Proving B_{r} is Open: Multivariable Proofs

[bobsmiters]

I have no luck with proofs...

Prove that B$$_{r}$$ ((x$$_{0}$$, y$$_{0}$$)) = {(x,y) : || (x,y) - (x$$_{0}$$, y$$_{0}$$)|| < r} is an open set in R.

Now I know that to be an open set if and only if each of its points is an interior point and if it contains no boundary points. I would consider trying to prove it for any (a,b) $$\in$$ B$$_{r}$$

Any ideas?

 

[HallsofIvy]

I have no luck with proofs...

Prove that B$$_{r}$$ ((x$$_{0}$$, y$$_{0}$$)) = {(x,y) : || (x,y) - (x$$_{0}$$, y$$_{0}$$)|| < r} is an open set in R.

Now I know that to be an open set if and only if each of its points is an interior point and if it contains no boundary points. I would consider trying to prove it for any (a,b) $$\in$$ B$$_{r}$$

You don't need to prove both of those. If all points are interior points, then none of them can be boundary points. Standard proofs that a given set are open show that every point is an interior point.

You are starting correctly. let (a,b) be some point in B_r. Then its distance from (x₀,y₀) is strictly less than r. You need to show that there exist some neighborhood of (a,b) consisting entirely of points that are in B_r: that is, that their distance from (x₀,y₀) is also strictly less than r.

It might be a good idea to draw a picture: B_r is, of course, the disk inside the circle around (x₀,y₀) of radius r. Hint: the triangle inequality is very helpful here!

By the way, I know a professor who says that it was being able to do precisely this proof as an undergraduate that convinced him he could be a mathematician!

 

[Kummer]

Let $$x\in B(a,r)$$. Then let $$\epsilon = r - |x-a|$$. Show that $$B(x,\epsilon) \subset B(a,r)$$. To show this you need to show given any $$y\in B(x,\epsilon)\implies y\in B(a,r)$$.