[proofplan] We prove both directions directly from the definitions. If $x_n \to x$, then the singleton $\{x\}$ is an open neighbourhood of $x$ because the topology on $X$ is discrete, so convergence forces all sufficiently late terms of the sequence to lie in $\{x\}$. Conversely, if all sufficiently late terms are equal to $x$, then every open neighbourhood of $x$ contains those sufficiently late terms, which is exactly topological convergence. [/proofplan] [step:Use the singleton neighbourhood to force eventual equality] Assume $x_n \to x$ in $X$. Since $X$ has the [discrete topology](/page/Discrete%20Topology), every subset of $X$ is open; in particular, the singleton $\{x\} \subset X$ is open. Also $x \in \{x\}$, so $\{x\}$ is an open neighbourhood of $x$. By the definition of convergence of a sequence in a [topological space](/page/Topological%20Space), applied to the open neighbourhood $\{x\}$ of $x$, there exists $N \in \mathbb{N}$ such that $x_n \in \{x\}$ for every $n \geq N$. Membership in the singleton $\{x\}$ is equivalent to equality with $x$, hence $x_n = x$ for every $n \geq N$. [guided] Assume $x_n \to x$ in $X$. To prove eventual equality with $x$, we need to find one index $N \in \mathbb{N}$ such that every later term satisfies $x_n = x$. The special feature of the discrete topology is that every subset of $X$ is open. Therefore the singleton set $\{x\}$ is open in $X$. Since $x \in \{x\}$, this singleton is an open neighbourhood of the proposed limit $x$. Now we unpack the definition of convergence in a topological space. The statement $x_n \to x$ means that for every open neighbourhood $U \subset X$ of $x$, there exists an index $N_U \in \mathbb{N}$ such that $x_n \in U$ for every $n \geq N_U$. We apply this definition to the particular open neighbourhood $U = \{x\}$. Thus there exists $N \in \mathbb{N}$ such that $x_n \in \{x\}$ for every $n \geq N$. Finally, the condition $x_n \in \{x\}$ says exactly that $x_n = x$. Hence there exists $N \in \mathbb{N}$ such that $x_n = x$ for every $n \geq N$, which is precisely eventual constancy with value $x$. [/guided] [/step] [step:Use eventual equality to verify convergence in every neighbourhood] Conversely, assume there exists $N \in \mathbb{N}$ such that $x_n = x$ for every $n \geq N$. Let $U \subset X$ be an open neighbourhood of $x$. Since $U$ is a neighbourhood of $x$, we have $x \in U$. Therefore, for every $n \geq N$, the equality $x_n = x$ implies $x_n \in U$. Thus for every open neighbourhood $U$ of $x$, there exists $N \in \mathbb{N}$ such that $x_n \in U$ for every $n \geq N$. By the definition of convergence of a sequence in a topological space, $x_n \to x$ in $X$. [/step]