The set of all limit points of a set Given a set $A \subseteq \mathbf{R}$, let $L$ be the set of all limit points of $A$. I recently worked through the proof that $L$ is closed. However, I have a few queries thinking deeper about this result.
Since $L$ is closed, this implies that if $x$ is a limit point of $L$ then $x \in L$, which further implies that $x$ must be a limit point of $A$. Therefore, if $x$ is a limit point of $L$ then it must also be a limit point of $A$. But consider the set
$$A = \{0\} \cup \{\frac{1}{n}: n \in \mathbf{N}\}$$
Clearly, $L = \{0\}$ but does not have any limit point itself, i.e., the set of all limit points of $L$ is $\emptyset$, but clearly $\emptyset$ is not a limit point of $A$, so isn't this a counter example to the above bolded statement? 
 A: Consider.  If $A = [0,1]$ then $L = [0,1] = A$ and set of limit points of $L = [0,1] = A$.  But $[0,1]$ is not a limit point of $A$!  Contradiction?  No.  A set is not a member of itself (by ZFC that can never happen) so the set of limit points is not a limit point.

Clearly, L={0} but does not have any limit point itself, i.e., the set of all limit points of L is ∅, but clearly ∅ is not a limit point of A

Which doesn't matter because $\emptyset \not \in \emptyset = \{$limit points of $L\}$.
But since there aren't any $x \in \{$limit points of $L\} = \emptyset$ that aren't limit points (because there aren't any $x \in \emptyset$ period) the all $x \in \emptyset$ (all ZERO of them) are limit points.  They are also all french pigs farting green sausages.  (Because there are any $x \in \emptyset$ that arent french pigs farting green sausages.)
A: From the comments above.

When you say that "$\emptyset$ is not a limit point of $A$", there seems to be some confusion: sets can anyway never be limit points, only points (in the metric space) can be limit points.
What you instead need to verify is that every element of $\emptyset$ is a limit point of $A$, and this is vacuously true.
