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

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 $$\forall epsilon >0$$, $$d(p_n, p) < \epsilon$$.

• I'm not really sure. It's just arbitrarily left as $\delta$ in the question, so I assume not. Edited my post to clarify that $(X, \delta)$ is described as a discrete metric space. Nov 13 '16 at 23:20
• @ZacharySelk If it helps, the problem says 'Consider a discrete metric space $(X, \delta)$ Nov 13 '16 at 23:22
• @ZacharySelk Sorry! :) And thanks for pointing that out! Nov 13 '16 at 23:22

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.