Proving that an Open Set Does Not Contain Its Boundary Points: A Comprehensive Guide

In this article, we will delve into the fundamental concepts of open sets, boundary points, and interior points in the context of metric spaces and topological spaces. We will provide a detailed proof to show that an open set does not contain any of its boundary points. This topic is crucial for understanding the intricacies of set theory and topology, which are foundational in advanced mathematics.

Definitions and Key Concepts

To begin, let's define some key terms that will be central to our discussion:

Open Set

A set ( U ) in a metric space or topological space is called open if for every point ( x in U ), there exists a radius ( r > 0 ) such that the ball ( B(x, r) { y : d(x, y)

Boundary Points

A point ( x ) is a boundary point of a set ( A ) if every neighborhood of ( x ) contains at least one point in ( A ) and at least one point not in ( A ). The boundary of ( A ) is denoted as ( partial A ).

Interior Points

A point ( x ) is an interior point of a set ( A ) if there exists a neighborhood around ( x ) that is entirely contained in ( A ).

Proof: Open Set Does Not Contain Its Boundary Points

Now let's prove that an open set does not contain its boundary points:

Step 1: Definition of Open Set and Boundary Points

Consider an open set ( U ). By definition, for any point ( x in U ), there exists a radius ( r > 0 ) such that the ball ( B(x, r) subseteq U ).

Step 2: Boundary Point Definition

A boundary point ( b ) of ( U ) satisfies the property that every neighborhood of ( b ) contains at least one point in ( U ) and at least one point not in ( U ).

Step 3: Proof by Contradiction

Suppose, for the sake of contradiction, that ( b in U ). Since ( U ) is open, there exists a radius ( r > 0 ) such that ( B(b, r) subseteq U ).

Step 4: Contradiction with Boundary Point Definition

Consider the neighborhood ( B(b, r) ). This neighborhood contains only points in ( U ) because ( B(b, r) subseteq U ). However, this contradicts the definition of ( b ) being a boundary point, which requires that ( B(b, r) ) contains points not in ( U ).

Conclusion

Our assumption that ( b in U ) must be false. Therefore, ( b ) cannot be in ( U ).

Additional Considerations and Examples

The statement that an open set does not contain any of its boundary points is correct in most standard contexts. However, it is important to note the nuance of certain specific cases, such as the open unit disc in the real plane.

Example: Open Unit Disc in the Real Plane

Consider an open disc with radius 1 centered at the origin in the real plane. This set includes all points inside the circle but excludes the boundary points on the unit circle:

The interior points of the open disc are points strictly inside the circle. These points are limits points of the set since every neighborhood of an interior point contains points from the set. However, the boundary points of the open disc, which lie on the unit circle, are not included in the open set. These boundary points are limit points but not interior points of the set.

Thus, the open unit disc is an open set that contains all its limit points but none of its boundary points.

Conclusion

In summary, we have shown that an open set does not contain any of its boundary points. This is a fundamental property of open sets in topological and metric spaces, and it holds in most standard contexts. However, special cases like the open unit disc in the real plane are notable exceptions, where limit points are included but boundary points are not.