# Interior point, neighborhood question

1. Nov 16, 2008

### kathrynag

1. The problem statement, all variables and given/known data
If D$$\subset$$R, then x$$\in$$D is said to be the interior point of D iff there is a neighborhood Q of x such that Q$$\subset$$D. Define $$D^{\circ}$$ to be the set of interior points of D. Prove that $$D^{\circ}$$ is open and that if S is any open set contained in D, then S$$\subset$$$$D^{\circ}$$. $$D^{\circ}$$ is called the interior of D.

2. Relevant equations

3. The attempt at a solution
So, we have a neighborhood [x-$$\epsilon,x+\epsilon$$].

2. Nov 16, 2008

### VeeEight

Take the real numbers as a general example since it is very simple.
Is a neighborhood open? Is the union of open sets open? These are things you should investigate in order to understand the interior of a set.

3. Nov 16, 2008

### kathrynag

neighborhood is open since it includes the real numbers and the real number line is open. union of open sets is open, I think.

4. Nov 16, 2008

### HallsofIvy

Staff Emeritus
The set of all real numbers is open but that was not the question! Is the interval: (a, b)= {x| a< x< b} open? If p is a point in (a, b), look at the distance, da, from p to a and the distance, db[/sup], from p to b. What can you say about $(d- \epsilon, p+ \epsilon)$ where d is the smaller of those distances?

Now your original post didn't say anything about the real numbers. Are you working only in the real numbers, or, more generally, in Rn, or, even more generally, in an abstract metric topology?

5. Nov 16, 2008

### kathrynag

I guess (a,b) would be closed. I'm not entirely sure what to say about this. $(d- \epsilon, d+ \epsilon)$ except this is the neighborhood Q of x.

We're only working wuth real numbers. That's what the R in my original post signified.

6. Nov 16, 2008

### VeeEight

The real number line is open but a subset of it does not have to be open. For example, the interval [0,1]. You may want to review the definition of open and picture it as a physical thing to help out your thinking process.

In R, the union of two open sets is open. How about more than two? You may want to try to do this as an exercise as it directly relates to the interior of a set. What is the interior of an open set?

7. Nov 16, 2008

### kathrynag

I would say more than 2 open sets will still be open.
Then the interior of an open set is all points included in that set.

8. Nov 16, 2008

### VeeEight

Okay. You may want to try proving your first statement (it is usually a result in real analysis). It is important to understand the proofs so you know the details and can understand the details about neighborhoods & open sets. Your second statement makes sense so you may try using these facts to solve your question.

9. Nov 16, 2008

### kathrynag

Ok, so I need to prove more than 2 sets will still be open.
A set A$$\subset$$R is open iff for each x$$\in$$A there is a neighborhood Q of x such that Q$$\subset$$A.
So for each open set, there is a neighborhood Q of x. Becuase for each set Q$$\subset$$A then there is a neighborhood. Thus, still open?

10. Nov 16, 2008

### VeeEight

Let O be the union of your open sets (union O_1, O_2, ...). Then for an arbitrary element of O, call it a, there is some positive integer i such that a neighborhood of a is contained in O_i since all of the O_i's are open. Thus the union is open.

11. Nov 16, 2008

### kathrynag

Ok, so we got that. Now do I go on to showing S$$\subset$$$$D^{\circ}$$?

12. Nov 16, 2008

### VeeEight

I believe that we have gone through the tools needed to solve the question so it is up to you to put them together.

13. Nov 19, 2008

### kathrynag

Ok, I sorta have an idea.
Let D be arbitrary.
Then Q$$\subset$$D will be the interior of D.
Let $$D^{\circ}$$=Q$$\subset$$D.
We wnat to show $$D^{\circ}$$ is open.

This is my start.

14. Nov 20, 2008

### HallsofIvy

Staff Emeritus
How did you get off onto proving that the union of open sets is open? The original problem was to prove that the interior of a set is open and that has nothing to do with a union of open sets.

15. Nov 20, 2008

### kathrynag

Ok, so I have to use facts about interior points.
The interior is Q$$\subset$$D where D is arbitrary, right?