Consider a discrete metric space $(X, \delta)$. Prove a sequence converges w.r.t $\delta$ if and only if it is eventually constant.

Question

I'm mostly confused with the forward direction. I'm trying to show that a sequence converges w.r.t $\delta$ if and only if it is eventually constant, i.e there exists a value $p \in X$ and $N \in \mathbb{N}$ such that $p_n = p \forall n \geq N$.

I found a proof online that went like this:

$\rightarrow$

Suppose $(x_n)_{x \in \mathbb{N}}$ converges to some $x \in M$. Then for $\epsilon = 1/2$, there is N so that $n \geq N$ guarantees $d_0 (p_n, p) < 1/2$. But then for $n \geq N$, $x_n = x$ since the metric is discrete. So we have taht $(x_n)_{x \in \mathbb{N}}$

My questions are:

1. Shouldn't we be trying to prove that $d_0 (p_n, p) < \epsilon \text{ }\forall \epsilon >0$?
2. How come we can conclude $x_n = x$ because the metric is discrete?

For the reverse direction, I have:

$\leftarrow$

Since $p_n = p$, then this implies $d(p_n, p) = 0 < \epsilon$ Hence, we have that $\forall epsilon >0$, $d(p_n, p) < \epsilon$.

Answer 1

Suppose $x_n \to x$. Then, for any neighborhood $U$ of $x$, there is some positive integer $N$ so $n\ge N \implies x_n \in U$.

Well in this case, $\{x\}$ is a neighborhood of $x$. Done.