We have the standard statement in analysis that
i) For all $\epsilon>0$ and two fixed $a,b\in\mathbb{R}$ it holds $|b-a|<\epsilon \Leftrightarrow$ $a=b$.
However, sometimes it is the case that $|b-a|<\epsilon$ holds while $a$ and $b$ vary with a decreasing $\epsilon$. So we have something like:
ii) "For all $\epsilon>0$ and two functions of $\epsilon$, $f_a(\epsilon),f_b(\epsilon)$ it holds $|f_b(\epsilon)-f_a(\epsilon)|<\epsilon$".
As an example: $f_b(\epsilon)$ might be a limit point or $f_a(\epsilon)$ and $f_b(\epsilon)$ might be Darboux sums.
I am not 100% sure if I have understood properly the difference between both statements. May be someone can tell me if the following two statements are right:
1.) I cannot apply the statement i) to ii) and say $f_b(\epsilon)=f_a(\epsilon)$. A simple explanation would be that in the context of ii) we don't have fixed numbers because $f_a(\epsilon),f_b(\epsilon)$ vary. A more rigourous explanation can be made if I apply the limit to $|f_b(\epsilon)-f_a(\epsilon)|<\epsilon$: $$\lim\limits_{\epsilon\to 0}|f_b(\epsilon)-f_a(\epsilon)|\leq\lim\limits_{\epsilon\to 0}\epsilon =0.$$ This means that the limit of $f_b(\epsilon)-f_a(\epsilon)$ exists. However, this doesn't automatically imply that from some small $\epsilon$ onwards we have $f_b(\epsilon)=f_a(\epsilon)$. It might be the case that $f_b(\epsilon)=f_a(\epsilon)$ but it is also possible that $f_b(\epsilon)\neq f_a(\epsilon)$ for all $\epsilon$.
2.) If I apply the explanation regarding the limit to statement i) it means that: $$\lim\limits_{\epsilon\to 0}|f_b(\epsilon)-f_a(\epsilon)|=\lim\limits_{\epsilon\to 0}|b-a|\leq\lim\limits_{\epsilon\to 0}\epsilon =0.$$ As $f_b(\epsilon)$ and $f_a(\epsilon)$ are constant functions and the limit of $f_b(\epsilon)-f_a(\epsilon)$ exists, it must be $f_b(\epsilon)=f_a(\epsilon)$ for all $\epsilon$, so $b=a$.
Are my thoughts in 1.) and 2.) correct? Any comments or further explanations are welcome :)