# Closed set equivalence theorem

1. Dec 10, 2009

### ssayan3

1. The problem statement, all variables and given/known data
Hi guys, this problem gave me some trouble before, but I'd like to know if I have it worked out now.....

"If S = S$$\cup$$BdyS, then S is closed (S$$_{compliment}$$ is open)

2. Relevant equations
S is equal to it's closure.

3. The attempt at a solution
1. Pick a point p in S$$^{compliment}$$.
2. For all points q$$^{n}$$ in S, let $$\delta$$ = min{|p-q$$^{n}$$|}
3. Dfine B(p,$$\delta$$)
4. For any point in S$$^{compliment}$$, we can produce $$\delta$$>0 such that B(p,$$\delta$$)$$\subset$$S$$^{compliment}$$. Therefore, all points in S$$^{compliment}$$ are interior points; therefore, S$$_{compliment}$$ is open, and S is closed.

2. Dec 10, 2009

### VeeEight

I am having a hard time figuring out what you mean by "S$$\cup$$BdyS"

3. Dec 10, 2009

### ssayan3

Oh! I'm sorry if I was unclear about something.....

"S$$\cup$$BdyS" refers to the union of the set S with its boundary, and is called Closure(S)

It can also be referred to as the union of the interior of S with the boundary of S.

(Someone tell me if I made an error!)

4. Dec 11, 2009

### HallsofIvy

Staff Emeritus
How do you know there exist such a minimum? Not every set has a minimum. Every set of real numbers, bounded below, has an infimum (greatest lower bound). But then how do you know it is not 0?

5. Dec 11, 2009

### ssayan3

Argh, that makes it even worse for me because now I really have no idea how to finish this one.

What I have so far is:
1. Pick a point z in Compliment(S)
2.Then, z is not in S, and is not in Closure(S) by hypothesis