Showing the Inclusion of Infimum and Supremum in the Closure of a Bounded Set

  • Context: Graduate 
  • Thread starter Thread starter cateater2000
  • Start date Start date
  • Tags Tags
    Bounded closure Set
Click For Summary

Discussion Overview

The discussion revolves around demonstrating that the infimum and supremum of a bounded subset of the real numbers belong to the closure of that set. The scope includes mathematical reasoning and technical explanation related to real analysis concepts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests that since the closure contains limit points, it seems obvious that infA and supA belong to the closure A* but seeks help in articulating this reasoning.
  • Another participant asks whether any open neighborhood of infA contains an element of A, implying a need for a more rigorous argument.
  • A different participant notes that it is possible to construct sequences converging to infA and supA, indicating that intervals around these points must contain elements of A.
  • A participant introduces a separate topic regarding a 6th-grade homework problem involving a sequence, indicating a shift in focus from the original mathematical discussion.
  • In response to the homework question, another participant suggests considering the differences between consecutive terms of the sequence as a method to approach the problem.
  • A participant revisits the original question about infA and supA, noting that their previous argument assumed these points were not in A, and acknowledges that if they are in A, the situation changes, but expresses uncertainty about the implications.
  • One participant questions the relevance of the homework problem to the original discussion, indicating a potential divergence in the conversation's focus.

Areas of Agreement / Disagreement

Participants express varying levels of certainty regarding the inclusion of infA and supA in the closure A*. While some provide constructive approaches, there is no consensus on the completeness or correctness of the arguments presented. Additionally, the introduction of a separate homework topic creates a divergence in the discussion.

Contextual Notes

There are unresolved assumptions regarding the definitions of limit points and the closure of sets. The discussion also reflects a lack of clarity on how the presence of infA and supA in A affects the initial arguments.

cateater2000
Messages
35
Reaction score
0
If A is a bounded subset of the reals, show that the points infA, supA belong to the closure A*.


At first the answer seems obvious to me since A* contains its limit points. I'm just having trouble putting it into words, any suggestions would be great, thanks.
 
Physics news on Phys.org
Can you show that any open neigborhood of inf A contains some element of A?
 
It is easy to construct two sequences of points which converge to infA and supA respectively. Take intervals of length 1/n above infA and below supA. Each of these must contain a point of A.
 
Help, anyone. I am very new to this forum and am really here trying to help my son with his wicked 6th grade homework. His teacher is in my opinion, assigning problems that are way too difficult. I was hoping someone here could maybe help. The problem is as follows: The students are given the sequence 1,5,13,25,41,61 and have to come up with an equation to solve the sequence. Any ideas? This should be easy for you all. But for me, who was good at math at one time, this is beyond what I can come up with. Any help would be appreciated!
 
There are many "natural" ways to "solve" this.

Why don't you consider the differences between consecutive terms?
 
If A is a bounded subset of the reals, show that the points infA, supA belong to the closure A*.

The answer I gave previously assumed that infA and supA were not in A. If they are in A there is nothing to prove, but my previous argument may not hold (surprise).

Question for debrawallenger - what has this got to do with the original question?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
9K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 29 ·
Replies
29
Views
4K
Replies
2
Views
6K