I know there is a possible duplicate, but I believe that question is asking for the proof to the right. I have already proved the right direction, but now I'm stuck on the left.

I'm trying to prove that given a discrete metric space $(X, \delta)$, a sequence converges with respet to $\delta$ if and only if it is eventually constant.

I've started with the assumption with eventually constant, or in other words $\exists p \in X$ and $N \in \mathbb{N}$ such that $p_n = p$ for all $n \geq N$. I know by definition of convergence that a sequence is said to converge if $\forall \epsilon > 0$, there is an interger N such that $n \geq N$ implies that $d(p_n, p) < \epsilon$. I feel like this problem is relatively simple, but I still don't see how I can prove that if a sequence is eventually constant then it converges.

  • $\begingroup$ Eventually constant means that for some $N\in\Bbb N$ the sequence is constant. Then any distance between two points is zero, then because $\epsilon>0$ for any $n\ge N$ the statement about convergence holds. $\endgroup$
    – Masacroso
    Oct 28 '16 at 3:17

To show convergence to a limit $L$, you need to show that: given any definition of "close enough" (given $\epsilon>0$), eventually (there exists $N\in\mathbb{N}$ so that for $n>N$) your sequence is "close enough" to the limit ($\delta(p_n, L)<\epsilon$).

In this case: if your sequence is always equal to $p$ for $n$ sufficiently large, then $p$ should certainly be the limit. And conveniently, we know that $\delta(p, p)=0$.

So, let any $\epsilon>0$ be given. We know that for $n>N$, $p_n=p$; so, for all $n>N$, what can you say about $\delta(p_n, p)$?

  • $\begingroup$ Oh. Would the distance just be 0 then? Then $epsilon$ can be 1? $\endgroup$
    – Nikitau
    Oct 28 '16 at 2:09
  • $\begingroup$ You don't get to choose $\epsilon$; you have to show that it works for EVERY $\epsilon>0$. But, that's easy in this case, given that the distance is eventually $0$. $\endgroup$ Oct 28 '16 at 3:07

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