# discrete metric converges iff it is eventually constant

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 $$x_{n_0}=x_{n_0+1}=...=x$$ ?

• You are not assuming anything. $x$ is what you choose to call the limit $x_{n_0} = \cdots$ Jun 8 '20 at 15:06

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 $$x$$.
• "The trivially converges to that $x$" is what I am looking for Jun 8 '20 at 15:18
• 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. Jun 8 '20 at 15:25
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.
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 $$n\geq N$$