[proofplan] Fix a point $x\in X$ and verify the definition of sequential convergence to $x$. An arbitrary neighbourhood $U$ of $x$ has finite complement because the topology on $X$ is cofinite. The finite-avoidance hypothesis applied to this complement gives an index after which all sequence terms lie in $U$. Since both $x$ and $U$ were arbitrary, the sequence converges to every point of $X$. [/proofplan] [step:Fix an arbitrary target point and neighbourhood] Let $x\in X$ be arbitrary. Let $U\subset X$ be an arbitrary open neighbourhood of $x$, meaning that $U$ is open in the [cofinite topology](/page/Cofinite%20Topology) and $x\in U$. Since $x\in U$, the set $U$ is not empty. By the definition of the cofinite topology, every nonempty open subset of $X$ has finite complement. Therefore the set $F_U := X\setminus U$ is finite. [guided] We want to prove that the sequence converges to the particular point $x$. By the definition of convergence of a sequence in a [topological space](/page/Topological%20Space), it is enough to show that every open neighbourhood of $x$ eventually contains all terms of the sequence. So fix an arbitrary open neighbourhood $U\subset X$ of $x$. The phrase “neighbourhood of $x$” means two things here: first, $U$ is open in the topology under consideration, and second, $x\in U$. Since $x\in U$, the set $U$ is not empty. Now we use the special form of the cofinite topology. In the cofinite topology, an [open set](/page/Open%20Set) is either $\varnothing$ or has finite complement in $X$. Our neighbourhood $U$ is not empty, so it is in the second case. Hence its complement $F_U := X\setminus U$ is a finite subset of $X$. [/guided] [/step] [step:Apply finite avoidance to the complement of the neighbourhood] Apply the finite-avoidance hypothesis to the finite set $F_U\subset X$. There exists $n_0\in\mathbb{N}$ such that $x_n\notin F_U$ for every $n\ge n_0$. For every $n\ge n_0$, the condition $x_n\notin F_U=X\setminus U$ implies $x_n\in U$, because each $x_n$ is an element of $X$. [/step] [step:Conclude universal convergence] We have shown that for the arbitrary open neighbourhood $U$ of the arbitrary point $x\in X$, there exists $n_0\in\mathbb{N}$ such that $x_n\in U$ for every $n\ge n_0$. This is exactly the definition that $(x_n)_{n\in\mathbb{N}}$ converges to $x$. Since $x\in X$ was arbitrary, the sequence converges to every point of $X$. [/step]