[proofplan] We verify the $T_1$ separation condition directly from the definition of the [cofinite topology](/page/Cofinite%20Topology). Given two distinct points, remove one of them from $X$; the complement is a singleton, hence finite, so the resulting set is open in the cofinite topology. This gives an open neighbourhood of either point which excludes the other point. [/proofplan] [step:Separate two distinct points by deleting one of them] Let $x,y\in X$ satisfy $x\ne y$. Define the subset $U_y\subset X$ by \begin{align*} U_y:=X\setminus\{y\}. \end{align*} Its complement in $X$ is \begin{align*} X\setminus U_y=\{y\}. \end{align*} The set $\{y\}$ is finite, so by the definition of $\tau_{\mathrm{cof}}$ we have $U_y\in\tau_{\mathrm{cof}}$. Since $x\ne y$, we also have $x\in U_y$ and $y\notin U_y$. Thus there is an [open set](/page/Open%20Set) containing $x$ and not containing $y$. [guided] Fix two points $x,y\in X$ with $x\ne y$. To prove the $T_1$ condition, we need an open set that contains one of these points while excluding the other. The cofinite topology is designed precisely for this kind of argument: a subset of $X$ is open when its complement is finite. Define \begin{align*} U_y:=X\setminus\{y\}. \end{align*} This set removes exactly the point $y$ from $X$. Its complement relative to the ambient set $X$ is \begin{align*} X\setminus U_y=\{y\}. \end{align*} Because a singleton is finite, the defining property of the cofinite topology gives \begin{align*} U_y\in\tau_{\mathrm{cof}}. \end{align*} The inequality $x\ne y$ implies $x\notin\{y\}$, so $x\in X\setminus\{y\}=U_y$. On the other hand, $y\notin X\setminus\{y\}=U_y$ by construction. Therefore $U_y$ is an open set containing $x$ and excluding $y$. [/guided] [/step] [step:Repeat the separation in the opposite direction] Define the subset $U_x\subset X$ by \begin{align*} U_x:=X\setminus\{x\}. \end{align*} Then \begin{align*} X\setminus U_x=\{x\}, \end{align*} which is finite. Hence $U_x\in\tau_{\mathrm{cof}}$. Since $x\ne y$, we have $y\in U_x$ and $x\notin U_x$. Thus there is an open set containing $y$ and not containing $x$. [/step] [step:Conclude that the cofinite space is $T_1$] For every pair of distinct points $x,y\in X$, the preceding steps produce an open set $U_y\in\tau_{\mathrm{cof}}$ with $x\in U_y$ and $y\notin U_y$, and an open set $U_x\in\tau_{\mathrm{cof}}$ with $y\in U_x$ and $x\notin U_x$. This is exactly the $T_1$ separation condition. Therefore $(X,\tau_{\mathrm{cof}})$ is a $T_1$ space. [/step]