Complete lattice proof I understand the defnition of complete lattice.
But often times the argument that is given for showing something is complete lattice is it has a upper bound and a lower bound.
For eg The partial order <= under a set of rational numbers from (0,1) is a complete lattice because it has a universal upper bound 1and lower bound 0
But complete lattice requires that every subset should have Greatest lower bound and least upper bound.Having Upper bound doesnt ensure it has Greatest Upper bound right?
How to prove a lattice is complete?
 A: The open interval of rationals $(0,1) \cap \Bbb Q$ is not complete as a lattice. The extrema (greatest lower bound / least upper bound) of many of its subsets are not in the interval. For example, $(0, 1/2] \cap \Bbb Q$ and $[1/2, 1) \cap \Bbb Q$ has a greatest lower bound of $0$ and least upper bound of $1$ respectively, but $0,1 \not \in (0,1) \cap \Bbb Q$. 
In fact, even the closed interval of rationals $[0,1] \cap \Bbb Q$ is not a complete lattice, even though its extrema are in the set. $e/3 \in [0,1]$ for instance, but the subset $[0,e/3) \cap \Bbb Q$ would not have a supremum in $[0,1] \cap \Bbb Q$.
More generally, a lattice $(S, \le)$ is complete when, for all $T \subseteq S$, the greatest lower bound $\inf(T)$ and the least upper bound $\sup(T)$ exist and are elements of $S$.
It is also clear after this that it is not at all sufficient for $(S,\le)$ to have upper/lower bounds as a whole if you wish to conclude completeness. In short: a lattice may be bounded, but that does not guarantee completeness.
