Are all complete lattices a pointed complete partial order, and vice versa?

Question

A friend of mine asked for my help in drawing a venn diagram that includes the notions of partial orders (PO) in general, complete partial orders (CPO), pointed complete partial orders (CPPO), total orders (TO), lattices and complete lattices.

Here are the relevant definitions:

• PO: A pair (W, R) where the relation R is reflexive, antisymmetric and transitive
• TO: A total (or linear) order is a PO (W, R) where all elements are comparable, i.e. for all x,y in W, either R(x,y) or R(y,x)
• Chain: A subset C of (W, R) is a chain if all elements in C are totally ordered
• CPO: A partial order (W, R) where every non-empty chain in W has a least upper bound (supremum) in W
• CPPO: A complete partial order with a least element, i.e. an element '0' such that for all x in W, we have R(0,x)
• Lattice: A PO where any two elements have a supremum and an infimum (greatest lower bound)
• Complete Lattice: A lattice where every subset of W has a sup and an inf.

This Venn Diagram needs to show which of these categories are included in which. Additionally, he needs to give an example for each intersecting or non-intersecting part of every set of relational structures.

Now, so far he knows that all categories fall within the notion of 'partial order', so everything else is included in that, and of course all complete lattices are lattices and all CPPOs are CPOs. Additionally, I helped him by writing a proof that every complete lattice is a cppo (see below). However, I am not 100% sure of this proof, and whether a CPPO that is a lattice is autimatically also a complete lattice. Neither of us can think of an example for a CPPO that is not a complete lattice, but still a non-complete lattice.

Answer 1

Here is my proof for showing that every complete lattice is a cppo:

Let F=(W, R) be a complete lattice. Then every subset of W including W itself has an infimum in W, hence F is pointed. It is also a partial order since F is a lattice. Finally, it is a complete partial order since if any subset has a supremum then every chain has a supremum.

Answer 2

Here is a proof of the fact that a lattice, in which every chain (including the empty chain) has a supremum, is a complete lattice.

A directed set is a poset $(D,\le)$ in which every pair of elements has an upper bound. The following lemma is copied from my answer to the old question Does closed under unions of chains imply closed under unions of upward directed families of sets?.

Lemma 1. If $D$ is a directed set of cardinality $\kappa$, an infinite cardinal, then $D$ is the union of a chain of directed subsets, each of cardinality less than $\kappa$.

Proof. We consider three cases.

Case 1. $\kappa=\aleph_0$.

Let $D=\{d_n:n\lt\omega\}$ be an enumeration of $D$. We will define a sequence of finite directed sets $D_n$. Let $D_0=\{d_0\}$. For $n\gt0$, having defined $D_{n-1}$, let $u_n$ be an upper bound for $D_{n-1}\cup\{d_n\}$, and let $D_n=D_{n-1}\cup\{d_n,u_n\}$. Then $\{D_n:n\lt\omega\}$ is a chain of finite directed sets whose union is $D$.

Case 2. $\kappa$ is an uncountable regular cardinal.

Let $D=\{d_\alpha:\alpha\in\kappa\}$. For $\beta\in\kappa$, let $D_\beta=\{d_\alpha:\alpha\lt\beta\}$. Then $B=\{\beta\in\kappa:D_\beta\text{ is directed}\}$ is unbounded in $\kappa$, so that $\{D_\beta:\beta\in B\}$ is a chain of directed sets whose union is $D$, and of course $|D_\beta|\lt\kappa$ for each $\beta$. (To see that $B$ is unbounded, given an ordinal $\beta_o\lt\kappa$, consider the limit of a sequence $\beta_0\lt\beta_1\lt\beta_2\lt\cdots\lt\beta_n\lt\cdots\lt\kappa$ where every finite subset of $D_{\beta_n}$ has an upper bound in $D_{\beta_{n+1}}$.)

Case 3. $\kappa$ is a singular cardinal.

Let $D=\bigcup_{\alpha\in\lambda}E_\alpha$ where $|E_\alpha|\lt\kappa$ and $\lambda=\operatorname{cf}\kappa$. Recursively define directed sets $D_\alpha\subseteq D\ (\alpha\in\lambda)$ so that $|D_\alpha|\lt\kappa$ and $E_\alpha\cup\bigcup_{\xi\lt\alpha}D_\xi\subseteq D_\alpha$. Then $\{D_\alpha:\alpha\in\lambda\}$ is a chain of directed sets whose union is $D$.

Lemma 2. If $P$ is a poset in which every chain has a supremum, then every directed subset of $P$ has a supremum in $P$.

Proof. Assuming the contrary, let $D$ be a directed subset of $P$ of minimum cardinality with no supremum in $P$. Of course $D$ must be infinite, since the empty set is a chain, and a nonempty finite directed set has a greatest element. By Lemma 1, we can write $D=\bigcup_{i\in I}D_i$ where $\{D_i:i\in I\}$ is a chain of directed sets, and $|D_i|\lt|D|$ for each $i\in I$. By the minimality of $|D|$, each $D_i$ has a supremum $\sup D_i=d_i\in P$. But then $\{d_i:i\in I\}$ is a chain in $P$, whose supremum is also the supremum of $D$.

Theorem. A lattice in which every chain has a supremum is a complete lattice.

Proof. Let $L$ be a lattice in which every chain has a supremum. By Lemma 2, every directed subset of $L$ has a supremum in $L$. If $X$ is any nonempty subset of $L$, then $\langle X\rangle$, the sublattice generated by $X$, is a directed subset of $L$, and $\sup\langle X\rangle=\sup X$.