[proofplan] We prove that no two distinct points of $X$ can be separated by disjoint open neighborhoods. Choose distinct points $x,y\in X$ and let $U,V\in \tau_{\mathrm{cof}}$ be open neighborhoods of $x$ and $y$, respectively. Since nonempty open sets in the [cofinite topology](/page/Cofinite%20Topology) have finite complements, disjointness of $U$ and $V$ would force $X$ to be the union of two finite sets, contradicting that $X$ is infinite. [/proofplan] [step:Choose two distinct points and arbitrary open neighborhoods] Since $X$ has at least two distinct points, choose $x,y\in X$ with $x\neq y$. Let $U,V\in \tau_{\mathrm{cof}}$ satisfy $x\in U$ and $y\in V$. Then $U$ and $V$ are nonempty because they contain $x$ and $y$, respectively. [/step] [step:Show that the two neighborhoods must intersect] Because $U$ is a nonempty [open set](/page/Open%20Set) in the cofinite topology, define the finite set $F_U\subset X$ by \begin{align*} F_U := X\setminus U. \end{align*} Because $V$ is a nonempty open set in the cofinite topology, define the finite set $F_V\subset X$ by \begin{align*} F_V := X\setminus V. \end{align*} Suppose, for contradiction, that $U\cap V=\varnothing$. Then every point of $X$ lies outside $U$ or outside $V$, so \begin{align*} X = (X\setminus U)\cup (X\setminus V) = F_U\cup F_V. \end{align*} The union of two finite sets is finite, so $X$ is finite. This contradicts the hypothesis that $X$ is infinite. Therefore $U\cap V\neq\varnothing$. [guided] The key point is that cofinite open sets in an infinite set are too large to be disjoint. Since $U$ is an open neighborhood of $x$, it is not empty. In the cofinite topology, every nonempty open subset has finite complement, so the set \begin{align*} F_U := X\setminus U \end{align*} is finite. Likewise, since $V$ is an open neighborhood of $y$, it is nonempty, and the set \begin{align*} F_V := X\setminus V \end{align*} is finite. Now assume that $U$ and $V$ are disjoint. We translate this disjointness into a statement about complements. If $z\in X$, then $z$ cannot belong to both $U$ and $V$. Hence either $z\notin U$ or $z\notin V$. This means \begin{align*} z\in (X\setminus U)\cup (X\setminus V). \end{align*} Since this holds for every $z\in X$, we have \begin{align*} X = (X\setminus U)\cup (X\setminus V) = F_U\cup F_V. \end{align*} But $F_U$ and $F_V$ are finite, and the union of two finite sets is finite. Therefore $X$ would be finite, contradicting the hypothesis that $X$ is infinite. The assumption $U\cap V=\varnothing$ is impossible, so every pair of open neighborhoods $U$ of $x$ and $V$ of $y$ must satisfy $U\cap V\neq\varnothing$. [/guided] [/step] [step:Conclude that the Hausdorff separation condition fails] The definition of a [Hausdorff space](/page/Hausdorff%20Space) requires that for every pair of distinct points there exist disjoint open neighborhoods of those points. We have found distinct points $x,y\in X$ such that every open neighborhood $U$ of $x$ and every open neighborhood $V$ of $y$ intersect. Therefore $(X,\tau_{\mathrm{cof}})$ is not Hausdorff. [/step]