Completeness axiom
Completeness axiom
Completeness is a property of the real numbers that, intuitively, implies that there are no "gaps" or "missing points" in the real number line. This contrasts with the rational numbers, whose corresponding number line has a "gap" at each irrational value.
The term "completeness" here should not be mistaken or mixed up with the common definition of completeness in the English dictionary, as it represents something entirely different.
Least-upper-bound property
We'll be defining the term "completeness" with the least-upper-bound property: Every nonempty subset \[A\] of \[\mathbb{R}\] that is bounded above has a least upper bound. That is, \[\sup A\] exists and is a real number.
Hence, we can form the statement "The real numbers are complete.", which means if you have a subset of real numbers that is bounded from above, you can always find its least upper bound by just stating "its complete thus there is a least upper bound".
The least-upper-bound property is a precise formulation of the completeness axiom. It guarantees that for any set of real numbers that doesn't extend to infinity and isn't empty, there is a definitive "boundary" within the real numbers itself. This means that the real number system is complete because it can accommodate the supremum of any such set, leaving no room for gaps. Without this property, certain bounded sets would lack a least upper bound within the real numbers, implying the existence of undefined points and disrupting the continuity of the number line.
Another example to this would be that "rational numbers are not complete". Let \[A=\left\{ x\in\mathbb{Q}\mid x^{2}<2 \right\}\]. Although this set has infinite upper bounds (such as \[1.42^{2}\] or \[1.45^{2}\]), it has no lowest rational upper bound, as the lowest upper bound would be \[\sqrt{2}\], which isn't a rational number. One could argue that the lowest upper bound for set \[A\] would be the smallest rational number larger than \[\sqrt{2}\], however, similar to the concept for the defintion of limits, given any rational larger than \[\sqrt{2}\] (e.g. \[1.42^{2}\]), you can always find a smaller number (e.g. \[1.415^{2}\]). Therefore, rational numbers are not complete.
Corollary
Every nonempty subset \[S\] of \[\mathbb{R}\] that is bounded below has a greatest lower bound, i.e. \[\inf S\] exists.
Proof: Assume that A is bounded below, \[A\subseteq \mathbb{R}\] and \[A\ne \emptyset\]. Define a new set \[B=\left\{ -a:a\in A \right\}\], i.e. a new set that contains the negative of every element in \[B\]. Since we know that \[B\] is bounded above (lower bound becomes upper bound after being reflected on the number line), by completeness axiom, set \[B\] will have a lowest upper bound. Call it \[m\]. Then by definition, \[m\ge b\] for any \[b\in B\]. Finally, multiply through by \[-1\], this reverses the inequality, \[-m\le -b\] for any \[b\in B\]. The set of numbers \[-b\] is precisely the elements of \[A\], therefore we've shown that \[-m\] is a lower bound of \[A\].
To prove that \[-m\] is the greatest lower bound of \[A\], let \[l\] be any lower bound of \[A\]. Consequently, \[l\le a\] for all \[a\in A\]. Multiply by \[-1\], \[-l\ge -a\implies -l\ge b\] for all \[a\in A\] and \[b\in B\]. Since \[m\] is the lowest upper bound of \[B\] and \[-l\] is any upper bound of \[B\], it follows that \[-l\ge m\]. Multiply by \[-1\] again, \[l\le -m\]. Thus, \[-m=\inf A\]. Therefore, we can conclude that any set that satisfies the conditions given for set \[A\] will have a greatest lower bound.