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Proving closure and boundary points

  • Thread starter muzak
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  • #1
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Homework Statement


Let S = {(x,y): x[itex]^{2}[/itex]+y[itex]^{2}[/itex]<1}. Prove that [itex]\overline{S}[/itex] is (that formula for the unit circle) [itex]\leq[/itex] 1 and the boundary to be x[itex]^{2}[/itex]+y[itex]^{2}[/itex]=1.


Homework Equations


Boundary of S is denoted as the intersection of the closure of S and the closure of S complement.
p [itex]\epsilon[/itex] boundary of S iff for every r > 0, B(p;r)[itex]\cap[/itex]S is non-empty and B(p;r)[itex]\cap[/itex]S complement is non-empty.


The Attempt at a Solution


I understand this conceptually and it's obvious that the boundary and closure are those equations respectively but I don't know how to translate that into a math proof. I wasn't exactly given any concrete examples and how to apply the theorems into a proof, was only presented with the theorems.
 

Answers and Replies

  • #2
Office_Shredder
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We know that the closure of S contains all of S, so you just need to show that it also contains points with [tex]x^2+y^2=1[/tex] as well. There are two parts to this:
1) Show if [tex]x^2+y^2=1[/tex] then [tex](x,y)\in \overline{S}[/tex]
2) Show if [tex]x^2+y^2>1[/tex] then [tex] (x,y)\notin \overline{S}[/tex]

It may help to write down different equivalent definitions of the closure of S when approaching this problem
 
  • #3
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Hi!

I am also stuck on this question. Could we also show the set [itex]\overline{S}=\{(x_1,x_2):x_1^2+x_2^2\le 1\}[/itex] is the closure of [itex]S=\{(x_1,x_2):x_1^2+x_2^2< 1\}[/itex] by showing that [itex](1)[/itex] [itex]\overline{S}[/itex] is closed, and [itex](2)[/itex] each point in [itex]\overline{S}[/itex] is in the closure of [itex]S[/itex]? To me, that would show that [itex]\overline{S}[/itex] is the smallest closed set that contains [itex]S[/itex], since adding more elements to [itex]\overline{S}[/itex] would result in a larger closed set that contains [itex]S[/itex]. Is this correct thinking?

Thanks!

Bijan
 
Last edited:
  • #4
HallsofIvy
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Yes, that would work.
 
  • #5
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Thanks, I believe I figured it out!
 

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