Prove that a set is an open set

[Andrés85]

Homework Statement

Show that {(x,y)ℝ^2/1<x^2+y^2<7} is an open set.

Homework Equations

The Attempt at a Solution

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[Chaos2009]

Sometimes a picture helps me a lot with these kinds of problems, so let's try and see what this set looks like.

Well, we know that x^2 + y^2 is the points distance from the origin squared, so it should look something like a donut with outer radius \sqrt{7} and inner radius 1. Also, since those are not equal, the boundaries there are not contained in this set.

There are a couple different definitions for open, but when talking about metric spaces, to say a set is open typically means that for any point in the set, you can find an open ball around that point that is contained within the set. So, if I were to give you a point \left( x, y \right) in this donut, could you find a radius small enough such that a ball of that radius centered at that point was entirely contained in the donut?

 

[Andrés85]

I think r < min {√7-|(x,y)| ; |(x,y)| - 1} works.

I tried to separate the problem in two sets, because the intersection of two open sets is an open set, and proved that x^2+y^2<7 is open, but the method I used with that proof don't work with the set 1 < x^2 + y^2

 

[Chaos2009]

That radius r is good. You may need to also prove that such an r > 0 exists for each (x, y) in the set, but that shouldn't be too bad.

Also, what method did you use to show that x^2 + y^2 < 7 is open?

 

[Andrés85]

if B(x,y) is an open ball with center (x,y) and radius r, i show that if (a,b) belongs to the ball B, then (a,b) belongs to the set, so a^2 + b^2 < 7

|(a,b)| = |(a,b) - (x,y) + (x,y)| = |(a-x, b-y) + (x,y)| ≤ |(a-x, b-y)| + |(x,y)|

|(a-x, b-y)| < r, so

|(a-x, b-y)| + |(x,y)| < r + |(x,y)|

r = √7 - |(x,y)|, so

r + |(x,y)| = √7 - |(x,y)| + |(x,y)| = √7

Finally, I get

|(a,b)| < √7

√(a^2 + b^2) < √7

a^2 + b^2 < 7

and that proves that if (a,b) is in the ball then (a,b) is in the set x^2 + y^2 < 7

 

[Chaos2009]

Okay, well the proof that \left\{ \left( x, y \right) \in \mathbb{R}^{2} : x^{2} + y^{2} &gt; 1 \right\} should be similar.

Perhaps if you consider that \left| \left( x, y \right) \right| &lt; \left| \left( x, y \right) - \left( a, b \right) + \left( a, b \right) \right|. It's similar to what you were using before, but you're using the triangle inequality in a different order.

 

[Andrés85]

I solved it that way! Thanks!