Let $E\subseteq\ \mathbb{R}$ Prove that:
a) $E$ is open iff $E^{o} = E$
b) If $O\subseteq\ E$ and $O$ is open, then $O \subseteq\ E^{o}$
c) If $E$ is dense in $\mathbb{R}$, then $\forall\ p\ \in\ \mathbb{R},\ \exists$ a sequence $p_{n} \in\ E$ s.t $\lim_{n \rightarrow \infty} p_{n} = p$

Question

Let $E\subseteq\ \mathbb{R}$ Prove that:
a) $E$ is open iff $E^{o} = E$
b) If $O\subseteq\ E$ and $O$ is open, then $O \subseteq\ E^{o}$
c) If $E$ is dense in $\mathbb{R}$, then $\forall\ p\ \in\ \mathbb{R},\ \exists$ a sequence $p_{n} \in\ E$ s.t $\lim_{n \rightarrow \infty} p_{n} = p$

anonymous · Answer

$E^{o}$ is the set of all interior points of E. No idea if that's the conventional notation or not.

zzr0ck3r · Answer

What is your definition of open?

zzr0ck3r · Answer

is the epsilon ball definition?

anonymous · Answer

Hmm, Rudin?

anonymous · Answer

Um
" A subset of $E$ of $\mathbb{R}$ is said to be open if $E = E^{o}$

Not Rudin, it's something my professor made on his own. He combined Rudin and two other textbooks and picked bits and pieces out of each, apparently.

anonymous · Answer

Any luck on any of these proofs?

anonymous · Answer

I havent looked at them as much as I would like. So I'm still in the brainstorming stage :P

anonymous · Answer

I guess the first one seems odd to me because it seems just be restating the definition I have given to me. By definition, E is open if $E = E^{o}$. So to ask me to prove that seems weird because it seems to be true simply by definition, not by using theorems and other knowledge I have access to.

anonymous · Answer

If that's the case, you have half the proof given to you.

anonymous · Answer

I suppose the other half would require me to show the converse?

anonymous · Answer

I should've caught on to that, the definition saying if, the proof asking iff, lol.

anonymous · Answer

Right. The way I had it introduced to me is that an open set is a set whose points are all interior points. This means if $E$ is open, then $E\subset E^o$. Since a set's interior points belong to the set, you also have $E^o\subset E$, and so $E=E^o$.

anonymous · Answer

I guess conceptually then, that's only possible for an infinite set then, right?

anonymous · Answer

Not necessarily, according to this link: http://mathworld.wolfram.com/OpenSet.html "... it is possible for a more general topological set to be both finite and open."

anonymous · Answer

Alright, so Ill make sure to keep that in mind before I just throw out that claim. Hm...I think I have an idea for the 2nd one, but it kind of revolves around the idea that if a set is infinite, then it is open. But again, not sure I can just claim that. If I can then Im sure I have #2.

anonymous · Answer

I suppose its similar to my previous question. Just instead of saying open implies infinite, I guess im hopign I can claim the opposite, infinite implies open.

anonymous · Answer

Do you have the converse of the first?

anonymous · Answer

What do you mean? As in do I know how to phrase it?

anonymous · Answer

How to prove it. It's fairly simple, straight from the definition of an open set.

anonymous · Answer

Well, if O is open, then it contains all of its interior points. And since O is a subset of E, all of the interior points of O are contained in E. But then the next step, I think, would require that E is always a subset of $E^{o}$ which I guess we just said above. But....it doesnt explicitly say it in my notes, but I guess that has to be true, eh? $E \subseteq\ E^{o}$ is always true?

anonymous · Answer

Like, I guess you can't ever say $|E^{o}| < |E|$?

anonymous · Answer

Hmm, I think you're going about it wrong. Try showing that $O^o\subset E^o$. Then you'll know that since $O$ is open, you have $O=O^o$ by the first proof, and so $O\subseteq E^o$.

anonymous · Answer

Right. I guess I thought that required me to be certain that $E \subseteq\ E^{o}$ was always true. I wasn't sure I could make the jump from $O \subseteq\ E$ to $O^{o} \subseteq\ E^{o}$ Without that n between step. Um....Oh, but you dont want me to go from $O$ to $O^{o}$, you think I need to start with $O^{o}$?

anonymous · Answer

Start off with some interior point $p$ of $O$. Using the definition of an interior point, you have a neighborhood of $p$, call it $N$, where $N\subset O$.

Now since $O\subset E$, you have $N\subset E$, so $p\in E^o$. This means $O^o\subset E^o$. The rest of the proof follows from here.

anonymous · Answer

Okay, using the definition of interior point let's me use the neighborhood idea, which will allow me to safely say my chosen element is in $E^{o}$ I think I feel more safely about that then. That probably outlines that well enough. 

I guess now I'm trying to understand the idea of "dense." The way it's defined for me is:
"Let E be a subset of $\mathbb{R}$. E is said to be dense in $\mathbb{R}$ if $E \cup\ E' = \mathbb{R}$."
So, since E' is the set of all limit points of E, I would think that all $E \cup\ E'$ really means is that you must include the endpoints of the set E. Which I guess reads to be as in you need a closed interval that equals $\mathbb{R}$. But I know this interpretation is wrong since it is noted that $\mathbb{Q}$ is dense in R. So maybe I'm misunderstanding limit points or something, but I'm missing something about the definition.

anonymous · Answer

I never really had a good idea of what dense meant myself, or maybe I did but it's been too long... in any case, I suggest watching these lectures, they really helped me out: http://www.youtube.com/watch?v=Ebnoxgp8mLM http://www.youtube.com/watch?v=6zs_PTUfKAk The prof doesn't really delve into the definition of dense until the last 4 or 5 minutes of the second lecture, but it might give you a better idea of what dense means. (The other parts of the videos make up a really nice comprehensive review of the other definitions.)

anonymous · Answer

Alright, Ill look them over and go from there. Thanks as always, Sith :)

anonymous · Answer

Happy to help!

anonymous · Answer

^_^