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)$\cap$S is non-empty and B(p;r)$\cap$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.

 

[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 $$x^2+y^2>1$$ 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 $S=\{(x_1,x_2):x_1^2+x_2^2< 1\}$ 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!