TITLE: Is this alternative notion of continuity in metric spaces weaker than, or equivalent to the usual one?
QUESTION [18 upvotes]: I will try to be as clear as possible. 
For simplicity I will assume that the function $f$ for which we define continuity at some point is real function of a real variable $f: \mathbb R \to \mathbb R$, although the same line of reasoning should be the same even if we talk about continuity of a function at a point that is defined at some metric space and that has values in some metric space. To define continuity of a function $f$ at some point $x_0$ from its domain we have the following standard and famous definition, the $\varepsilon$-$\delta$ definition of continuity, which goes like this:

Definition 1: $f$ is continuous at the point $x_0$ from its domain if for every $\varepsilon>0$ there exists $\delta>0$ such that when $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\varepsilon$. 

It is clear that we could write $\delta (\varepsilon)$ instead of $\delta$ because there really is dependence of $\delta$ on $\varepsilon$.
Now, I was thinking about alternative definition of continuity that would go like this:

Definition 2: $f$ is continuous at the point $x_0$ from its domain if there exist two sequences $\varepsilon_n$ and $\delta_n$ such that for every $n \in \mathbb N$ we have $\varepsilon_n>0$ and $\delta_n>0$ and $\lim_{n\to\infty}\varepsilon_n=\lim_{n\to\infty}\delta_n=0$ and when $|x-x_0|<\delta_n$ then $|f(x)-f(x_0)|<\varepsilon_n$.

We could also write here $\delta_n(\varepsilon_n)$ instead of $\delta_n$ because obviously there is dependence of $\delta_n$ on $\varepsilon_n$. 
It is clear that second definition does not require that for every $\varepsilon$ there exist $\delta(\varepsilon)$ (which includes in itself an uncountable number of choices for $\varepsilon$) but instead requires that for every member of the sequence $\varepsilon_n$ there is some $\delta_n$ (and this includes in itself only countable number of choices because the set $\{\varepsilon_n : n \in \mathbb N\})$ is countable).
It is clearly obvious that definition 1 implies definition 2, and the real question is is the converse true, in other words:

If the function is continuous at some point according to definition 2, is it also continuous at the same point according to definition 1?

REPLY [14 votes]: Definition 2 indeed implies Definition 1,
However:
There is a superfluous requirement in Definition 2. 
Namely, the requirement the $\delta_n\to 0$ is not needed. 
The most important part in the Definition of continuity is the fact that $\varepsilon>0$ can be as small as possible. On the other hand, apart from the fact that $\delta$ depends on $\varepsilon$, the size of $\delta$ is irrelevant.
So, it could be reworded as follows:

Definition 2. The function $f:A\to\mathbb R$ is continuous at $x_0\in A$ if there exist two sequences $\{\varepsilon_n\}_{n\in\mathbb N}$ and $\{\delta_n\}_{n\in\mathbb N}$ of positive numbers, with  $\lim_{n\to\infty}\varepsilon_n=
0$, such that, whenever $\,\lvert x-x_0\rvert<\delta_n$, then $\,\lvert\,f(x)-f(x_0)\rvert<\varepsilon_n$.