Why is it that in a discrete metric space only eventually constant sequences are convergent? I just read this result and was wondering what is the intuitive idea behind this ?
 A: A sequence $(x_n)$ in a metric space $(X, d)$ converges to $x \in X$ if for every $\varepsilon > 0$, there is $N\in\mathbb{N}$ such that $d(x_n, x) < \varepsilon$ for all $n \geq N$. Now consider the case where $d$ is the discrete metric and $\varepsilon \in (0, 1]$.
As for intuition, a sequence $(x_n)$ has limit $x$ if the terms of the sequence get closer and closer to $x$ as $n$ increases. If $(x_n)$ is not an eventually constant sequence, then the terms of the sequence are not getting closer to any fixed element $x$; the terms $x_n$ keep moving away from $x$ to create a distance of one between them.
A: In a discrete metric space, suppose a sequence $(x_n)$ converges to $x$, then for $\epsilon = 1/2$, there must be $n_0 \in \mathbb{N}$ such that
$$
d(x_n, x) < 1/2 \quad\forall n\geq n_0
$$
But if $x_n\neq x$, then $d(x_n,x) =1$, and hence
$$
x_n = x \quad\forall n\geq n_0
$$
A: If a sequence $\{x_n\}$ converges to a point $x$ not in this sequence, then the set of terms of the sequence $\{x_n\colon n\in\mathbb N\}$ is not closed. But in a discrete space every set is closed.
