Riesz interpolation property

The interpolation property, in its most general form, may be interpreted as follows: given a set $S$ and a transitive relation $⪯$ defined on $S$, we say that $(S,⪯)$, or $S$ for short, has the interpolation property if for any $a,b\in S$ with $a⪯b$, there is a $c\in S$ such that $a⪯c⪯b$.

Let $P$ be a poset. Let $𝒜$ be the set of all finite subsets of $P$. Define $⪯$ on $𝒜$ as follows: for any $A,B\in 𝒜$, $A⪯B$ iff $a\le b$ for every $a\in A$ and every $b\in B$. It is not hard to see that $⪯$ is a transitive relation on $𝒜$. The following are equivalent:

1. $(𝒜,⪯)$ has the interpolation property

2. for every pair of doubletons $\{a_1,a_2\}$ and $\{b_1,b_2\}$ with $a_i\le b_j$ for $i,j\in \mathrm{𝟐}$, there is a $c\in P$ such that $a_i\le c\le b_j$ for $i,j\in \mathrm{𝟐}$.

3. for every pair of finite sets $\{a_1,\mathrm{\dots },a_n\}$ and $\{b_1,\mathrm{\dots },b_m\}$ with $a_i\le b_j$ for $i\in 𝐧$ and $j\in 𝐦$, there is a $c\in P$ such that $a_i\le c\le b_j$ for $i\in 𝐧$, and $j\in 𝐦$.

Here, $𝐧$ denotes the set $\{1,\mathrm{\dots },n\}$.

Proof.

Clearly $1\Rightarrow 2$ and $3\Rightarrow 1$. To see that $2\Rightarrow 3$, we use induction twice:

if $𝐧=\mathrm{𝟐}=𝐦$, then we are done. Now, fix $𝐧=\mathrm{𝟐}$ and induct on $𝐦$ first. Let $i\in \mathrm{𝟐}$. If $a_i\le b_j$ for $j\in 𝐦+\mathrm{𝟏}$, then $a_i\le b_j$ for $j\in 𝐦$ in particular, so there is a $c\in P$ such that $a_i\le c\le b_j$ for $j\in 𝐦$ (induction step). This means $a_i\le c$ and $a_i\le b_{m+1}$. Apply $2$ to get a $d\in P$ with $a_i\le d$ and $d\le c$ and $d\le b_{m+1}$. As a result, $a_i\le d\le b_j$ for $j\in 𝐦+\mathrm{𝟏}$.

Next, fix $𝐦$ and induct on $𝐧$. Let $j\in 𝐦$. If $a_i\le b_j$ for $i\in 𝐧+\mathrm{𝟏}$, then $a_i\le b_j$ for $i\in 𝐧$ in particular, so there is an $e\in P$ such that $a_i\le e\le b_j$ for $i\in 𝐧$ (induction step). This means $a_{n+1}\le b_j$ and $e\le b_j$. Apply the result from the previous induction step, we find an $f\in P$ such that $a_{n+1}\le f$ and $e\le f$ and $f\le b_j$. As a result, $a_i\le f\le b_j$ for $i\in 𝐧+\mathrm{𝟏}$.∎

Definition. A poset is said to have the Riesz interpolation property if it satisfies any of the three equivalent conditions above.

In other words, if one finite set, say $A$, is bounded above by another finite set $B$, then there is an element $c$ that serves as an upper bound for $A$ and a lower bound for $B$. One readily sees that any lattice has the Riesz interpolation property. In fact, a poset having the Riesz interpolation property can be thought of as an intermediate concept between an arbitrary poset and a lattice.

A poset having the Riesz interpolation property can be illustrated by the following Hasse diagrams:

Remark. One can generalize the Riesz interpolation property on a poset $P$ to the countable interpolation property, if $𝒜$ is to be the set of countable subsets of $P$, or a universal interpolation property, if $𝒜=2^P$, the powerset of $P$.