Proof: open sphere is an open set

In summary: Thank you for pointing it out!In summary, we want to prove that an open ball in ##\mathbb{R}^m## is an open set. To do this, we need to show that any point inside the ball is an interior point. Let ##B(P_0, r)## be an open ball with center ##P_0## and radius ##r##, and let ##P## be an arbitrary point inside the ball. This means that there exists an ##\epsilon > 0## such that ##B(P, \epsilon) \subset B(P_0, r)##. Since we have assumed ##B(P_0, r)## to be open, it does not contain its boundary.
  • #1
nuuskur
Science Advisor
866
923

Homework Statement


Prove that an open sphere in [itex]\mathbb{R}^m[/itex] is an open set.

Homework Equations

The Attempt at a Solution

To show that an open sphere is an open set, any point inside the sphere has to be an interior point:
Let us have a sphere [itex]B(P_0, r), r > 0[/itex], where [itex]P_0[/itex] is the centerpoint and r is the radius of the sphere and also an arbitrary point [itex]P[/itex] inside the sphere. Hence [itex]\exists\varepsilon>0\colon B(P,\varepsilon)\subset B(P_0, r) [/itex] and as we assumed the [itex]B(P_0, r)[/itex] to be open, then it does not contain its boundary.

The idea is to use the triangle inequality by taking another random point in a sphere around the already random point P in the large sphere: [itex]S\in B(P, \varepsilon)[/itex]. The distance between two points [itex]d(A,B)[/itex] (Pythagorean)
[itex]d(S, P_0)\leq d(S,P) + d(P,P_0) <\varepsilon + d(P,P_0)[/itex]. This is a shaky part - how can I justify this inequality is always correct? I can visualize it in my head, but is it accurate?

Per that assumption I would fix [itex]\varepsilon\colon = r - d(P,P_0) > 0[/itex] since [itex]P[/itex] is an interior point and its distance cannot be greater than or equal to the radius of the large sphere, therefore [itex]\varepsilon[/itex] is always strictly positive and [itex]\forall S\in B(P, \varepsilon_1)[/itex] I can state that the distance [itex]d(S,P_0) < \varepsilon + d(P,P_0) = r[/itex] which would mean that the open sphere is an open set as all its points are interior points.
 
Physics news on Phys.org
  • #2
The argument is essentially correct, but things are a bit mixed up and it doesn't flow. I suggest trying to tidy it up. I would start with something like this:

Let ##B(P_0, r)## be an open sphere in ##\mathbb{R}^m##

Let ##P \in B(P_0, r)##

##d(P_0, P) < r## hence ##\exists \epsilon > 0## ...
 
  • #3
A silly comment, but I don't like the term "open sphere". The terminology "open ball" is standard, because a sphere is usually the name given to the boundary of the ball, not something in the interior.
 
  • Like
Likes nuuskur
  • #4
micromass said:
A silly comment, but I don't like the term "open sphere". The terminology "open ball" is standard, because a sphere is usually the name given to the boundary of the ball, not something in the interior.
I knew there was something off in the translation. I'll use ball to describe the set and sphere to describe its boundary, then.
 

Related to Proof: open sphere is an open set

1. What is an open sphere?

An open sphere is a set of all points that are a certain distance away from a center point. The distance is typically referred to as the radius, and the center point is known as the center of the sphere. An open sphere does not include the boundary points, meaning the points on the surface of the sphere are not included.

2. How is an open sphere defined as an open set?

An open set is a set of points where every point has a neighborhood (a set of points that are close to it) that is entirely contained within the set. In an open sphere, every point has a neighborhood that is contained within the sphere, making it an open set.

3. What is the significance of an open set in mathematical analysis?

Open sets are important in mathematical analysis because they allow for the definition of important concepts such as continuity, differentiability, and limits. They also play a crucial role in topology, which is the study of the properties of spaces that are preserved under continuous deformations.

4. How is the openness of an open sphere proven?

The openness of an open sphere can be proven using the definition of an open set. This involves showing that every point in the sphere has a neighborhood that is completely contained within the sphere, meaning no points on the boundary are included. This can be done using mathematical notation and logical reasoning.

5. Can an open sphere also be a closed set?

No, an open sphere cannot be a closed set. A closed set includes all of its boundary points, while an open sphere does not. In fact, an open sphere and its complement (the set of points not in the sphere) together make up a closed set. This is known as the closure of the open sphere.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
989
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
Replies
2
Views
946
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
Replies
29
Views
5K
  • Calculus and Beyond Homework Help
Replies
1
Views
5K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top