# Closed Sets and Convergent Sequences

• Aug 10th 2010, 08:41 AM
RedBarchetta
Closed Sets and Convergent Sequences
I was reading this theorem in my analysis text and came across something I could not quite follow:

Theorem: Let S be a subset of the metric space E. Then S is closed if and only if, whenever $p_1, p_2, ...$ is a sequence of points of S that is convergent in E, we have $\lim_{n\to\infty} p_n \in S$.

I understand the "only if" part. Now for the "if" part the text states:

"Suppose S is not closed. Then the complement is not open, and there exists a point p in the complement of S such that any open ball of center p contains points of S. *Hence for each positive integer n we can choose $p_n \in S$ such that $d(p, p_n)<\frac{1}{n}$.Then $\lim_{n\to\infty} p_n = p$, with each $p_n \in S$ and $p \notin S$.*This shows that if the hypothesis on convergent sequences holds, then S must be closed, completing the proof. "

^^ I've bolded the text that I don't understand. I've thought about this and I can't figure out where the 1/n comes from as well as the conclusion $\lim_{n\to\infty} p_n = p$. Thanks for the help!
• Aug 10th 2010, 08:49 AM
Ackbeet
The $1/n$ is like an epsilon. If a set is not open, then there is a point in that set such that every $\varepsilon$-neighborhood around the point contains points not in the set, correct? In this case, we're considering $S^{c}$, which by hypothesis is not open. The point $p\in S^{c}.$ But now the author wants to create a sequence of points in a sequence of index-able neighborhoods, so instead of using an $\varepsilon,$ he uses the equivalent sequence $1/n$ in order to be able to index them. The sequence $p_{n}\in S,$ and hence is not in $S^{c},$ by construction.

Does that make sense?
• Aug 10th 2010, 09:07 AM
RedBarchetta
I think I am getting confused on the use of n. Specifically, are we choosing a positive integer n and then considering all p_n such that $d(p, p_n)<\frac{1}{n}$ ?
• Aug 10th 2010, 09:22 AM
Plato
Quote:

Originally Posted by RedBarchetta
I think I am getting confused on the use of n. Specifically, are we choosing a positive integer n and then considering all p_n such that $d(p, p_n)<\frac{1}{n}$ ?

Actually you are constructing a sequence of points in $S$ such that the sequence converges to $p\notin S$ which contradicts the given.

Now why? If $S^c$ is not open then for some $p\in S^c$ there is no ball at $p$ which is a subset of $S^c$.
For each $n$ the ball $B\left( {p;\frac{1}{n}} \right)$ is not a subset of $S^c$.
Thus $\left( {\exists p_n \in \left( {B\left( {p;\frac{1}
{n}} \right) \cap S} \right)} \right)$

Does that help?
• Aug 10th 2010, 10:14 AM
RedBarchetta
I guess my issue is taking these sets and using them to prove that this point p is the limit of the sequence of $p_n$.

So by definition: a point $p \in E$ is called a limit of the sequence $p_1, p_2, ...$ if, given any real number $\epsilon > 0$, there is a positive integer $N$ such that $d(p, p_n)< \epsilon$ whenever $n>N$.

How do we take this collection of open balls and prove using the definition that the limit is indeed p? So why can we use just 1/n rather than just some arbitrary real number? how do we know we can find big N? Can S be a finite set? or is an infinite set? I guess I have too many questions!

So are you saying when you consider a set B(p, 1/n) for some n, then there exists some p_n in S that is in that open ball? Say we let n=1, then are you saying that p_1 is in B(p, 1)? or just some arbitrary p_n? could it be just one p_n say p_3? or possibly multiple p_n's say p_3, p_4?

Sorry for not getting this, I feel like i'm missing something. (Headbang)
• Aug 10th 2010, 10:38 AM
Plato
Quote:

Originally Posted by RedBarchetta
So are you saying when you consider a set B(p, 1/n) for some n, then there exists some p_n in S that is in that open ball? Say we let n=1, then are you saying that p_1 is in B(p, 1)? or just some arbitrary p_n? could it be just one p_n say p_3? or possibly multiple p_n's say p_3, p_4?
Sorry for not getting this, I feel like i'm missing something.

You are missing a great deal. Or maybe you don’t understand sequential limits.

First, did you notice we said for each n?
That means for each n we get a $p_n$.

Moreover, for each $\varepsilon >0$, $\left( {\exists n} \right)\left[ {\frac{1}{n} < \varepsilon } \right]$.
Thus $d\left( {p,p_n } \right) < \frac{1}{n} < \varepsilon$.
That means $\left( {p_n } \right) \to p \notin S$.