I'm fairly new to formal proof, so when I started learning about real analysis it has been a huge source of confusion to me. Not too long ago I was introduced to the least-upper-bound property, or, what my teacher calls it, the axioma de completez, meaning "axiom of completeness", which states "any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers."
I know that this property somehow sets the rational numbers and the real numbers apart by making the real numbers complete, meaning there are no gaps whatsoever between any element of the set, whereas the rational numbers have gaps (which correspond to irrational numbers); but I don't understand why. To me, it just states that if there's any number that's bigger than all of those in a set (called a bound), then there must be one bound that's smaller than any of the others, called the least upper bound or supremum. Could anybody explain this to a newbie to real analysis and formal proof?
Best Answer
The proof that $\mathbb{R}$ does indeed have the Least Upper Bound property really depends upon how you're defining the real numbers; for example, if $\mathbb{R}$ is constructed using Dedekind cuts, then the proof is rather straight-forward and easy. If you're constructing $\mathbb{R}$ using equivalence classes of Cauchy sequences, then it's involved (at least the proofs I've seen/done).
Thus, I'll use this answer to address the concerns of why the Least Upper Bound property needs to be mentioned at all, and why it doesn't trivially hold true for everything.
Firstly, the Least Upper Bound property is essentially the reason calculus can be done; as we shall see, there are "gaps" in the rational numbers. The ability to take limits, which is central to everything done in Real Analysis, is closely related to the Least Upper Bound property. To show why this is the case, I'll quote some equivalences:
All of the above are equivalent, and all are central to Real Analysis.
As far as the Least Upper Bound property not holding more generally in, for example, $\mathbb{Q}$, consider the following set: $$\{ q\in \mathbb{Q} \mid q>0 \wedge q^2<2 \}$$ It is clearly bounded above; 2 is an upper bound, for example. Does this set have a least upper bound? In $\mathbb{R}$ it certainly would, and would be $\sqrt{2}$; since $\mathbb{Q}$ is dense in $\mathbb{R}$, if there were a least upper bound in $\mathbb{Q}$ for this set, then it would also be a least upper bound of the set in $\mathbb{R}$. But we know that $\sqrt{2}$ is irrational, so this cannot be the case. Thus, $\mathbb{Q}$ cannot have the Least Upper Bound property. This is why the Real numbers are necessary.