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

[KataKoniK]

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.

 

[EnumaElish]

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.

 

[HallsofIvy]

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

Now show that $\frac{x_n}{2^n}$ is a decreasing sequence of numbers in U₀ and so has lower bounds and, by monotone convergence, converges to some number, call it $\alpha$. Finally, show that $\alpha$ is the least upper bound of A.

 

[KataKoniK]

Alright, I'll ditch the compact idea.
Thanks a lot HallsOfIvy

 

[HallsofIvy]

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.