I was recently writing out a proof that a certain real valued function is continuous, and something bothered me along the way.
Suppose we have a function $f: D \subset \mathbb{R} \to \mathbb{R}$, and we want to prove that $$\lim_{x\to a} f(x) = L.$$
In order to do so we show that for all positive real $\epsilon$, there exists a positive real $\delta$ such that for all $x \in D$ such that $0< |x-a| < \delta$, we have $|f(x) - f(a)| < \epsilon$.
Suppose we are confident that we have found such a $\require{enclose} \enclose{horizontalstrike}{\delta}$ for which the above holds for any $\require{enclose} \enclose{horizontalstrike}{\epsilon > 0}$ given any $\epsilon >0$, we can find a $\delta$ such that the above holds and want to prove the statement using it.
We let $\epsilon >0$, and define our $\delta$.
Then, we let $x \in D$ such that $|x-a|<\delta$. But hold on, how do we know that such an $x$ exists? Have we implicitly assumed that such an $x$ exists, and do not consider the case where one doesn't exist, since the statement "for all $x \in D$ such that $|x-a|<\delta$, we have $|f(x) - f(a)| < \epsilon$" would be vacuously true?
I'm confident that it can be proved that such an $x$ would exist (although I'm not sure exactly how), but I am pretty sure that it is not necessary to include such proofs when proving statements about limits.
My main question is: is it okay to assume the existence of a real number in the domain of $f$ such that $|x-a| < \delta$, and if so, why?
Generally speaking, the definition of the functional limit at a point $x=a$ necessitates that $f(x)$ has the definition over some one deleted neighborhood of $x=a$，which is a premise. Of course, in some more advanced calculus courses, $x=a$ may be the accumulating point.