[proofplan] We prove convergence to an arbitrary point $x\in X$ by checking the definition of net convergence. Given any neighbourhood $U$ of $x$, the [cofinite topology](/page/Cofinite%20Topology) implies that $X\setminus U$ is finite, because $U$ is a nonempty [open set](/page/Open%20Set). The finite-avoidance hypothesis applied to this finite complement says that the net is eventually outside $X\setminus U$, hence eventually inside $U$. Since $x$ was arbitrary, the same argument proves convergence to every point of $X$. [/proofplan] [step:Fix a point and reduce convergence to eventual membership in each neighbourhood] Let $x\in X$ be arbitrary. To prove that $(x_d)_{d\in D}$ converges to $x$, it is enough to prove that for every open set $U\subset X$ with $x\in U$, there exists $d_0\in D$ such that for every $d\in D$ with $d\ge d_0$, one has $x_d\in U$. [guided] We fix a point $x\in X$ because the conclusion says that the net converges to every point of $X$. Since no particular property of $x$ will be used, proving convergence to this arbitrary $x$ will prove universal convergence. The definition of convergence of a net says the following: the net $(x_d)_{d\in D}$ converges to $x$ if, for every neighbourhood $U$ of $x$, the net is eventually in $U$. In the present topological language, this means that for every open set $U\subset X$ satisfying $x\in U$, there must exist an index $d_0\in D$ such that whenever $d\in D$ and $d\ge d_0$, we have $x_d\in U$. Thus the task is reduced to proving eventual membership in an arbitrary open neighbourhood of $x$. [/guided] [/step] [step:Apply finite avoidance to the finite complement of the neighbourhood] Let $U\subset X$ be open and suppose $x\in U$. Since $U$ is nonempty and $X$ has the cofinite topology, the complement \begin{align*} F:=X\setminus U \end{align*} is finite. By the finite-avoidance hypothesis applied to this finite set $F$, there exists $d_0\in D$ such that for every $d\in D$ with $d\ge d_0$, one has $x_d\notin F$. For such $d$, the condition $x_d\notin X\setminus U$ is equivalent to $x_d\in U$. Therefore there exists $d_0\in D$ such that for every $d\ge d_0$, one has $x_d\in U$. [/step] [step:Conclude universal convergence] We have shown that for the arbitrary point $x\in X$ and every open neighbourhood $U$ of $x$, the net $(x_d)_{d\in D}$ is eventually in $U$. Hence $(x_d)_{d\in D}$ converges to $x$. Since $x\in X$ was arbitrary, $(x_d)_{d\in D}$ converges to every point of $X$. [/step]