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?
Best Answer
Indeeed definition $2$ implies definition $1$:
Let us assume that $f$ satisfies definition $2$, and try to prove it is continuous by definition $1$.
Let $\epsilon > 0$. Then, because $\lim_{n\to\infty} \epsilon_n = 0$, there exists such a $N$ that $\epsilon_N < \epsilon$.
Now, let us set $\delta=\delta_N$, and let $|x-x_0|<\delta$. Because $\delta=\delta_N$, this means that $|x-x_0|<\delta_N$, and by definition $2$, this means that $|f(x)-f(x_0)|<\epsilon_N$.
This further means that $|f(x)-f(x_0)|<\epsilon_N<\epsilon$, meaning that $f$ is continuous at $x_0$.