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

In summary, the points infA and supA belong to the closure A* because A* contains its limit points and any open neighborhood of infA and supA must contain a point of A. If infA and supA are already in A, then there is nothing to prove. However, if they are not in A, we can construct two sequences that converge to infA and supA respectively, and each of these sequences must contain a point of A. Therefore, the points infA and supA must belong to the closure A*. Additionally, there are multiple ways to solve the given sequence of numbers, such as considering the differences between consecutive terms. The original question does not appear to be related to the discussion about the sequence of
  • #1
cateater2000
35
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.
 
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  • #2
Can you show that any open neigborhood of inf A contains some element of A?
 
  • #3
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.
 
  • #4
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!
 
  • #5
There are many "natural" ways to "solve" this.

Why don't you consider the differences between consecutive terms?
 
  • #6
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?
 

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

What is the definition of closure of a bounded set?

The closure of a bounded set is the smallest closed set that contains all the points in the original set. It includes all the limit points of the set and can be thought of as the set's "boundary".

How is the closure of a bounded set different from the closure of an unbounded set?

The closure of a bounded set is a finite set, while the closure of an unbounded set can be an infinite set. This is because a bounded set has a finite limit and therefore its closure is also finite.

Can a bounded set have an empty closure?

Yes, a bounded set can have an empty closure if the set itself is empty. For example, the set of real numbers between 1 and 2 has an empty closure because it does not include the points 1 and 2.

What is the relationship between the closure of a bounded set and its interior?

The interior of a bounded set is the largest open set that is contained within the bounded set. The closure of a bounded set, on the other hand, includes the boundary points of the set as well. This means that the closure of a bounded set is equal to the union of its interior and its boundary.

How is the closure of a bounded set used in mathematical analysis?

The closure of a bounded set is used to define concepts such as continuity, convergence, and compactness in mathematical analysis. It helps to identify the boundary points of a set and determine whether a given limit point is contained within the set or not.

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