Proving the open ball of radius 1 is open in R^2

[rman144]

I need to prove that the following is an open subset of R^2:

$$\left\{(x,y)\in$$R$$^{2}$$|$$\sqrt{x^2+y^2}$$<1}


I think the substition r=min{sqrt[x^2+y^2],1-sqrt[x^2+y^2]} works, but I'm stuck on how to take it from that to showing that the distance between X0 and X1 is less that r, and more importantly, proving that this means that the subset is open. Any help would be must appreciated.

 

[snipez90]

Isn't this just an open disc centered at the origin of radius 1? The basic idea is to take a point (a,b) in that set, and note that its distance from the origin, which we can denote d((a,b),(0,0)) is less than the radius of the original set, which is 1. If we then take an open disc of radius r = 1 - d((a,b),(0,0)) centered at (a,b), then intuitively this disc will lie entirely in the original set. Proving this will probably be a bit more annoying, since you're dealing with the Euclidean metric instead of a general metric.

 

[HallsofIvy]

What is your definition of "open set"? How you prove a set is open depends strongly on exactly what the definition is.