This is with reference to Principles of Mathematical Analysis, Third Edition by Walter Rudin. The theorem establishes that every ordered set with the least-upper-bound property also has the greatest-lower-bound property.
I understand the general logic and flow of the proof in the book. However, I must have some fundamental misunderstanding of the definitions because I can't find what's wrong with this "counterexample".
$S = \{x \in \mathbb{Q}: x \leq 3 \}$
$B = \{x \in \mathbb{Q}: x^2 < 2 \}$
As far as I can tell, $S$ has the LUB property (the $\sup$ being 3), and $B$ is also bounded below in $S$. However, I don't see how the infimum of $B$ exists in this case.
Maybe it is because $L \notin S$ and this would imply the theorem conditions are not met?