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

[cateater2000]

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.

 

[NateTG]

Can you show that any open neigborhood of inf A contains some element of A?

 

[mathman]

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.

 

[debrawallenger]

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!

 

[matt grime]

There are many "natural" ways to "solve" this.

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

 

[mathman]

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?