# Difference between pre-lattice and complete posets?

As they are defined, they seem to be the same thing to me, other than that a pre-lattice involves finite subsets... Is there another important difference?

These are the definitions in the text:

We say that a partially ordered set $( S, \leq)$ has the largest-lower- bound property if $\inf E$ exists for every subset $E \subseteq S$ which is nonempty and bounded below.

Dually, we say that $S$ has the least-upper-bound property if $\sup E$ exists for subset $E \subseteq S$ which is nonempty and bounded above.

Partially ordered sets with the largest-lower-bound property are said to be inf-complete, and those with the least-upper-bound property are said to be sup-complete.

A partially ordered set $( S, \leq)$ is called a pre-lattice if every nonempty finite subset $E \subseteq S$ has supremum and infimum.

• It's a good idea to include context in your post, so we can know how to help! Could you add your definitions of pre-lattice and complete posets so that we know exactly your problem? – Isaac Browne Sep 5 '18 at 18:47
• Your definition of pre-lattice seems to coincide with the usual definition of lattice. Also, you don't define complete posets. Are they the posets which are both inf-complete and sup-complete? – amrsa Sep 6 '18 at 8:25

If you take e.g. $\mathcal P(\{a,b\})\smallsetminus\{\emptyset\}$ with respect to $\subseteq$, by your definition it is not a pre-lattice because $\{\{a\},\{b\}\}$ does not have $\inf$. But it is $\inf$-complete because you don't ask for $\{\{a\},\{b\}\}$ to have $\inf$ since it is not bounded from below.
Also if you take $\mathcal F$ to be the family of finite and co-finite subsets of $\mathbb N$ with respect to $\subseteq$, all finite sets have $\inf$ and $\sup$, but not all bounded from below sets have $\inf$ and bounded from above have $\sup$. E.g. $\{\{2n\}\mid n\in\mathbb N\}$ doesn't have $\sup$ in $\mathcal F$.