I've been using Steven R. Lay's book, Analysis with an Introduction to Proof as a self-study for real analysis. I thought I understood the definitions of a neighborhood (that contains its center), a deleted neighborhood, and then Lay's shift towards an arbitrary neighborhood (the neighborhood could either be one that contains its center as one of its members or not), but based on the following excerpt and a problem that I'm stuck on, I'm not sure if I do understand these concepts as well as I initially thought.
After defining the ideas of a neighborhood $N(x;\epsilon) = (x-\epsilon,x+\epsilon)$ and a deleted neighborhood $N^*(x;\epsilon) = N(x;\epsilon)-\{x\}$, he writes
If for every neighborhood $N$ of $x$, $N\cap S \neq \varnothing$ and $N \cap (\mathbb{R}-S) \neq \varnothing$, then $x$ is called a boundary point of $S$ [, where $S \subseteq \mathbb{R}$].
brackets are mine
This is the first instance where he mentions an arbitrary neighborhood $N$, and I always thought that $N$ could either be $N(x;\epsilon)$ or $N^*(x;\epsilon)$. Thus, a statement concerning every neighborhood, in my mind, would pertain to both $N(x;\epsilon)$ and $N^*(x;\epsilon)$, like the definition of a boundary point.
Thus, is it possible that my understanding is not correct? That is, for example, could a point $x$ be a boundary point if an arbitrary $N(x;\epsilon)$ satisfies the definition of a boundary point, but for some $\delta > 0$, the deleted neighborhood $N^*(x;\delta)$ does not? Another example might be if $x$ is an isolated point, then $x$ is a boundary point, because every $N(x;\epsilon)$ satisfies the definition, whereas there are some deleted neighborhoods that do not. Or if a statement includes every neighborhood it actually means every neighborhood?
I am not familiar with Lay's text, but the standard definition uses neighborhoods (which include their center). That is almost certainly the intended meaning here.
The only difference between using a neighborhood or a deleted neighborhood is when $S$ contains an isolated point or its complement does. If $S$ consists of a single point $p \in \mathbb{R}$, then $p$ belongs to the boundary of $S$ using the standard interpretation. Using deleted neighborhoods instead would imply that $p$ does not belong to the boundary of $S$, since no deleted neighborhood of $p$ meets $S$.
It should also be mentioned that a common definition of a neighborhood of $x \in \mathbb{R}$ is a set $N$ that contains $N(x,\epsilon)$ for some $\epsilon > 0$. But this may not be the definition used in Lay's text.