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I have read previous answers about the proof but there is a small point I want to be sure of.

First if we assume that the sequence converges $x_n \to x$ there for all $N\leq n$ there is $\varepsilon >0$ s.t $d(x_n,x)< \varepsilon$

Let $\varepsilon = \frac{1}{2}$, because $x_n$ converges there is $N\leq n_0$ such that $d(x_{n_0},x)< \frac{1}{2}$, now if $x_{n_0}\neq x$ then by definition of the discrete metric $d(x_{n_0},x)= 1$ with is a contradiction and therefore it must be that $x_{n_0}=x$ which means that $x_n$ is eventually constant.

Second if $x_n$ is eventually constant that means that there is $N\leq n_0$ such that $x_{n_0}=x_{n_0+1}=...$ why can we assume that $x_{n_0}=x_{n_0+1}=...=x$ ?

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  • $\begingroup$ You are not assuming anything. $x$ is what you choose to call the limit $x_{n_0} = \cdots$ $\endgroup$ Jun 8 '20 at 15:06
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You know there is $N$ such that $\forall n \ge N$ you have $d(x_n, x) <1/2$ and you know that this implies $x_n=x$ because the metric is discrete. Thus you proved that not only $x_N=x$, but that FOR ALL $n \ge N$ it happens that $x_n=x$.

If the sequence is eventually constant, then it literally means that there are $x$ and $N$ such that $x_n=x$ for every $n \ge N$. So it trivially converges to that $x$.

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  • $\begingroup$ "The trivially converges to that $x$" is what I am looking for $\endgroup$
    – newhere
    Jun 8 '20 at 15:18
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    $\begingroup$ Well for every $n \ge N$ you have $x_n=x$. So if you pick any $\varepsilon >0$, you get $d(x_n,x)=0<\varepsilon$ for every $n \ge N$. It is your definition of convergence if you look at it. $\endgroup$
    – Dunnò000
    Jun 8 '20 at 15:25
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Let $x_n$ be a sequence of points of the metric space, converging to a point $x$. By definition of converging sequence, that means that for every $\epsilon$ you fixed, greater than $0$, you can find a positive integer $N_{\epsilon}$ such that $d(x_n,x)<\epsilon$ for all indexes $n$ greater than $N_{\epsilon}$. That is, the distance of $x_n$ from the limit point $x$ is eventually smaller than $\epsilon$. If you choose $\epsilon$ smaller that $1$, this means that you can find $N_1$ such that $d(x_n,x)$ is smaller than $1$ for all indexes $n$ greater than $N_1$. As you said, in the discrete metric, two distinct points have distance $=1$. Therefore, having distance smaller than $1$ is equivalent to being the same point. This means that, for $n$ big enough, the distance between the $n$-th term of the sequence and the limit point $x$ will be smaller than $1$, that in discrete metric is equivalent to say that for $n$ big enough$, the $n$-th term of the sequence and the limit point will be the same point.

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Dude, as u said earlier $x_n$ converging to $x$. Now in discrete metric (or in any metric) $d(x,y)=0$ iff $x=y$ hence now for all $n\geq N$ $d(x_{n},x)<1/2$ so in this discrete metric $d(x,y)=1$ or $0$ so $d(x_n,x)=0$ so $x_n=x$ for all $n\geq N$

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