I'm having a difficult time understanding what an accumulation point is and how to find it. My textbook uses the following definition:
"An accumulation point of a nonempty set $E$ of real numbers is an extended real number a such that every neighborhood of a contains a point of $E$ not equal to $a$."
I've searched online for definitions that are easier to understand, but still cannot conceptualize. And how are accumulation points different from limit points? Lastly, how would I go about finding APs? Specifically for the set:
$$ \left\{\frac{2}{m} - \frac{3}{n} \colon m, n \in \mathbb{N}\right\} $$ I think $-1$ and $0$ would be APs but I have no idea how to prove that?
A point $x$ is a limit point of $E$ if whenever you draw a ball around $x$, you can find another point in $E$ that is not equal to $x$ inside the ball. Hence by drawing multiple balls by shrinking the radius, you can construct a sequence of points in $E$ not equal to $x$ that converges to $x$ ,which is an intuitive reason why it is called a limit point. Also, inside each such ball, you should be able to find infinitly many points of $E$.
The set of limit points should be
$$\{ 0\} \cup \left\{ \frac2m : m \in \mathbb{N}\right\} \cup \left\{ -\frac3n : n \in \mathbb{N}\right\}. $$
$0$ is a limit point since $$\lim\limits_{m \to \infty} \left[\lim\limits_{n \to \infty} \left(\frac2m-\frac3n\right)\right] = 0$$
Also, for any $m \in \mathbb{N}$, $\frac2m$ is a limit point since $$\lim\limits_{n \to \infty} \left(\frac2m - \frac3n \right)=\frac2m.$$
For any $n \in \mathbb{N}$, $-\frac3n$ is a limit point since
$$\lim_{m \to \infty} \left(\frac2m - \frac3n \right)=-\frac3n.$$
For points that are bigger than $2$ or smaller than $-3$, it should be clear that they are not limit points.
Now suppose a point $x$ satisfies $$\frac2{m+1} < x< \frac2m$$ for some $m \in \mathbb{N}$.
Notice that $\left( \frac2{m+1} , \frac{x + \frac2m}{2} \right)$ and $\cup_{i=1}^m \left[\cup_{n=1}^\infty\{\frac2i-\frac3n\}\right]$ can only intersect at finitely many points since for each $i$, $\lim\limits_{n \to \infty}\left(\frac2i - \frac3n \right)=\frac2i$ and we only have to consider finitely many $i$. This implies that $\left( \frac2{m+1} , \frac{x + \frac2m}{2} \right)$ intersect with your original set finitely many times since the other points are less than $\frac{2}{m+1}$. Hence, such $x$ cannot be a limit point. Similar argument can be used for the other negative values.