[proofplan] We prove both directions directly from the definitions. If $(X,\tau)$ is Hausdorff and $X$ had two distinct points, Hausdorffness would provide disjoint open neighbourhoods of those points; but in the [indiscrete topology](/page/Indiscrete%20Topology) every nonempty [open set](/page/Open%20Set) is all of $X$, so the two neighbourhoods must coincide with $X$ and cannot be disjoint. Conversely, if $X$ has at most one point, there are no distinct pairs of points to separate, so the Hausdorff condition is satisfied vacuously. [/proofplan] [step:Use Hausdorff separation to rule out two distinct points] Assume that $(X,\tau)$ is Hausdorff. Suppose, for contradiction, that there exist points $x,y\in X$ with $x\neq y$. By the definition of Hausdorffness, there exist open sets $U,V\in\tau$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. Since $x\in U$, the set $U$ is nonempty. The only nonempty open set in the indiscrete topology $\tau=\{\varnothing,X\}$ is $X$, so $U=X$. Similarly, since $y\in V$, the set $V$ is nonempty, and hence $V=X$. Therefore \begin{align*} U\cap V=X\cap X=X. \end{align*} Because $x\in X$, the set $X$ is nonempty, so $U\cap V\neq\varnothing$, contradicting the choice of $U$ and $V$ as disjoint open sets. Hence no two distinct points of $X$ exist, so $X$ has at most one point. [guided] Assume that $(X,\tau)$ is Hausdorff. To show that $X$ has at most one point, we must show that it is impossible to find two different points in $X$. Suppose, toward a contradiction, that such points exist; that is, let $x,y\in X$ satisfy $x\neq y$. The Hausdorff condition says that distinct points can be separated by disjoint open neighbourhoods. Applying this condition to the distinct points $x$ and $y$, there are open sets $U,V\in\tau$ such that $x\in U$, $y\in V$, and \begin{align*} U\cap V=\varnothing. \end{align*} Now we use the special form of the indiscrete topology. Since $\tau=\{\varnothing,X\}$, the only open sets are $\varnothing$ and $X$. Because $x\in U$, the open set $U$ cannot be $\varnothing$, so it must be $X$. Likewise, because $y\in V$, the open set $V$ cannot be $\varnothing$, so it must also be $X$. Thus the two open neighbourhoods forced by Hausdorffness are actually the same set: \begin{align*} U=X \end{align*} and \begin{align*} V=X. \end{align*} Consequently, \begin{align*} U\cap V=X\cap X=X. \end{align*} Since $x\in X$, the set $X$ is nonempty, so this intersection is not empty. This contradicts the Hausdorff requirement that $U$ and $V$ be disjoint. Therefore two distinct points cannot exist in $X$, and $X$ has at most one point. [/guided] [/step] [step:Verify Hausdorffness when there is no distinct pair to separate] Conversely, assume that $X$ has at most one point. To prove that $(X,\tau)$ is Hausdorff, we must verify that for every pair of points $x,y\in X$ with $x\neq y$, there exist disjoint open sets $U,V\in\tau$ with $x\in U$ and $y\in V$. There is no pair $x,y\in X$ with $x\neq y$, because $X$ has at most one point. Therefore the universal condition in the definition of Hausdorffness has no counterexample. Hence $(X,\tau)$ is Hausdorff. Combining the two directions, $(X,\tau)$ is Hausdorff if and only if $X$ has at most one point. [/step]