I am trying to understand the following definition of a limit point:
A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_\epsilon(x)$ of $x$ intersects the set $A$ at some point other than $x$.
I am trying to understand what "other than $x$" means. I know that a limit point $x$ need not be in $A$. So, suppose we know somehow that $x \notin A$. Then, would it be valid to modify the definition of a limit point as follows?
A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_\epsilon(x)$ of $x$ intersects the set $A$ at some point in $A$.
EDIT: I found a different definition of a limit point:
A point $x$ is a limit point of a set $A$ iff $x = \lim a_n$ for some $(a_n) \subseteq A$ satisfying $a_n \neq x$ $\forall n \in \mathbb{N}$
Does this definition fail when it comes to the sequence $(a, a, \dots)$ whose limit point clearly is $a$?
Best Answer
Let $\mathbb{R}$ be given the usual metric.
The idea of limit points of a subset $A \subset \mathbb{R}$ is to say that in the neighbourhood of such points, I can find at least one other point different from that point which is also in $A$.
So if $x \notin A$, then $x$ is a limit point of $A$ iff in any $\varepsilon$-neighbourhood of $x$, I can find at least one point of $A$. This is precisely your modified definition. So yes, it should be valid.
Addendum:
For the edited part: Note the condition $a_n \neq x$ for all $n \in \mathbb{N}$ carefully. This condition precisely prevents the constant sequence that you mentioned.
If this condition is not present, then every point in a set $A$ could be a limit point of $A$. This is not necessarily true. Consider the following set \begin{equation} S = \{ 1/n : n \in \mathbb{N}\}. \end{equation} Then none of the points in $S$ is a limit point of $S$. Moreover, the only limit point of $S$ is $0$ which is not in $S$.
Finally, the definition does not fail with the constant sequence as it is sufficient to find just one such sequence that satisfies $x = \lim a_n$.