Does Every Bounded Set in R Have a Least Upper Bound?

Your idea of using compactness seemed to be the possibility of using the Heine-Borel theorem but that would require the set be closed.) You are trying to show that if A is a bounded set of real numbers then it has a least upper bound. The least upper bound property is equivalent to the axiom of completeness and so to the convergence of every Cauchy sequence. (It is a theorem in real analysis that a sequence has a limit if and only if it is Cauchy.) I would suggest you look at the discussion of the least upper bound property to see what you have available to you. You have to prove a number of things here. First, you have to show that
  • #1
KataKoniK
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Q: Show that every bounded set in R has a least upper bound. Using either

"Every monotonic and bounded sequence is convergent" or
"Every bounded sequence has an accumulation point" or
"Every bounded sequence has a convergent subsequence"

I'm not really sure how to start this out, but would showing the statement true using a bounded compact set in R that has properties of those three statements be valid enough to show that R has a least upper bound (if this is alright, how would you show it mathematically?)? If not, then what would be the best way to show the statement is true? Thanks.
 
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  • #2
You have thrown in "compact" so my guess is that won't be okay. It would be okay if you make the connection between accumulation point and closedness, hence compactness (I think).

PS: R doesn't have bounds. But that's not what the question asked, so you're okay.
 
  • #3
Wow, that's some problem! I did precisely this myself some years ago (I wanted to give my own proofs that all of (1) Least Upper Bound Property, (2) Monotone Convergence, (3) Cauchy Criterion, (4) The set of all real numbers, with the "usual" metric, is connected, (5) every closed and bounded set of real number, with the usual metric, is compact, are "equivalent". That is, given anyone you can prove the others.) Most calculus books prove that the least upper bound property implies monotone converges- that's easy because any sequence is a set. The other way- that monotone convergence implies the least upper bound property is harder- a set is not generally a sequence so you have to construct the sequence to which monotone convergence applies.

Here's an outline of my method.

Suppose A is a non-empty set having upper bound b. Let U0 be the set of all upper bounds of A. That is non-empty since it contains b. In fact, it contains every number larger than B and so contains integers. It has every member of A as lower bound and so ("well ordered" property of integers) contains a smallest integer. Call it x0.

For n a positive integer, define Un= { 2nx where x is in U0}. Since U0 contains integers and 2n times an integer is an integer, Un contains integers. Since 2n times any lower bound on U0 is a lower bound on Un and U0 has lower bounds, Un has lower bounds and so contains a smallest integer: call it xn.

Now show that [itex]\frac{x_n}{2^n}[/itex] is a decreasing sequence of numbers in U0 and so has lower bounds and, by monotone convergence, converges to some number, call it [itex]\alpha[/itex]. Finally, show that [itex]\alpha[/itex] is the least upper bound of A.
 
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  • #4
Alright, I'll ditch the compact idea.
Thanks a lot HallsOfIvy
 
  • #5
Don't thank me yet! That's a non-trivial proof. If you can work out the details, you should feel very pleased with yourself.

Dropping compactness is probably a very good idea since the problem says nothing about compactness nor about the set being closed which is necessary for compactness.
 

FAQ: Does Every Bounded Set in R Have a Least Upper Bound?

1. How do you prove that a set is bounded?

To prove that a set is bounded, we need to show that there exists a number M such that all the elements in the set have an absolute value less than or equal to M. This means that the set cannot have any elements with extremely large values, and therefore, it is bounded.

2. What is the definition of a bounded set?

A bounded set is a set that has a finite upper and lower limit. This means that there exists a maximum and minimum value in the set, and all the elements in the set fall within this range.

3. Can a set be both bounded and unbounded?

No, a set cannot be both bounded and unbounded at the same time. A set is either bounded or unbounded, based on the definition of boundedness.

4. What is an example of a bounded set?

An example of a bounded set is the set of all real numbers between -1 and 1, denoted as [-1, 1]. This set has a finite upper and lower limit, and all the elements in the set fall within this range.

5. How does a bounded set differ from an unbounded set?

A bounded set has a finite upper and lower limit, while an unbounded set does not have a finite limit. This means that the elements in an unbounded set can have infinitely large values, while the elements in a bounded set are limited to a certain range.

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