Subset of Rationals with No Supremum

Take a subset of the rational numbers Q such that $x^2 < 2$ for all x in Q. Show that this set does not have a supremum in Q.

In this problem, we are exploring a fascinating aspect of the real numbers through the lens of rational numbers: the concept of supremum. A supremum, or least upper bound, is an important concept in real analysis, particularly when discussing bounded sets. The essence of this problem lies in the completeness of real numbers, a property lacking in the rational numbers, which is why you will show that a set can be bounded without having a rational supremum.

When tackling this problem, consider why the square root of 2 is irrational and how this fact plays a critical role in constructing sets of rational numbers where no rational number can adequately be the supremum. The key strategy here involves demonstrating that, however close a rational candidate for a supremum might get to the square root of 2, there will always be another rational number in your set exceeding this candidate but still satisfying the condition $x^2 < 2$.

This problem exemplifies the concept of density and the ordered, but not complete, nature of rational numbers. It also ties into deeper discussions on the Dedekind cuts and how the real numbers can be viewed as the completion of the rationals. Such problems deepen your understanding of the structural differences between various number sets and solidify foundational concepts related to limits, bounds, and completeness in real analysis.