# Let $(\mathbb{R},d)$ with $d$ discrete metric, then limit of sequences exist?

Consider $\mathbb{R}$ with the discrete metric $d$. Let $\{x_n\}_{n\in\mathbb{N}}$ a sequence in $X$ such that $x_n\neq x_m$ for $n\neq m$. Then exist $\lim_{n\to\infty}{x_n}$?

There exist a sequence (without the condition $x_n=a,\forall n\in\mathbb{N}$) that converges in $(\mathbb{R},d)$

My approach: If limit exist, then $\lim{x_n}=a\implies$ for every $\epsilon>0$, we can find a $n_0\in\mathbb{N}$ such that $n>n_0\implies d(x_n,a)<\epsilon$, i.e. $x_n=a$. Thanks!

Your are on the right way. Consider $\varepsilon\in(0,1)$. Then you can conclude:

If $d(x_n,a)<\varepsilon<1$ then $d(x_n,a)=0$ and therefore $x_n=a$.

So you see, that a sequence is convergent in $(\mathbb R, d)$ iff there exists $n_0\in \mathbb N$ such that $x_n=a$ for all $n\geq n_0$.

If $x_n\neq x_m$ for all $n\neq m$ the sequence $(x_n)_n$ is not convergent in $(\mathbb R, d)$, i.e. $\left(\frac1n\right)_n$ is such an example.