[proofplan] We prove non-metrizability by contradiction. First we record the key separation failure of the [cofinite topology](/page/Cofinite%20Topology) on an infinite set: any two nonempty open subsets must intersect. If a metric induced the cofinite topology, then two distinct points would have disjoint metric balls around them, producing two disjoint nonempty cofinite-open sets. This contradicts the intersection property. [/proofplan] [step:Show that any two nonempty cofinite open sets intersect] Let $U,V\in\tau_{\mathrm{cof}}$ be nonempty open subsets of $X$. Since $U$ and $V$ are nonempty and open in the cofinite topology, the complements $X\setminus U$ and $X\setminus V$ are finite subsets of $X$. Hence their union \begin{align*} (X\setminus U)\cup(X\setminus V) \end{align*} is finite. By De Morgan's law, \begin{align*} X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V). \end{align*} If $U\cap V=\varnothing$, then $X\setminus(U\cap V)=X$, so $X$ would be finite. This contradicts the hypothesis that $X$ is infinite. Therefore $U\cap V\neq\varnothing$. [/step] [step:Assume a metric induces the cofinite topology] Suppose, for contradiction, that $(X,\tau_{\mathrm{cof}})$ is metrizable. Then there exists a metric \begin{align*} d:X\times X\to[0,\infty) \end{align*} such that the topology induced by $d$ is exactly $\tau_{\mathrm{cof}}$. Since $X$ is infinite, choose two distinct points $a,b\in X$. Because $d$ is a metric and $a\neq b$, the number $d(a,b)$ is positive. Define \begin{align*} r=\frac{d(a,b)}{3}. \end{align*} Then $r>0$. [guided] We now assume the opposite of what we want to prove. Metrizability means that there is a genuine metric \begin{align*} d:X\times X\to[0,\infty) \end{align*} whose open sets are exactly the sets in $\tau_{\mathrm{cof}}$. The word metric is important here: for distinct points, the distance is strictly positive. Because $X$ is infinite, it contains at least two distinct points. Choose $a,b\in X$ with $a\neq b$. The metric separation axiom gives \begin{align*} d(a,b)>0. \end{align*} Set \begin{align*} r=\frac{d(a,b)}{3}. \end{align*} Then $r>0$. The purpose of choosing one third of the distance is to create two metric balls around $a$ and $b$ that cannot overlap, by the triangle inequality. [/guided] [/step] [step:Construct two disjoint nonempty metric open balls] Define the metric open balls \begin{align*} B_d(a,r)=\{x\in X:d(a,x)<r\} \end{align*} and \begin{align*} B_d(b,r)=\{x\in X:d(b,x)<r\}. \end{align*} These sets are nonempty because $a\in B_d(a,r)$ and $b\in B_d(b,r)$. They are open in the topology induced by $d$, hence they belong to $\tau_{\mathrm{cof}}$. We claim that $B_d(a,r)\cap B_d(b,r)=\varnothing$. If $x\in B_d(a,r)\cap B_d(b,r)$, then the triangle inequality gives \begin{align*} d(a,b)\leq d(a,x)+d(x,b)<r+r=2r. \end{align*} Since $r=d(a,b)/3$, this says \begin{align*} d(a,b)<\frac{2}{3}d(a,b), \end{align*} which is impossible because $d(a,b)>0$. Therefore the two balls are disjoint. [/step] [step:Derive the contradiction and conclude non-metrizability] The previous step produced two nonempty sets $B_d(a,r),B_d(b,r)\in\tau_{\mathrm{cof}}$ with \begin{align*} B_d(a,r)\cap B_d(b,r)=\varnothing. \end{align*} This contradicts the fact that any two nonempty open sets in the cofinite topology on an infinite set must intersect. Hence no metric on $X$ induces $\tau_{\mathrm{cof}}$, and therefore $(X,\tau_{\mathrm{cof}})$ is not metrizable. [/step]