[proofplan] We prove hyperconnectedness directly from the definition: take two nonempty open subsets $U,V\subset X$ and show that their intersection is nonempty. In the [cofinite topology](/page/Cofinite%20Topology), every nonempty [open set](/page/Open%20Set) has finite complement, so $X\setminus U$ and $X\setminus V$ are finite. The complement of $U\cap V$ is the finite union $(X\setminus U)\cup (X\setminus V)$; if $U\cap V$ were empty, this would force the infinite set $X$ to be finite. [/proofplan] [step:Take two nonempty open subsets in the cofinite topology] Let $U,V\in \tau_{\mathrm{cof}}$ be nonempty open subsets of $X$. By the definition of hyperconnectedness, it suffices to prove that \begin{align*} U\cap V\neq \varnothing. \end{align*} Since $\tau_{\mathrm{cof}}$ is the cofinite topology on $X$, every open subset of $X$ is either $\varnothing$ or has finite complement in $X$. Because $U$ and $V$ are nonempty, define \begin{align*} A := X\setminus U \end{align*} and \begin{align*} B := X\setminus V. \end{align*} Then $A$ and $B$ are finite subsets of $X$. [/step] [step:Compute the complement of the intersection] By De Morgan's law applied inside the ambient set $X$, \begin{align*} X\setminus (U\cap V) = (X\setminus U)\cup (X\setminus V). \end{align*} Using the definitions of $A$ and $B$, this becomes \begin{align*} X\setminus (U\cap V)=A\cup B. \end{align*} Since $A$ and $B$ are finite, their union $A\cup B$ is finite. Therefore $X\setminus (U\cap V)$ is finite. [guided] We want to show that $U\cap V$ contains at least one point. In a cofinite topology, the useful information about a nonempty open set is not usually the set itself, but its complement. Since $U$ is nonempty and open in $\tau_{\mathrm{cof}}$, its complement \begin{align*} A := X\setminus U \end{align*} is finite. Similarly, since $V$ is nonempty and open in $\tau_{\mathrm{cof}}$, its complement \begin{align*} B := X\setminus V \end{align*} is finite. Now compute the complement of the intersection. For any point $x\in X$, the condition $x\notin U\cap V$ is equivalent to saying that $x\notin U$ or $x\notin V$. Hence De Morgan's law gives \begin{align*} X\setminus (U\cap V) = (X\setminus U)\cup (X\setminus V). \end{align*} Substituting the definitions of $A$ and $B$, we obtain \begin{align*} X\setminus (U\cap V)=A\cup B. \end{align*} The set $A\cup B$ is finite because it is the union of two finite sets. Thus the complement of $U\cap V$ in $X$ is finite. This is the key point: the only way for $U\cap V$ to be empty would be for its complement to be all of $X$, which cannot happen because $X$ is infinite while the complement has just been shown to be finite. [/guided] [/step] [step:Use infinitude of $X$ to force a nonempty intersection] Suppose, for contradiction, that \begin{align*} U\cap V=\varnothing. \end{align*} Then \begin{align*} X\setminus (U\cap V)=X\setminus \varnothing=X. \end{align*} But the previous step showed that $X\setminus (U\cap V)$ is finite. Hence $X$ would be finite, contradicting the hypothesis that $X$ is infinite. Therefore \begin{align*} U\cap V\neq \varnothing. \end{align*} Since every pair of nonempty open subsets $U,V\subset X$ intersects, $(X,\tau_{\mathrm{cof}})$ is hyperconnected. [/step]