Problem 1144

25.45. Let's conduct the proof only for the exact upper bound. Let $x_{0}$ and $x_{1}$ be two exact upper bounds, with $x_{1} > x_{0}$. Then for $\varepsilon = \frac{x_{1} - x_{0}}{2}$, there exists $x$ (a number from the considered set) such that $x + \frac{x_{1} - x_{0}}{2} > x_{1}$, i.e., $x > \frac{x_{1} + x_{0}}{2} > x_{0}$. But this contradicts the fact that $x_{0}$ is the exact upper bound.

Now let's prove the existence of the exact upper bound. Construct a non-decreasing sequence $\left\{a_{n}\right\}$ and a non-increasing sequence $\left\{b_{n}\right\}$ such that for each $n$ there exists $x \geqslant a_{n}$ and no $x \geqslant b_{n}$. Specifically, take $a_{1}$ as any number from the considered set, and $b_{1}$ as the number $c$ from the definition of a set bounded above. Let $c_{2}$ be the midpoint of the interval $\left[a_{1}, b_{1}\right]$. Take $a_{2}$ as the number $c_{2}$ if it fits; otherwise, take $a_{1}$. Take $b_{2}$ as the number $c_{2}$ if it fits; otherwise, take $b_{1}$. Subsequent terms are chosen similarly. The constructed sequences have a common limit $x_{0}$. It is easy to see that $x_{0}$ is the exact upper bound.

CHAPTER 26