Confusion about least upper bound for 2 elements in set theory I have this example in a set of notes i found about set theory and i cant quite reconcile the definition of least upper bound of the book i'm using and the example in the notes.
the set the problem is based on is the following:
The poset of the divisors of 60, ordered by divisibility 
Given that problem b states that the greatest lower bound between 2 and 20 is 2 instead of 1, it would seem that the GLB of two elements on a set A is treated as the GLB of a subset of A containing those 2 elements, so the GLB will be found outside the substet but inside A.
given that, why is the GLB not 4 instead?  or the LUB not 30 instead of 60?

 A: With $S=\{6,20\}$ we see that among the elements of $L=\{1,2,3,4,5,6,10,12,20,30,60\}$, only $60$ is an upper bound (and hence is the least upper bound) for $S$.
In particular, $u:=20$ fails to be an upper bound because we need $s\le u$ for all $s\in S$, but $6\not\le 20$ in the partial order relation considered (divisibility).
A: The number 30 is not even an upper bound for 20, since 20 does not divide 30. Such elements are said to be incomparable. That's why divisibility is only a partial order rather than a total one. Some elements cannot be compared to one another.
Perhaps you are considering the fact that, in the picture, 30 appears to be slightly higher than 20, but this is an unfortunate aspect of the drawing. The important thing to notice is that there is no path of strictly upward edges that leads from 20 to 30.
A: The greatest lower bound of two numbers $x<y$ is $x$, hence GLB(2,20)=2 because 2 divides 20 (2<20). Remember the definition of order you have. Keep in mind how you define things in a particular problem and you should be fine.
