Trouble understanding the concept of an accumulation point. I am having some trouble grasping the intuition behind the concept of an accumulation point. The book that I am reading gives the following definitions:
Definition 1.1.20 A neighbourhood of $a\in \mathbb{R} $ is any open interval containing $a$. If the neighbourhood is of the form $(a-\delta, a+\delta), \forall\delta \in \mathbb{R},$ with $\delta \gt0$, it is called neighbourhood of radius $\delta$ of $a$ and denoted by $V_{\delta}(a)$.
Definition 1.1.21 We say that $a\in \mathbb{R}$ is an accumulation point of $B\subset \mathbb{R} $ if every neighbourhood of $a$ contains a point of $B$ distinct of $a$.
My questions are these:
($1$) Why does the interval of a neighbourhood need to be open?
($2$)The book states that there are no accumulation points in $\mathbb{Z}$, but I don't understand why. Wouldn't it be possible to create an element $a$ such that $3 \lt a \lt 6$ ( for example ) and as such this interval would contain a point of $B$ distinct of $a$, like four or five?
($3$) The book also states that finite subsets of $\mathbb{R}$ cannot have accumulation points, but once again I don't really understand the reason.
Please note that I have no background in analysis or topology. The book is an introduction to single variable calculus. Also please excuse me if I'm asking too many questions, it's just that I'm having a hard time understanding this concept.
 A: An accumulation point of a set is a point which is, vaguely speaking, arbitrarily close to the set without necessarily being on the set. For example, $0$ is an accumulation point of $(0,1)$ because for every $\delta > 0$, we have that $(-\delta,\delta)\cap (0,1) \neq \varnothing$ (for instance, $\min\{\delta/2,1/2\}$ is in the intersection). As a second example, if $b < 0$, then $b$ is not an accumulation point of $(0,1)$, because there exists a specific $\delta >0$ such that $(b-\delta,b+\delta)\cap (0,1) = \varnothing$ (for instance, $\delta = |b|/2>0$). Similarly, $1$ is an accumulation point of $(0,1)$, while any $b>1$ is not.
As for why open intervals and neighborhoods are considered, is because given a point $x_0$ in such an open neighborhood, both sided limits $x\to x_0^+$ and $x\to x_0^-$ make sense. You need some room around $x_0$ to be able to approach it from both sides.
So, if we have a subset $A$ and an accumulation point $x$ of $A$, it is possible to "tend to $x$ along $A$". There are more specific definitions of "left accumulation point" and "right accumulation point" which are tailor-made for one-sided limits (I'll let you guess the definition).
About $\mathbb{Z}$: you need to pay attention to quantifiers. To show that some point $x$ is an accumulation point of $\mathbb{Z}$, every open neighborhood of $x$ must intersect $\mathbb{Z}$ in a point different from $x$, while to show that $x$ is not an accumulation point, it suffices to exhibit one open neighborhood of $x$ which intersects $\mathbb{Z}$ at most in $x$. To see that $\mathbb{Z}$ has no accumulation points, I claim that $\delta = 1/2$ works for every point. Namely, we have that $(x-\delta,x+\delta) \cap \mathbb{Z}$ is either $\{\lfloor x\rfloor\}$, $\{\lceil x\rceil\}$, or $\varnothing$, where $\lfloor\cdot\rfloor$ and $\lceil\cdot\rceil$ are the "floor" and "ceil" functions. Since the intersection always has size $0$ or $1$ and $x \not\in \{\lfloor x\rfloor,\lceil x\rceil\}$ if $x\not\in \mathbb{Z}$, and $x=\lfloor x\rfloor=\lceil x\rceil$ if $x\in\mathbb{Z}$, we conclude that $$((x-\delta,x+\delta)\setminus \{x\})\cap \mathbb{Z} = \varnothing.$$
For finite subsets, the reason is similar. Write $A = \{a_1,\ldots, a_k\}$ and assume that $a_1 < \cdots < a_k$. If $x<a_1$, then $\delta = (a_1-x)/2>0$ shows that $x$ is not an accumulation point of $A$. If $x>a_k$, then $\delta = (x-a_k)/2$ does the job. If $a_i<x<a_{i+1}$ for some $i\in \{1,\ldots, k-1\}$, then $\delta = (1/2)\min\{x-a_i, a_{i+1}-x\}$ does it.
