What is an effective approach to proving that interior points are open?

  • Thread starter Thread starter pokemonsters
  • Start date Start date
  • Tags Tags
    Interior Points
pokemonsters
Messages
2
Reaction score
0

Homework Statement


For S \subset Rn, prove that S° is open.

Homework Equations


S° are all interior points of S.

The Attempt at a Solution


My class has only learned how to use balls to solve these types of problems (no metric spaces). So I need to choose an ε > 0 so that Bε(x) \subset S°, where x is any arbitrary point in S°. To show this is true, let y \subset Bε(x) be arbitrary. (then I don't know how to progress further...how do I show that the neighbourhood contains only points in S°?)
 
Physics news on Phys.org
pokemonsters said:

Homework Statement


For S \subset Rn, prove that S° is open.

Homework Equations


S° are all interior points of S.

The Attempt at a Solution


My class has only learned how to use balls to solve these types of problems (no metric spaces). So I need to choose an ε > 0 so that Bε(x) \subset S°, where x is any arbitrary point in S°. To show this is true, let y \subset Bε(x) be arbitrary. (then I don't know how to progress further...how do I show that the neighbourhood contains only points in S°?)
Start by choosing an arbitrary x \in S^o. By definition this is an interior point of S, so there exists \epsilon > 0 such that B_\epsilon(x) \subset S. Now, if you can show that every point y \in B_\epsilon(x) is an interior point of S then you're done. To do this, it certainly suffices to show that you can fit a smaller ball B_\delta(y) around y which is entirely contained within B_\epsilon(x), because then you will have y \in B_\delta(y) \subset B_\epsilon(x) \subset S. Try drawing a picture to see how to define \delta, the radius of the smaller ball.
 
jbunniii said:
Start by choosing an arbitrary x \in S^o. By definition this is an interior point of S, so there exists \epsilon > 0 such that B_\epsilon(x) \subset S. Now, if you can show that every point y \in B_\epsilon(x) is an interior point of S then you're done. To do this, it certainly suffices to show that you can fit a smaller ball B_\delta(y) around y which is entirely contained within B_\epsilon(x), because then you will have y \in B_\delta(y) \subset B_\epsilon(x) \subset S. Try drawing a picture to see how to define \delta, the radius of the smaller ball.

Thank you, I did not notice that you can put another ball inside the ball to make the proof work.

Using this, I was able to make a series of inequalities using the triangle inequality, and managed to prove that y \in B_\delta(y) \subset B_\epsilon(x) \subset S.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top