This may seem trivial but, the following definitions of continuity of a function $f(x)$ on a metric space are equivalent -
- $\forall \epsilon > 0$, $\exists \delta > 0$ such that for every point $x$, $d(x,y) < \delta \implies d(f(x), f(y)) < \epsilon $
- $x_n \rightarrow x \implies f(x_n) \rightarrow f(x)$
I will just give the proof for $2) \implies 1)$, since the other way round is easier.
To find the $\delta$ for any $\epsilon > 0$ in $1)$, Consider the set $\mathcal{A}$ of all sequences $x_n \rightarrow x$ . Now, from the definition of continuity in $2)$ we know that for all sequences $f(x^i_n) \rightarrow f(x)$ for $\{x^i_n\} \in \mathcal{A}$. Hence, $\exists N^i$ such that $\forall n>N^i$ , $d(f(x^i_n), f(x)) < \epsilon$. Let
$ \delta = \arg_{x^i_N} \min_{x^i_n \in \mathcal{A}} d(x,x^i_N) $
Thus this value of $\delta$ can be used to find the neighbour-hood in $2)$.