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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 ?

Josh
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Kingkong
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    It's good attempt but change the order. If $p\in F^c$, since $F^c$ is open, there exists $N_{\epsilon}(p)$ such that contained in $F^c$. It contradicts as you see. – Arturo Dec 07 '13 at 01:23
  • is $N_]epsilon \in F^c$ this mean $F^c$ is open . and F is close. this does not contradict the original statement or do I misunderstood u – Kingkong Dec 07 '13 at 02:22
  • Complement of open set is closed set and complement of closed set is open. It's just trivial truth. Then, as I say, since $F^c$ is open, there exists $\epsilon$ such that $N_{\epsilon}(p)$ is contained in $F^c$. But since $p$ is limit point, it's contradiction since there exists infinitely many $p_n$ near $p$. – Arturo Dec 07 '13 at 02:27

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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 $p$?

Josh
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  • $N_\epsilon \in F^c$ since F^c is open. then this imply $$p \cap F = empty $$.But since the statement implies that p is the limit of the sequence in F. statement $$p \cap F = empty $$ is wrong i guess – Kingkong Dec 07 '13 at 04:21
  • Do you mean to say $N_\epsilon(p) \cap F = \varnothing$? Yes, that is how you get the contradiction. – Josh Dec 07 '13 at 04:28