[proofplan] We verify the definition of sequential convergence in a [topological space](/page/Topological%20Space). Fix a sequence and a candidate [limit point](/page/Limit%20Point). Any [open set](/page/Open%20Set) containing that point is nonempty, hence in the [indiscrete topology](/page/Indiscrete%20Topology) it must be all of $X$. Therefore every term of the sequence lies in every open neighborhood of the point, so the sequence is eventually in each such neighborhood. [/proofplan] [step:Fix a sequence and an arbitrary candidate limit point] Let $s:\mathbb{N}\to X$ be a sequence in $X$, and write $x_n=s(n)$ for each $n\in\mathbb{N}$. Let $x\in X$ be arbitrary. We will prove that $(x_n)_{n\in\mathbb{N}}$ converges to $x$ in the topology $\tau=\{\varnothing,X\}$. [/step] [step:Show every open neighborhood of the point is the whole space] Let $U\in\tau$ be an open set such that $x\in U$. Since $x\in U$, the set $U$ is not empty. The only elements of $\tau$ are $\varnothing$ and $X$, so $U\neq\varnothing$ implies $U=X$. [guided] Let $U\in\tau$ be an open neighborhood of $x$, meaning that $U$ is open in the topology $\tau$ and contains the point $x$. Because $x\in U$, the set $U$ cannot be $\varnothing$. The topology is indiscrete, so its only open sets are exactly $\varnothing$ and $X$. Therefore the only possible open set in $\tau$ that contains $x$ is $X$ itself. Hence every open neighborhood of $x$ is equal to $X$. [/guided] [/step] [step:Verify the eventual containment condition for convergence] Let $U\in\tau$ be any open set with $x\in U$. By the previous step, $U=X$. Since $x_n\in X$ for every $n\in\mathbb{N}$, we have $x_n\in U$ for every $n\in\mathbb{N}$. Choose $N=1$. Then for every $n\in\mathbb{N}$ with $n\ge N$, we have $x_n\in U$. This verifies that for every open neighborhood $U$ of $x$, the sequence is eventually contained in $U$. By the definition of convergence of a sequence in a topological space, $(x_n)_{n\in\mathbb{N}}$ converges to $x$. Since both the sequence $s:\mathbb{N}\to X$ and the point $x\in X$ were arbitrary, every sequence in $X$ converges to every point of $X$. [/step]