How do I prove the closure and the boundary of a concrete example?

1. Sep 25, 2011

amanda_ou812

1. The problem statement, all variables and given/known data
Let X = R2 with the Euclidean metric and let S = {(x1, x2) : x1^2+x2^2 <1}.Prove that Closure of S ={(x1,x2):x1^2+x2^2<= 1} and that the Boundary of S= { (x1, x2) : x1^2 +x2 ^2=1 } .

2. Relevant equations

3. The attempt at a solution
I was able to prove all my theorems but I don't know how to prove this concrete example. All my theorems just talked about one point in the closure or boundary. I can clearly see from a picture that these are the answers but how do I prove it without a picture?

2. Sep 25, 2011

Dick

Pick P=(x,y) such that x^2+y^2=1. Now take the ball B(P,r). (1-r/2)*P is in the ball and S and (1+r/2)*P is in the ball and outside of S. Can you prove that? That would show P is in boundary of S, right? Does that help?

3. Sep 25, 2011

amanda_ou812

What does (1-r/2)*P and (1+r/2)*P mean. Its probably something really simple but I am just not seeing it.

Yes, I can see how showing that an arbitrary point that is both in S and in its complement would show that it is a boundary point. Would this method suffice if my teacher uses the definition of boundary as the intersection of the closure of S with the complement of the closure of S?

I suppose I could show that this point is in the closure of S by showing it is in the intersection of all closed sets containing S (which is my teacher's definition of closure). Ok, let me think of how to do that...

4. Sep 25, 2011

Dick

P*(1-r/2)=(x,y)*(1-r/2)=(x*(1-r/2),y*(1-r/2)). Just a vector product. Showing points P that are in the boundary of S are the points where every open neighborhood of P contains points that are both in S and outside of S is equivalent to your teachers definition of boundary is a 'proof' question. You said you could do those.

Last edited: Sep 25, 2011
5. Sep 25, 2011

amanda_ou812

So, how did you choose those numbers? (1-r/2) and (1+r/2). I would like to know so that can learn how to do this type of problem.

And, I show that (1-r/2)P is in B(p, r) by showing that d((1-r/2)p, y)<r and I show that it is in S by showing that (1-r/2)P and y satisfies the condition of S (x1^2+x2^2 <1). And, I show that (1+r/2)P is in B(p, r) by showing that d((1+r/2)p, y)<r and I show that it is in S-complement by showing that (1+r/2)P and y satisfies the condition of S-complement (x1^2+x2^2 >= 1).

6. Sep 25, 2011

Dick

The distance between P and (1-r/2)*P=|P-(1-r/2)*P|=|P||1-(1-r/2)|=(r/2)|P|. |P|=1 since x^2+y^2=1. r/2<r. That means (1-r/2)*P is in B(P,r). I'm just using vector arithmetic on R^2 with the usual norm (i.e. the Euclidean metric). And |(1-r/2)*P|=|1-r/2|*|P| and since |P|=1 and |1-r/2|<1, then (1-r/2)*P is in S. You don't have to square everything out. Just picture multiplying vectors by constants.

Last edited: Sep 25, 2011