Suppose $A \subset \mathbb{R}$ let L be the set of limit points of A, show that doesn't produce new limit points I just wanted to check if my proof is correct.
Given $A \subset \mathbb{R}$, let L be the set of all limit points of A.
Show that L is closed set.
Suppose that $\{x_n\}_{n = 1}^{\infty} \subset L$ s.t $x_n \longrightarrow x$ we wish to show that $x \in L$. Since each $x_i \in L$ is a limit point of some sequence $\{y_j^{i}\}_{j = 1}^{\infty}$ so that $y_j^i \longrightarrow x_i \longrightarrow x \implies y_u^u \longrightarrow x$ as u goes to infinity. As $\{y_m^m\}_{m = 1}^{\infty} \subset A \implies x \in L$.  I am actually stuck in the step where I want to show that $y_m^m \longrightarrow x$ as m approaches infinity. Any help in that step would be nice.
I wanted to check with everyone if my proof is correct. 
 A: I assume you are trying to prove the below statement:
Given $A \subset \mathbb{R}$, let L be the set of all limit points of A. Show that L is closed set.
You want to show that $L$ is a closed set. One way to do this is to show that $x$ is also a limit point of $A.$ Let $\epsilon > 0 $ be arbitrary. 
Let $\{x_n\}_{n = 1}^{\infty} \subset L$ be such that $x_n \longrightarrow x$.  Show that $(x - \epsilon, x+ \epsilon)$ intersects $A$ in a point other than $x$. Choose $i$ such that $\forall j > i$, $|x_j - x| < \frac{\epsilon}{2}$. $ U = (x_j -\frac{\epsilon}{2}, x_j + \frac{\epsilon}{2}) $ is a subset of $$(x - \epsilon, x+ \epsilon)$$
that intersects $A$ in $x^{*}$ due to $x_j$ being a limit point. If $x^* \ne x$, we are done. If $x^* = x$, then shrink $U$ to exclude $x$ and call this interval $U'$. $U'$ intersects $A$ in a point other than $x$ and $x_j$ and is a subset of $(x - \epsilon, x+ \epsilon)$. Thus, $x$ is a limit point of $A$. Hence $x\in L$.
A: Tyring to follow your proof, I cannot find a way to showing that $y_u^u \to x$.
But there is an argument quite close to your approach:
For any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\forall i \geq N$, 
$|x_i - x| < \epsilon/2$.
Furthermore, for a fixed $i$, there exists an $M_i \in \mathbb{N}$ such that $\forall j \geq M_i$, 
$|y_i^j - x_i| < \epsilon/2$.
So, there exists a sequence $y_i^{M_i} \in A$, for which we know, when $i \geq N$
$$
|y_i^{M_i} - x| \leq |y_i^{M_i} - x_i| + |x_i - x| < \epsilon.
$$
That is $x$ is a limit point of $A$.
A: You don't need to  show that $y_n^n\to x$. You only need to find a sequence $z_{k}$ in $A$ such that $z_k\to x$. Here is one way you may use. We first select 
a subsequence of $\{x_n\}_{n=1}^\infty$, which is still denoted by  $\{x_n\}_{n=1}^\infty$,, such that $|x_n-x|\leq \frac{1}{2^n}$. Then for each $x_n$, we can select $z_n\in A$ such that $|z_n-x_n|\leq \frac{1}{2^n}$. This is possible since each $x_n$ is the limit of some sequence $\{y^n_k\}$. You can just select one $y^n_k$ such that $|y^n_k-x_n|\leq \frac{1}{2^n}$ and denote it by $z_n$.
Then the sequence $\{z_n\}\subset A$ and 
$$
|z_n-x|\leq |z_n-x_n|+|x_n-x|\leq \frac{1}{2^{n-1}}.
$$
So it converges to $x$. 
A: Given that $L$ is the set of limit points of $A\subset\mathbb{R}$ show that there is no point $x$ which is a limit point of $L$ which is not also a limit point of $A$.
Suppose such an $x$ exists.
Let $x_1,x_2,x_3,\cdots \longrightarrow x$ where each $x_i\in L$.
Since by assumption $x$ is not a limit point of $A$ there is an $\epsilon$ neighborhood $N_\epsilon(x)$ of $x$ which contains no point of $A$. But since $x$ is a limit point of the sequence $\{x_i\}$, $N_\epsilon(x)$ contains a point $x_k$ of the sequence. But $x_k$ is a limit point of $A$ so $N_\epsilon(x)$  must contain a point of $A$.
This contradicts the assumption that such a point $x$ exists.
Thus every limit point of $L$ must also be a limit point of $A$.
