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Least upper bound axiom: Every non-empty subset of $\mathbb R$ that has an upper bound must have a least upper bound.

This sounds too obvious as it works for both closed and open subsets of $\mathbb R$

A newcomer naive person will say:

Every subset of $\mathbb R$ must have a least upper bound

Then he realizes the first limitation that the subset of $\mathbb R$ must be non-empty.

Then after some time he realizes the second limitation that subset of $\mathbb R$ must have an upper bound.

Only then he states the least upper bound axiom completely.

No person who have gone through this axiom have found other limitations. This doesn't mean that there are no other limitations.

So how can we say that we have now stated the least upper bound axiom completely? And how can we justify its applications in several areas of mathematics (as there is a probability that the lub axiom is incomplete)?

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    $\begingroup$ This is not at all obvious. For example the set $\mathbb {Q} $ of rationals does not possess this property (try to think why it is so). In fact this is the only property which differentiates $\mathbb {R} $ from $\mathbb {Q} $. $\endgroup$ – Paramanand Singh Jul 22 at 17:49
  • $\begingroup$ Then why is it made as an axiom in the very beginning of elementary real analysis course? Are not the students getting deceived by this? $\endgroup$ – Joe Jul 22 at 18:19
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    $\begingroup$ IMHO every introductory real analysis text/course must introduce the readers to the thing called real numbers. There is some amount of deception going on in most textbooks in the name of sound pedagogy championed primarily by Walter Rudin and his followers. I would suggest you to grab a copy of G. H. Hardy's A Course of Pure Mathematics and study real numbers in a systematic and straightforward way. $\endgroup$ – Paramanand Singh Jul 22 at 18:26
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    $\begingroup$ Just as one is aware of the fact that a rational number can be described using two integers, a real analysis student must be aware of how a real number can be described in terms of rationals. Unfortunately this is somewhat difficult to understand but not as hard as most textbooks would lead you to believe. $\endgroup$ – Paramanand Singh Jul 22 at 18:29
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    $\begingroup$ While there is a lot of deception that goes on during most of mathematics education, there's nothing deceptive about choosing to take a statement as an axiom. At worst, it suggests (but does not imply) that the statement couldn't otherwise be derived. Depending on the framework within which you are working this may or may not be true. $\endgroup$ – Derek Elkins Jul 22 at 18:41
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In my opinion, your question is: what is $\mathbb{R}$?

If you follow an axiomatic approach, you define $\mathbb{R}$ as an ordered field (being an ordered field means to fulfill some axioms) that is Dedekind-complete i.e. that satisfies the least upper bound axiom: every non-empty subset of $\mathbb{R}$ with an upper bound in $\mathbb{R}$ has a least upper bound in $\mathbb{R}$ (see here). Therefore, $\mathbb{R}$ is any object that fulfill these axioms. It can be proved that such an object is unique, up to isomorphism. If you change the axioms (in particular, if you replace the least upper bound axiom with something else), likely you are talking of another mathematical object, which is not $\mathbb{R}$.

It turns out that the set of axioms I sketched above is not the only way to define $\mathbb{R}$: there are several equivalent axiomatic definitions of $\mathbb{R}$, which means that there are several sets of axioms that describe the same object $\mathbb{R}$. For instance, you can define $\mathbb{R}$ as an ordered field such that every Cauchy sequence converges (Cauchy axiom). Then you can prove that, in a ordered field, the least upper bound axiom follows from the Cauchy axiom, and vice-versa. This corroborates the idea that the least upper bound axiom (or other equivalent axioms such as the Cauchy one) formalizes correctly our intuition about $\mathbb{R}$.

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Because we can prove that that axiom holds after having constructed $\mathbb R$ by some method (Dedekind cuts, Cauchy sequences of rational numbers, …). And, since it is proved for every non-empty subset of $\mathbb R$ with an upper bound, there can be no other conditions missing.

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  • $\begingroup$ Thanks a lot sir. I am studying real analysis from this website. As you can see there are three chapters of sequences. Will I get to study the construction of $\mathbf{R}$ and the proof of (lub axiom) therin? $\endgroup$ – Joe Jul 22 at 7:51
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    $\begingroup$ No, it doesn't have that. You can find it in, say, the final chapters of Michael Spivak's Calculus. $\endgroup$ – José Carlos Santos Jul 22 at 8:28
  • $\begingroup$ @Joe: the simplest construction is by Dedekind and you should read his pamphlet Continuity and irrational numbers or its exposition in first chapter of Hardy's A Course of Pure Mathematics. $\endgroup$ – Paramanand Singh Jul 22 at 17:52

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