Open Balls and metrics

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In summary, the conversation discusses verifying that for all x in Re and any epsilon greater than 0, the set (x-epsilon, x+epsilon) is an open neighborhood for x. The relevant definitions of a neighborhood and open set are also mentioned. The conversation includes an attempt at a solution and a question about whether the proof is complete. The final response clarifies the definition of an open set and suggests reproducing Rudin's theorem to prove that neighborhoods are open.
  • #1


First of all, sorry if the notation is hard to read, I'm still getting used to this text entry.

Consider [tex]\Re[/tex] with metric [tex]\rho[/tex] (x,y) = |x-y|. Verify for all x [tex]\in[/tex] [tex]\Re[/tex] and for any [tex]\epsilon[/tex] > 0, (x-[tex]\epsilon[/tex], x+[tex]\epsilon[/tex]) is an open neighborhood for x.

Relevant Definitions:
Neighborhood/Ball of p is a set Nr(p) consisting of all q s.t. d(p,q)<r for some r>0.

Attempt at solution:

Take [tex]\alpha[/tex] > 0, [tex]\alpha[/tex] < [tex]\epsilon[/tex]. Take [tex]\rho[/tex](x, x-[tex]\alpha[/tex]) = |x-(x- [tex]\alpha[/tex] )| = [tex]\alpha[/tex] < [tex]\epsilon[/tex].
Take [tex]\rho[/tex](x, x+[tex]\alpha[/tex]) = |x-(x+[tex]\alpha[/tex])| = [tex]\alpha[/tex] < [tex]\epsilon[/tex].
Therefore, any positive [tex]\alpha[/tex] < [tex]\epsilon[/tex] is in N[tex]\epsilon[/tex](x).

So, I think that this proof is ok, but I also feel that it is missing something.
Thanks in advance for any comments or suggestions.
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  • #2
You've shown that [itex](x-\epsilon, x+\epsilon)[/itex] is a "neighborhood" of radius [itex]\epsilon[/itex] about [itex]x[/itex], but you haven't shown that it's open.
  • #3
Ok, there is a theorem (2.19 in Rudin) that says: "Every neighborhood is an open set." I must be misinterpreting this theorem then?

(x - [tex]\epsilon[/tex]) is a limit point of the set, but (x - [tex]\epsilon[/tex]) [tex]\notin[/tex] N[tex]\epsilon[/tex] (x), and (x + [tex]\epsilon[/tex]) is a limit point of the set, but (x + [tex]\epsilon[/tex]) [tex]\notin[/tex] N[tex]\epsilon[/tex] (x). Therefore the set is open.

Does this complete the proof?
  • #4
Sorry for misplacing this post, this belongs in Calculus and Beyond
  • #5
Rudin defines an open set as one which contains only interior points -- that is, points which have neighborhoods contained within the set. Prove that neighborhoods are open, given this definition. (Rudin does it for you in the theorem you mentioned, but it would be helpful for you to reproduce it yourself.)

Does the logic you just provided show that the neighborhood fits this definition?

1. What is an Open Ball?

An Open Ball in mathematics is a set of points that are all within a certain distance from a given point. It is denoted as B(x,r), where x is the center point and r is the radius. The points within the set are not included in the set itself.

2. How is an Open Ball different from a Closed Ball?

An Open Ball does not include the boundary points, while a Closed Ball includes the boundary points. In other words, an Open Ball is an open set, while a Closed Ball is a closed set.

3. What is the importance of Open Balls in topology?

Open Balls are an essential concept in topology as they help define open sets, which are the building blocks of topological spaces. They also play a crucial role in defining continuity and convergence in metric spaces.

4. How are metrics used in Open Balls?

Metrics are used to measure the distance between points in a metric space. In the context of Open Balls, metrics are used to determine which points are within a given distance from a center point, thus defining the set of points within the Open Ball.

5. Can Open Balls be applied in real-world situations?

Yes, Open Balls can be applied in various real-world situations. For example, in physics, they are used to define the range of a force field around a point, and in economics, they are used to define the range of influence of a certain market or business. They also have applications in computer science, specifically in the study of algorithms and optimization problems.

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