Let $x \in \mathbb{R}$. Let $c$ be a real number constant, and $\delta > 0$ and also a real number. Consider the following:
$$ |x - c| < \delta \label{1}\tag{1}$$
$$0 \leq | x - c | < \delta \label{2}\tag{2}$$
Is \eqref{1} equivalent to \eqref{2}?
I think it is, because any absolute value must be $\geq 0$ by definition.
The question came up while trying to figure out the solution set for:
$$ 0 < | x - c | < \delta \label{3}\tag{3}$$
I know the solution set for \eqref{1} can be found as follows:
$$ |x - c| < \delta \Longleftrightarrow - \delta < x - c < \delta \Longleftrightarrow c - \delta < x < c + \delta \Longleftrightarrow (c - \delta, c + \delta) \label{4}\tag{4}$$
If \eqref{1} and \eqref{2} are equivalent, then the solution set for \eqref{3} can be found by subtracting $\{ c \}$ from the solution found in \eqref{4}:
$$( c - \delta, c + \delta) \setminus \{ c \} = (c - \delta, c) \cup (c, c + \delta)$$
Is my reasoning correct? Is there a better or more efficient way to find the solution set of \eqref{3}?
Your reasoning is correct. $(1)$ and $(2)$ are equivalent (by definition). Indeed, the solution set of $(3)$ thus equals the solution set of $(2)$ minus $\{c\}$ (i.e. $]c-\delta,c+\delta[\setminus\{c\}$.)
Note that the same thing also holds for complex numbers (or more generally also for Banach spaces where you use a general norm in the place of the absolute value.)