I was thinking about the definition of a limit point and couldn't find a definition that contained any information if there is a minimum amount of points that need to be $\epsilon$-close to a point $x_0$ to call this point a limit point.
So I tried to come up with a sequence that could look something like that for example:
$$a_n:=\left\{\begin{array}{ll} \frac{1}{n}, & n \not \in [m-1,m] \\ x_0, & n= m \\ x_0-\epsilon, & n=m-1\end{array}\right. \qquad n,m \in \mathbb{N}$$ (First of all I am not sure if this is even a legitimate definition of a sequence, especially the $\epsilon$ in the defintion, not to mention the correctness of the notation.)
If a sequence like this exists then this would be a null-sequence outside of the interval $[m-1,m]$ but would have a limit point at $x_0$.
Is this a correct statement or did I miss some details of the definition?
The definition I used was:
A number $p \in \mathbb{R}$ is a limit point of a set $M$, if for every $\epsilon > 0$ an element $k \in M$ exists such that $k \neq p$ and $|k-p| < \epsilon$
Alternatively I also thought about defining a $\xi>0$ which is infinitely small and then write my sequence as $a_n:=\left\{\begin{array}{ll} ... \\ x_0-\xi, & n=m-1\end{array}\right .$, so that I can get rid of the $\epsilon$ in my sequence definition. I just wanted to see if the usage of $\epsilon$ inside of a sequence is even allowed
The issue is that your sequence depends on $\epsilon$, and therefore isn't a limit point. Here's a fact:
Why is this true? We'll prove the contrapositive. Suppose that $x_0$ is a limit point and there is some $\epsilon$ so that only finitely many members of $S$ are within $\epsilon$ of $x_0$. Let's enumerate them, $s_1,\ldots,s_n$. We may then set $\delta = \min_{1 \leq j \leq n} |x_0 - s_j|$ and note that $S$ has no elements within distance $\delta/2$ from $x_0$, i.e. $x_0$ is not a limit point.