Prove that $\lim_{n\to\infty}x_n=x_\infty$ if and only if $\lim_{n\to\infty} d_X(x_n,x_\infty)=0$ Let $(X,d_X)$ be a metric space and let $(x_n)_{n\in\mathbb N}$ be a sequence in $X$.
Let $x_\infty\in X$.
Prove that $\lim_{n\to\infty}x_n=x_\infty$ if and only if $\lim_{n\to\infty} d_X(x_n,x_\infty)=0$.
This is what I have so far:
$(\Rightarrow) \lim_{n\to\infty}x_n=x_\infty$ means that $\forall \epsilon>0, N\in\mathbb N$ s.t. $n\in\mathbb N, n>N$
then $|x_n-x_\infty|<\epsilon$.
(Unfinished, I don't know where to go from here)
$(\Leftarrow)$ $\lim_{n\to\infty}d_X(x_n,x_\infty)=0$ means that
$\lim_{n\to\infty} |x_n -x_\infty|=0$
As $\epsilon>0$, $|x_n -x_\infty|<\epsilon$. Therefore, $\forall \epsilon>0, N\in\mathbb N$ s.t. $n\in\mathbb N, n>N$
then $|x_n -x_\infty|<\epsilon$.
(Finished, would appreciate it if it could be checked)
 A: As far as I have understood, you are not necessarily working on $\mathbb{R}$.
Let $(X,d_{X})$ be a metric space and $\{x_{n}\}$ be a sequence in $X$.
We say that $x_{n}$ converges to $x$ iff for every $\varepsilon > 0$ there corresponds $n_{\varepsilon}\in\mathbb{N}$ s.t. for every $n\in\mathbb{N}$
\begin{align*}
n\geq n_{\varepsilon}\Rightarrow d_{X}(x_{n},x) < \varepsilon
\end{align*}
Notice that $d_{X}(x_{n},x)$ is a sequence itself.
Having said that, we can rewrite the proposed definition of convergence as follows
\begin{align*}
(\forall\varepsilon > 0)(\exists n_{\varepsilon}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{\varepsilon} \Rightarrow |d_{X}(x_{n},x) - 0| < \varepsilon)
\end{align*}
which corresponds exactly to the result proposed.
Hopefully this helps!
A: Note that we do not have any structure of addition in a general metric space. This means that $|x_n-x_\infty|<\varepsilon$ is undefined. What we have however is that $d(x_n,x_\infty)<\varepsilon$.
Suppose first that $\lim_{n\to \infty}x_n=x_\infty$. Then for each $\varepsilon$ we can choose $N$ such that for $n>N$ we have, $d(x_n,x_\infty)<\varepsilon$. But then for every $\varepsilon$ we can use this same $N$ such that for $n>N$, $|d(x_n,x_\infty)|=d(x_n,x_\infty)<\varepsilon$ which shows that the sequence converges to $0$. The converse is a similar argument.
