Closed set equivalence theorem

[ssayan3]

Homework Statement

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\cupBdyS, then S is closed (S_{compliment} is open)

Homework Equations

S is equal to it's closure.

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)\subsetS^{compliment}. Therefore, all points in S^{compliment} are interior points; therefore, S_{compliment} is open, and S is closed.

 

[VeeEight]

I am having a hard time figuring out what you mean by "S\cupBdyS"

 

[ssayan3]

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

"S\cupBdyS" 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!)

 

[HallsofIvy]

Homework Statement

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\cupBdyS, then S is closed (S_{compliment} is open)

Homework Equations

S is equal to it's closure.

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}|}

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?

3. Dfine B(p,\delta)
4. For any point in S^{compliment}, we can produce \delta>0 such that B(p,\delta)\subsetS^{compliment}. Therefore, all points in S^{compliment} are interior points; therefore, S_{compliment} is open, and S is closed.

 

[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