# Are there necessary and sufficient conditions so that every element in a partially ordered set is either the least element or in the upset of an atom?

This is the first of three self-answered questions which will culminate in a proof of necessary and sufficient conditions for Krein-Milman type conclusions.

The second question is here.

The third question is here.

A partially ordered set will be the underlying object for this work.

Suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set with least element $\bot$. For all $a \in P$ we will say that $a$ is an \emph{atom} if and only $\bot < a$ for all $p \in P$ the inequality $p \leq a$ implies $p = \bot$ or $p = a$.

Are there necessary and sufficient conditions so that for all $p \in P \smallsetminus \{ \bot \}$ there is an atom $a$ with $a \leq p$?

Suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set with least element $\bot$. For all nonempty $A \subseteq P$ and all $g \in P$ we will write $\wedge A = g$ if and only if $A$ has a greatest lower bound in $P$ and that greatest lower bound is $g$. We do not assume that every subset of $P$ has a greatest lower bound. If we write $\wedge A = g$ then we will be asserting that $A$ has a greatest lower bound and that greatest lower bound is $g$.

If $A \subseteq P$ has a least element we will denote that element by $\bot_{A}$.

For all $p \in P$ we define

$$\downarrow p = \{ a \in P \colon a \leq p \} .$$

Suppose that $B \subseteq P$ has a least element $\bot_{B}$. For $b \in B$ we will say that $b$ is a $B$*atom* if and only if for all $a \in B$; if $a \leq b$ then $a = \bot_{B}$ or $a = b$. We will say that $B$ satisfies the finite meet property if and only if for all nonempty $A \subseteq B$; if $\wedge A = \bot_{B}$ then there exists a finite $F \subseteq A$ with $\wedge F = \bot_{B}$. We will say that $B$ satisfies the eventual finite meet property if and only if for all $b \in B$ there exists an $a \in B \cap (\downarrow b)$ for which the set $B \cap (\downarrow a)$ satisfies the finite meet property.

1. Proposition. Suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set and $B \subseteq P$ has a least element $\bot_{B}$. If $B$ satisfies the finite meet property then for all $b \in B$ there exists a $B$-atom $a \leq b$.

Proof. Select $b \in B \smallsetminus \{ \bot_{B} \}$. Define $$\mathcal{A} = \{ T \subseteq (B \cap \downarrow b) \smallsetminus \{ \bot_{B} \} \colon b \in T \text{ and } T \text{ is totally ordered by } \leq \} \text{.}$$ The set $\mathcal{A}$ is partially ordered by inclusion. Suppose that $\mathcal{T} \subseteq \mathcal{A}$ is nonempty and totally ordered by inclusion. We wish to prove that $\cup \mathcal{T} \in \mathcal{A}$. Since $\mathcal{T}$ is not empty we have $b \in \cup \mathcal{T}$. Select $t_{0}, t_{1} \in \mathcal{T}$. For some $L_{0}, L_{1} \in \cup \mathcal{T}$ we have $t_{0} \in T_{0}$ and $t_{1} \in T_{1}$. Since the elements of $\mathcal{T}$ are totally ordered by inclusion for some $i \in \{ 0, 1 \}$ we have $T_{1 - i} \subseteq T_{i}$. But then $t_{0}, t_{1} \in T_{i}$, a totally ordered set. Thus $\cup \mathcal{T} \in \mathcal{A}$ and $\cup \mathcal{T}$ is an upper bound for $\mathcal{T}$. Applying Zorn's lemma the set $\mathcal{A}$ has a maximal element $T^{*}$. There are three cases:

1. We have $\wedge T^{*} = \bot_{B}$.
2. The set $T^{*}$ does not have a greatest lower bound in $B$.
3. For some $a \in B \smallsetminus \{ \bot_{B} \}$ we have $\wedge T^{*} = a$.

In the following three paragraphs we will show that the first two cases can not happen. Then we will finish the proof for the third case.

In this paragraph we will suppose that case 1 holds. Using the finite meet property there is a finite $F \subseteq T^{*}$ satisfying $\wedge F = \bot_{B}$. Since $F$ is finite and totally ordered $\bot_{B} \in F$, contradicting $\bot_{B} \notin T^{*}$.

Suppose that case 2 holds. Since $T^{*}$ does not have a greatest lower bound in $B$ there exist $b_{0}, b_{1} \in B$ satisfying $\bot_{B} < b_{0} < b_{1} \leq t$ for all $t \in T^{*}$. Observing that $b_{0} \notin T^{*}$ and $T^{*} \cup \{b_{0} \} \in \mathcal{A}$ we contradict the maximality of $T^{*}$.

Thus case 3 holds. The maximality of $T^{*}$ implies that $a \in T^{*}$. Select $c \in B$ with $\bot_{B} \leq c \leq a$. If $\bot_{B} < c$ then the maximality of $T^{*}$ ensures that $c = a$. But then $a$ is a $B$-atom.

2. Theorem. Suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set and $B \subseteq P$ has a least element $\bot_{B}$. If $B$ satisfies the eventual finite meet property then for all $b \in B \smallsetminus \{ \bot_{B} \}$ there exists a $B$-minimal $a \leq b$.

Proof. In this paragraph we will suppose that the eventual finite meet property holds for $B$. Select a $b \in B \smallsetminus \{ \bot_{B} \}$ and then, using the eventual finite meet property, a $b^{*} \in (B \cap {\downarrow} b) \smallsetminus \{ \bot_{B} \}$ for which the set $B \cap {\downarrow} b^{*}$ satisfies the finite meet property. Recalling proposition 1 we see there is a $B$-atom $a \leq b^{*} \leq b$.

Now suppose that for all $b \in B \smallsetminus \{ \bot_{B} \}$ there exists a $B$-atom $a \leq b$. Now choose $b \in B \smallsetminus \{ \bot_{B} \}$. The finite set $B \cap {\downarrow} a = \{ \bot_{B}, a \}$ satisfies the finite meet property.

Postscript In the third question, when we discuss extreme elements we will call atoms minimal elements in keeping with the usual terminology for extreme subsets.