Closed set equivalence theorem

  • Thread starter ssayan3
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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[tex]\cup[/tex]BdyS, then S is closed (S[tex]_{compliment}[/tex] 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[tex]^{compliment}[/tex].
2. For all points q[tex]^{n}[/tex] in S, let [tex]\delta[/tex] = min{|p-q[tex]^{n}[/tex]|}
3. Dfine B(p,[tex]\delta[/tex])
4. For any point in S[tex]^{compliment}[/tex], we can produce [tex]\delta[/tex]>0 such that B(p,[tex]\delta[/tex])[tex]\subset[/tex]S[tex]^{compliment}[/tex]. Therefore, all points in S[tex]^{compliment}[/tex] are interior points; therefore, S[tex]_{compliment}[/tex] is open, and S is closed.
 
I am having a hard time figuring out what you mean by "S[tex]\cup[/tex]BdyS"
 
Oh! I'm sorry if I was unclear about something.....

"S[tex]\cup[/tex]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!)
 

HallsofIvy

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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[tex]\cup[/tex]BdyS, then S is closed (S[tex]_{compliment}[/tex] 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[tex]^{compliment}[/tex].
2. For all points q[tex]^{n}[/tex] in S, let [tex]\delta[/tex] = min{|p-q[tex]^{n}[/tex]|}
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,[tex]\delta[/tex])
4. For any point in S[tex]^{compliment}[/tex], we can produce [tex]\delta[/tex]>0 such that B(p,[tex]\delta[/tex])[tex]\subset[/tex]S[tex]^{compliment}[/tex]. Therefore, all points in S[tex]^{compliment}[/tex] are interior points; therefore, S[tex]_{compliment}[/tex] is open, and S is closed.
 
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
 

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