How does adding $0$ to the set $\mathbf A = \bigl\{\frac{1}{n}: n \in \mathbb N \bigr\}$ make it a closed set? By definition, a closed set is a set that contains its limit points. However, by the time the closed set contains its limit points, those points are no longer limit points and become isolated points. For example:
$\mathbf A = \{\frac{1}{n}: n \in \mathbb N \}$. The limit of this set (set $\mathbf A$) is clearly equal to $0$. This is because the $\epsilon$ -neighborhood $\mathbf V_{\epsilon}(0) \cap \mathbf A = \{\frac{1}{n} \}$, and $\frac{1}{n} \neq 0$. However, when $0$ is included, the $\epsilon$ -neighborhood $\mathbf V_{\epsilon}(0) \cap \mathbf A = \{0 \}$ for $\mathbf A=[0,\frac{1}{n} ]$. This will contradicts the definition of limit point of set A and hence $0$ must be an isolated point.
Another example: $\left(a,b\right)$ is an open interval with limit $a$ and $b$. Then its closure $\bar A$ will be $\left[a,b\right]$. By definition, the $\epsilon$ -neighborhood of any point in $\left[a,b\right]$ intersects the closure $\bar A$ at that same point, and hence, no points in that closure set is a limit point: A contradiction that closure sets are closed sets.
Also, I am trying to prove the lemma: If x is a limit point of $A \subseteq A'$, then x is limit point of $A'$.
Proof: Suppose x is a limit point of $A$, then there exists a sequence $(a_n)$ $\subset A \subseteq A'$: lim($a_n$)=x with $a_n$ $\neq x \forall n \in \mathbb N$. Then since $(a_n)$ $\subset A'$, it follows that x must be a limit point of $A'$.
So my questions are:
1. What is wrong with my contradiction in the 2 examples? Please explain them to me.
2. Is my proof for the lemma correct? I am going to use it for the proof that closure set is closed.
My background: I am studying elementary Real Analysis by starting with Abbot. I thank you very much for your help.
Extra question: We have this theorem: x is a limit point of set $A$ if and only if there exists a sequence $(a_n) \subset A$ such that $\lim (a_n)=x$ $\forall a_n \neq x$. So, for some finite $n \in \mathbb N$ such that $a_n = x$, x is still a limit point of set A . Is this correct? I thought that x would be an isolated points since we need $a_n \neq x \forall n \in \mathbb N$
I thank you again for your answers.

In the first case your neighbourhood of zero includes $\frac 1{n+1}, \frac 1{n+2} \dots$
In the second case I'm not sure what you mean. Any neighbourhood of a point in $[a,b]$ (we take $b\gt a$) intersects $[a,b]$ in (a set containing) an interval around the point. And any point within $[a,b]$ is a limit point.
