Here is a problem from the book Analysis by Terence Tao:
(Vol 1, Exercise 5.5.2)
Let $E$ be a non-empty subset of $R$, let $n \geq 1$ be an integer, and let $L < K$ be integers. Suppose that $K/n$ is an upper bound for $E$, but that $L/n$ is not an upper bound for $E$. Without using the theorem EoLUB (stated below), show that there exists an integer $L < m \leq K$ such that $m/n$ is an upper bound for E, but that $(m-1)/n$ is not an upper bound for E. (Hint: prove by contradiction, and use induction. It may also help to draw a picture of the situation.)
Theorem EoLUB: Let $E$ be a non-empty subset of $R$. If $E$ has an upper bound, then it must have exactly one least upper bound.
Any idea how to solve this?
Moreover, is it possible to give a constructive proof of this, other than one using contradiction? Because I think it is this exercise that illustrates the existence of least upper bound, I will be very happy to see a constructive proof.
Thanks.
