Least Upper Bound Property

[amcavoy]

Let \left\{x_{n}\right\} be a nonempty sequence of monotonically increasing rational numbers bounded from above. Prove that \left\{x_{n}\right\} has a least upper bound in \mathbb{R}.

If we choose a monotonically decreasing sequence of upper bounds \left\{b_{n}\right\} with the property that 2^{n-1}\left(b_{n}-x_{n}\right)=b_{1}-x_{1}, can we show that there exists an accumulation point and conclude that it is the least upper bound? Given the choice of \left\{b_{n}\right\}, we can say that \left(x_{n+1},b_{n+1}\right)\subset\left(x_{n},b_{n}\right), but how do we show from here that there is a least upper bound?

Thank you.

 

[morphism]

Personally I would go about finding an appropriate sequence of upper bounds and proving it's Cauchy.

After that one can use the fact that x \leq b_n \, \forall n implies x \leq \lim_n b_n (because if b = \lim b_n < x, then there can be no b_N within (x-b)/2 of b, so the sequence (b_n) cannot possibly converge to b).

 

[HallsofIvy]

What property of the real numbers are you using to prove this? If you can use the "Heine-Borel" property, that every closed and bounded set of real numbers is compact, it's relatively straight forward. If you can use "monotone convergence", that an increasing sequence of real numbers, having upper bound, converges, what I would do is first use that to prove Heine-Borel, then prove the existence of a least upper bound. You could also start with the Cauchy Criterion, or the fact that the set of real numbers is connected. You are going to have to use one of those because those, like the Least Upper Bound Property are "defining" properties of the real numbers.

I notice that you specified a sequence of "rational" numbers. If you already have "Least Upper Bound" for real numbers, then it's trivial- rational numbers are real numbers.

 

[rudinreader]

It's kind of a vacuous problem.. reading out of Chapter 1 in Rudin:


***
An ordered set S is said to have the "least upper bound property" if the following is true:
If E \subseteq S, E is not empty, and E is bounded above, then sup E exists in S.

Theorem 1.19: There exists an ordered field R which has the least upper bound property. Moreover, R contains Q as a subfield.
***


So the existence of the least upper bound in R above is a direct consequence of the "completeness axiom (theorem)" for R.

While there are equivalent formulations of the completeness axiom, the "least upper bound property" is what is pertinent in this example...

 

[HallsofIvy]

If the "least upper bound property" is taken as an axiom for the real numbers, then, yes, the original statement is trivial. That was why I asked "What property of the real numbers are you using to prove this". If we take as the "fundamental" property, the Cauchy Criterion, then the method morphism suggested would work. I count 6 equivalent "fundamental" properties: The Least Upper Bound Property, Monotone Convergence, the Bolzano-Weierstrasse property, the Cauchy Criterion, the connectedness of the set of real numbers (with the usual topoloy), and the Heine-Borel property.

Of course, you could also define the real numbers using the Dedekind Cut and then easily prove the least upper bound property. Until we hear back from amcavoy, we'll never know!

 

[amcavoy]

I remember someone showing me an outline of this proof and saying that eventually we look for an accumulation point \ell such that \ell =\sup\left\{x_{n}\right\}=\inf\left\{b_{n}\right\} given the choice of the upper bound sequence. I know how to prove this given that a bounded monotonically increasing sequence converges as does a bounded monotonically decreasing sequence, but as far as I know the logic is circular since I used the least upper bound property to prove that a bounded monotonic sequence converges to its infinium or supremum.

I know the proof using Dedekind Cuts but was just wondering about this to see if there is a quick way of doing it given the choice of the upper bound sequence I posted above.

Just out of curiosity, how does one prove the least upper bound property using the Bolzano-Weierstrass property? Can we choose an upper bound b and some x_{k} so that I=\left[x_{k},b\right] is compact and thus the sequence \left\{x_{n}\right\}_{n=k}^{\infty}\subset I has a convergent subsequence? If so, how do we procede to show that there is a least upper bound?

Thank you.

 

[rudinreader]

amcavoy, instead of responding with more questions, why don't you follow Hall's suggestion just solve the problem:

True or false? (Prove your answer) The following properties are equivalent:

(i) For any closed interval [a,b] in R, every open cover of [a,b] has a finite subcover
(ii) Every bounded infinite set has a limit point in R
(iii) Every Cauchy sequence converges
(iv) R has the least upper bound (or equivalently greatest lower bound) property
(v) Every bounded sequence has a convergent subsequence
(vi) Every monotonic bounded sequence of rational numbers converges in R
(vii) R is isomorphic/isometric to the "metric space completion" of the rational numbers - i.e. the field of Cauchy sequences of rationals
(viii) R is isomorphic/isometric to the set of Dedekind cuts of rational numbers
(viiii) and almost forgot: R is connected

That's all I can think of off the top of my head. Thanks Halls for actually pointing that out, I hadn't actually thought about the fact that there are so many places where you can take as an axiom.

**Edit: OK perhaps you are not interested in all of the above properties. But at least identify which of the ones you are interested in, and solve the equivalence of those. In any case, it's not that it should take that long to solve the whole thing.. although I would mention that the definition of a "completion" is a little tricky to prove is actually "complete", and also that the problem of showing "limit point compact" spaces are compact (in any metric space) is also a little tricky and requires tinkering around with a countable basis I think.

 

[nmichalo]

It is not true that all of the statements are equivalent. For example, the least upper bound axiom is not equivalent to the convergence of Cauchy sequences.