Basic question on set ordering I have some difficulties with a question from Jech's book (Introduction to Set Theory). Any help would be appreciated.
Give examples of a finite ordered set $(A, \leq)$ and a subset $B$ of $A$ so that


*

*$B$ has no greatest element.

*$B$ has no least element.

*$B$ has no greatest element, but $B$ has a supremum.

*$B$ has no supremum.
The only set I can think of is the empty set. What am I missing ?
 A: If you got "finite" correct, then you need $A$ to be a partial order (not total) for some parts to have a solution. Clearly you can satisfy (a) if you can satisfy (c), which is easy as long as you have two elements in $B$ that are incomparable but both less than something outside $B$. You only need a partial order with 3 elements for this to be possible. Similarly for (b). For (d) the empty set does not necessarily work, depending on what $A$ is. If $A$ has a least element, then $\emptyset$ does have a supremum. If not, then it does not. A one element subset in a partial order always has a supremum, so if you want a bigger example you need two elements. Can you figure out what kind of $A$ can give you an example?
A: Thanks for your help.I think I got it right now.For (a), (b) and (d) we could use: $A = \{a, b\}$ and $\leq$ as $ a \leq a, b \leq b$ and choosing $B = A$. For (c) we could choose: $A = \{a, b, c\}$ and $\leq$ as $a \leq a, b \leq b, a \leq c, b \leq c$. Thanks all for your help.
