I know I must be overlooking something here, but I currently am stuck with the following issue.
Consider the set $\boldsymbol{S = (a, b) \cup [c, d)}$, with $a, b, c, d \in \mathbb{R}, c > b$.
I know that in order to claim that $S$ does satisfy the greatest lower bound property we must verify that for every non-empty subset $B \subseteq S$ bounded below in $S$ the maximum of the set of all lower bounds of $B$ is in $S$.
On the other hand, since there is a theorem that establishes that both greatest lower bound property (glb) and least upper bound property (lub) necessarily imply each other, observing that $S$ fails to satisfy lub —and it does— should suffice to claim that it also fails to satify glb.
Allow me to show that lub fails to be satisfied.
Consider $E = (a, b)$. $E$ is a non-empty subset of $S$. Does it have any upper bound in $S$? Yes: $[c, d)$ is the set of upper bounds of $E$ that also happen to lie in $S$. The minimum of these upper bounds is $c$.
Now, the set of all upper bounds of $E$ —not restricting ourselves to just the ones that are in $S$— is $[b, +\infty)$. The minimum is $b$. There you have it: $\boldsymbol{c \neq b}$, and thus lub does not hold.
Then, glb does not hold either, and it should be possible to find a non-empty subset of $S$ bounded below in $S$ such that the maximum of the set of all lower bounds of that subset is not in $S$. The problem is that I just can not find such a subset of $\boldsymbol{S}$.
I tested on $[c, d)$. The set of its lower bounds in $S$ is $(a, b) \cup \{c\}$. Maximum: $c$. The set of all of its lower bounds is $(-\infty, c]$. Maximum: $c$. No discrepancy.
Am I not getting the concepts right?