Re: Hasse Diagram confusion

In any poset (partially ordered set), an element u is an upper bound of a set S iff $\displaystyle s\leq u$ for all s in S. In particular for {a,b,c} both l and m are upper bounds from your diagram. j is not an upper bound since j is not greater than or equal to c. You can thus determines all upper bounds for {a,b,c}. Now the least upper bound of a set S is an upper bound v of S such that $\displaystyle v\leq u$ for any upper bound u of S. (By the anti symmetry property of $\displaystyle \leq$, if a least upper bound exists, it is unique.) Once you have listed all upper bounds, it is clear that k is the least upper bound. Similar analysis for lower bounds and greatest lower bound.