Show a sequence $(x_n)^{\infty}_{n=1}$ converges to a point of S if and only if it is eventually constant Suppose a set $S$ is given the discrete metric $d_0$. Show a sequence $(x_n)^{\infty}_{n=1}$ converges to a point of $S$ if and only if it is eventually constant; there exists $N \in \mathbb{N}$ such that $x_n=x_N$ for all $n \geq N$.
pf: If $(x_n)_{n\in \Bbb N}$ converges to an element $x \in M$, there exists a positive integer $k$ such that $d_0(x_n, x) < 1/2$ for all $n \ge k$. How to show $x_n=x$ for all $n \geq k$?
Conversely, if $(x_n)_{n\in \Bbb N}$ is eventually constant, there exists an $x \in M$ and a positive integer $k$ such that $x_n = x$ for all $n \ge k$. Thus, for any $\varepsilon > 0$, $d_0(x_n, x) = 0 < \varepsilon$ for all $n \ge k$. Consequently, $(x_n)_{n\in \Bbb N}$ converges to $x$.

The discrete metric is defined by
$$d(x, y) = \begin{cases} 0 & \text{if } x = y \\ 1 & \text{otherwise.}\end{cases}$$
So, if $d(x, y) < 1/2$, then $d(x, y) \neq 1$, and hence $x = y$.
