Proving closure and boundary points

[muzak]

Homework Statement

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

Homework Equations

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

 

[Office_Shredder]

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

It may help to write down different equivalent definitions of the closure of S when approaching this problem

 

[Post-its]

Hi!

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

Thanks!

Bijan

 

[HallsofIvy]

Yes, that would work.

 

[Post-its]

Thanks, I believe I figured it out!