If I was given that $\lim_{x\to x_0}f(x)=L$, for some $f:\mathbb R\to\mathbb R$.
This means that for every $\epsilon>0$, there is some $\delta>0$, such that $|x-x_0|<\delta\implies|f(x)-L|<\epsilon$.
These $\delta$ values for corresponding $\epsilon$ exist, but are unknown.
Given this can I make a set like:
$$S=\{(\epsilon,\delta)\in\mathbb R\times\mathbb R\mid (\epsilon>0)\wedge(\delta>0:(\forall x:|x-x_0|<\delta\implies|f(x)-L|<\epsilon))\}$$
(I didn't know how to write it in a way for everyone to understand).
Essentially I want to make a set (or show it exists) of all possible $\epsilon$-$\delta$ pairs satisfying the limit definition. Is the set possible? Why / why not?
I can think of the Zorn's lemma, as this seems like a 'maximal set', but I don't know how to apply it. What kind of ordering would it be? What are the chains?
Maybe I can make a family: $F=\{U_i\mid i\in\mathbb R\}:U_i=\{\delta\mid\forall x: |x-x_0|<\delta\implies |f(x)-L|<i\}$ ?
I've only ever seen Zorn's lemma in a proof showing every vector space has a basis.
One small correction in your definition of limit: the implication should be $0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < \epsilon$.
Yes, your set $S$ is fine. You don't need to do anything more to prove that it exists--you have defined it.