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

[pokemonsters]

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 B_ε(x) $\subset$ S°, where x is any arbitrary point in S°. To show this is true, let y $\subset$ B_ε(x) be arbitrary. (then I don't know how to progress further...how do I show that the neighbourhood contains only points in S°?)

 

[jbunniii]

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 B_ε(x) $\subset$ S°, where x is any arbitrary point in S°. To show this is true, let y $\subset$ B_ε(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.

 

[pokemonsters]

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 $y \in B_\delta(y) \subset B_\epsilon(x) \subset S$.