Prove that the limit of a sequence in a closed set $F$ is in $F$ Say $F$ is a closed subset of $\mathbb{R}$
and let ${p_n}$ be a sequence in $F$ converging to $p \in \mathbb{R}$.
We want to prove $p \in F$.
I think I did some point wrong but can not figure out:
If ${p_n}$ converges then there exist $\epsilon$ >0, and there exist $p_n \in N_\epsilon(p)$  for all $n\geq n_0$
Assume p $\notin F $, then $p \in F^c $.
Then since $p$ is the limit point of sequence. Can i say $N_\epsilon(p) \in F^c$ ???
The $F^c$ is open. It is concluded that $F$ is closed. 
Failed contradiction. 
Which part should I fix ? 
 A: Okay, let's first fix some slight problems with your quantifiers (i.e. your "there exists" statements).

"if $p_n$ is converges then there exist $\epsilon >0$, and there exist $p_n \in N_\epsilon(p)$ for all $n\geq n_0$"

This is not quite what it means for the sequence $p_n$ to converge. What you mean to say is:

For all $\epsilon >0$, there exists $n_0$ such that $p_n \in N_\epsilon(p)$ for all $n\geq n_0$.

(I'm guessing you might have seen this written as
$$\forall \epsilon>0\ \exists n_0 \in \mathbb{N} \ \text{such that} \ \forall n>n_0, \ p_n \in N_\epsilon(p),$$
but that's just notation.)
Your approach after that is right: assume $p\notin F$, so $p \in F^c$.
What comes after is slightly off though.
$p$ is a limit point of $p_n$, but this is not what implies $N_\epsilon(p) \subset F^c$.
Remember what was important about $F$: it is closed. So $F^c$ is ___? (Fill in the blank yourself.)
Since $p\in F^c$, and $F^c$ is open, this tells us that there is an open neighborhood around $p$ contained entirely in $F^c$, call it $N_\delta(p)$ for some $\delta >0$.
Now you can hopefully finish your contradiction. There is a neighborhood of $p$ that only contains points in $F^c$, how does this contradict the fact that $p_n$ converges to $p$?
