ such that one of the following alternatives holds: \begin{align*} x\in U \text{ and } y\notin U. \end{align*} or \begin{align*} y\in U \text{ and } x\notin U. \end{align*} Thus the $T_0$ assumption produces an open set that contains exactly one of the two chosen points. [/guided] [/step] [step:Contradict separation using the only open sets in the indiscrete topology] Since $\tau=\{\varnothing,X\}$, the open set $U$ is either $\varnothing$ or $X$. If $U=\varnothing$, then neither $x$ nor $y$ belongs to $U$, so $U$ cannot contain exactly one of $x$ and $y$. If $U=X$, then both $x$ and $y$ belong to $U$, so again $U$ cannot contain exactly one of $x$ and $y$. This contradicts the open set supplied by the $T_0$ condition. Hence $X$ cannot contain two distinct points, so $X$ has at most one point. [/step] [step:Conclude the equivalence] We have shown that if $X$ has at most one point, then $(X,\tau)$ is $T_0$, and that if $(X,\tau)$ is $T_0$, then $X$ has at most one point. Therefore $(X,\tau)$ is $T_0$ if and only if $X$ has at most one point. [/step]