Understanding analysis definition from Abbott on limit points There is this definition in Abbott's understanding analysis:

"Definition 3.2.4: A point $x$ is a limit point of a set $A$ if every
  $\epsilon$-neigborhood $V_{\epsilon}(x)$ of $x$ intersects the set $A$
  in some point other than $x$."

I am bit unclear about what intersects means. If we consider some set $A$ of the form:
$$A = \{ \frac{1}{n}: n \in \mathbb{N} \}$$
and select $\epsilon = \frac{1}{n} - \frac{1}{n+1}$. Then the epsilon neighborhood of $1/n$ is:
$$V_{\epsilon}(1/n) = \{ x \in \mathbb{R}: |x-1/n| < 1/(n+1) \} = \{ x: x \in (1/(n+1),(n+2)/n(n+1)\}$$
Then we can see that at least for one $\epsilon$ the $\epsilon$-neighborhood does not intersect (does it mean it does not include?) any of the points in set $A$ other than $1/n$. Is that what intersect means? That the neighborhood does not include the points from set $A$?
 A: If $x$ is a limit point of $A$ then it means that for every $\epsilon$-neighborhood exists a point $y\in A$ and different of $x$ such that
$$y\in V_\epsilon(x)\cap A$$
Notice that $x$ doesnt need to belong to $A$ to be a limit point of $A$.
The concept of limit point is topological. In this context we are using the standard topology of $\Bbb R$.
A: Use this definition instead:
x is a limit point of A iff for each $\delta$>0 there exists a$\in$A with a$\neq$x and $\left |x-a \right |<\delta$
and interpret $\left |x-a \right |<\delta$ as "the distance of x from a."
I don't like to use of the term "limit point." The phrase doesn't capture the nature of the concept: Other textbooks call them "accumulation points," or "cluster points." The reason is because the definition literally means there are infinitely many points from A clustered around and touching x.
Think about it as an "accumulation point of A." Remember the archimedian principle of $\mathbb R$ that states, if $\delta>0$, there exists n in $\mathbb N$ such that 0<$\frac{1}{n}<\delta$?
Go as close as you want to 0, a tiny tiny tiny positive distance of $\delta_1$ from 0. There is a very big $n_1$ in $\mathbb N$ so that $0<\frac{1}{n}<\delta_1$. So there is $\alpha_1$ ($=\frac {1}{n}$) from A between 0 and $\delta_1$. Now shrink $\delta$ to $\delta_2$ so that $\delta_2$ is between 0 and $\alpha_1$ Then there is $\alpha_2$ from A between 0 and $\delta_2$.
Do this again and again and again to get $\alpha_1,\alpha_2,\alpha_3,...\alpha_k,...$ all from A so that
$0<....<\alpha_k<\delta_k<\alpha_{k-1}<\delta_{k-1}<.....<\alpha_1<\delta_1$
All of the $\alpha_i$'s are from A. No matter how close you get, there's always more there. They are literally clustered around 0. It's an accumulation point of A.
Now prove it with the definition.
If the set was A={$\frac{(-1)^n}{n}: n\in N$} You'd even have the $\alpha_i$'s on both sides. Try to figure it out why; and then do the same thing for { 5 + $\frac{(-1)^n}{n}: n\in \mathbb N$}. Prove each with the definition.
