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

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  • $\begingroup$ 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. $\endgroup$
    – Nikitau
    Nov 13 '16 at 23:20
  • $\begingroup$ @ZacharySelk If it helps, the problem says 'Consider a discrete metric space $(X, \delta)$ $\endgroup$
    – Nikitau
    Nov 13 '16 at 23:22
  • $\begingroup$ @ZacharySelk Sorry! :) And thanks for pointing that out! $\endgroup$
    – Nikitau
    Nov 13 '16 at 23:22
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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.

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