Least Upper Bound

Calculus Level 3

The greatest lower bound of an ordered set $S \subseteq U$ is the largest number $x$ in $U$ such that $x \leq y$ for all $y$ in $S$ .

For example, let $U$ be the set of all real numbers, and let $S = (0,1)$ . Then $0$ is the greatest lower bound of $S$ . Notice how the greatest lower bound differs from the minima of a set in that the greatest lower bound of a set might not even belong to the set.

An ordered set $X$ is said to have the greatest-lower-bound property if the following holds true:

If $E \subseteq X,$ $E$ is not empty, and $E$ is bounded below, then $E$ has a greatest lower bound.

Which of the following ordered sets (under their usual order) does not have the greatest-lower-bound property?

The set of rational numbers The set of positive integers The set of real numbers

\{1 , 2 , 3 , 4 , 5 \}

2 solutions

Anthony Holm
Jun 9, 2017

Consider the intersection of the rational numbers with the open interval (e, 3). Presume that a greatest lower bound exists. Then it must be a rational q with either q>e or q<e. If q>e, then there exists a rational number p, because the rationals are dense in the reals, such that q>p>e, meaning that q is not in fact a lower bound. If q<e, then again by the density of the rationals in the reals there exists a rational k such that q<k<e and because k<e, k must also be a rational lower bound on the set and thus q is not the greatest lower bound. Because the presumption that a greatest lower bound exists leads to a contradiction, there must be no greatest lower bound.

Very nice! Can you prove or explain why the other sets have the greatest-lower-bound property?

Shourya Pandey - 4 years ago

Hana Wehbi
Jun 13, 2017

Because the rational numbers have bounds which are the irrational numbers or bounded by the irrational numbers and we have infinity of irrationals. That is why the rational numbers can't have a greatest lower bound.

Least Upper Bound

2 solutions

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