How can we prove that a given set is open in R^n without making any assumptions?

[rumjum]

Homework Statement

In general, in R^n, what is the best way to approach the problem - a given set is open?
The given set E is such that for all x,y that belong to the given set, d(x,y) < r.

Homework Equations

The Attempt at a Solution

let x be the center of the sphere and y be any point such that d(x,y) < r. Now, let z be any boundary point such that d(x,z) = r.
Also let d(y,z) < epsilon. We can make a neighborhood N with epsilon as radius and y as point such that all points of N are subset of the given set. In general we can construct a neighborhood N of smallest (of all possible neighborhoods with the same center) radius r ,
such that N is a subset of E. Hence, all points of the given open set are internal points. Hence, the given set is open.

Is it an okay proof? Or should I be proving that the complacent of the open set in a given universe is closed. Hence, the set is open?.

I am somewhat new to the method of writing proofs, and so want to know that which is a better way to prove?

 

[HallsofIvy]

Homework Statement

In general, in R^n, what is the best way to approach the problem - a given set is open?
The given set E is such that for all x,y that belong to the given set, d(x,y) < r.

In the most general situation, go back to the definition of "open". You didn't say what definition you are using (there are several) but most often the definition is "all points in the set are interior points. Of course the definition of "p is an interior point of set A" (in metric spaces which Rⁿ is) is that there exist some neighborhood of p which is completely contained in A. I'm not entirely sure how your set is defined/

Homework Equations

The Attempt at a Solution

let x be the center of the sphere and y be any point such that d(x,y) < r. Now, let z be any boundary point such that d(x,z) = r.
Also let d(y,z) < epsilon. We can make a neighborhood N with epsilon as radius and y as point such that all points of N are subset of the given set. In general we can construct a neighborhood N of smallest (of all possible neighborhoods with the same center) radius r ,
such that N is a subset of E. Hence, all points of the given open set are internal points. Hence, the given set is open.

Is it an okay proof? Or should I be proving that the complacent of the open set in a given universe is closed. Hence, the set is open?.

I am somewhat new to the method of writing proofs, and so want to know that which is a better way to prove?[/QUOTE]
I have a friend who is a math professor. He tells me that it was when he was able to prove exactly the theorem you mention that he knew he could be and wanted to be a mathematician! I'm concerned that if I help you too much you might miss that thrill!

Let y be a point in the set. One of the things you should think about is how close y is to the boundary set. I strongly recommend you draw a picture. In R² of course. Draw a circle with center x and mark a point y in the circle. How large can a neighborhood about y be and still be in the set? Now mark a point z in that neighborhood. How far is z from x? Use the triangle inequality.

 

[rumjum]

Thanks for your reply. I did not want to state any assumptions in the problem because there could be several ways to approach. I wanted to leave it as an open ended question... but I am beginning to find that thrill although I get stuck many times!