Suppose $d$ is the discrete metric, that is $d(x_1, x_2) = 1$ for all distinct points $x_1$, $x_2$ in X. Prove that if $(x_n)$ converges to $y$ then there exists a natural number $N$ such that $x_n = y$ for all $n > N$.
I understand why this is true, but I don't know how to prove it, please help.
It is really a matter writing out the definitions of the notions involved.
For contra positive, assume there does not exist such an $N$ as described i.e. assume that for every $N \in \mathbb{N}$ there exists some $n > N$ s.t. $x_n \neq y$. Then we want to show that $(x_n)_{n \in \mathbb{N}}$ does not converge to $y$ i.e. we want to show that there is some $\varepsilon> 0$ such that there does not exist an $N \in \mathbb{N}$ with the property that $\forall n > N: x_n \in B_{\varepsilon}(y)$. Take here $\varepsilon := 1/2$. Then indeed, the above gives us precisely that, since whenever $x_n \neq y$ we have $x_n \not\in B_{\varepsilon}(y)$.
We can also show this directly. Assume that $x_n \rightarrow y$. That means that for every $\varepsilon > 0$, there exists some $N \in \mathbb{N}$ s.t. for any $n > N$ we have $x_n \in B_{\varepsilon}(y)$. Now choose $\varepsilon := 1/2$. Then this gives you that after some index $N$, all the $x_n$ have distance strictly less than $1/2$ to the point $y$. But the only point in the entire space with that property is $y$ itself. Hence after this index, we have $x_n = y$.
