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# Riesz interpolation property

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

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

1. $(\mathcal{A},\preceq)$ has the interpolation property

2. for every pair of doubletons $\{a_{1},a_{2}\}$ and $\{b_{1},b_{2}\}$ with $a_{i}\leq b_{j}$ for $i,j\in\mathbf{2}$, there is a $c\in P$ such that $a_{i}\leq c\leq b_{j}$ for $i,j\in\mathbf{2}$.

3. for every pair of finite sets $\{a_{1},\ldots,a_{n}\}$ and $\{b_{1},\ldots,b_{m}\}$ with $a_{i}\leq b_{j}$ for $i\in\mathbf{n}$ and $j\in\mathbf{m}$, there is a $c\in P$ such that $a_{i}\leq c\leq b_{j}$ for $i\in\mathbf{n}$, and $j\in\mathbf{m}$.

Here, $\mathbf{n}$ denotes the set $\{1,\ldots,n\}$.

###### Proof.

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

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

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

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:

$\xymatrix@!=40pt{b_{1}\ar@{-}[rd]|!{"2,1";"1,2"}\hole\ar@{-}[d]&b_{2}\ar@{-}[% ld]\ar@{-}[d]\\ a_{1}&a_{2}}\xymatrix@!=7pt{&&\\ &\mbox{ implies }&\\ &&}\xymatrix@!=7pt{b_{1}\ar@{-}[rd]&&b_{2}\ar@{-}[ld]\\ &c\ar@{-}[rd]\ar@{-}[ld]&\\ a_{1}&&a_{2}}$ |

## Mathematics Subject Classification

06F15*no label found*06A99

*no label found*06F20

*no label found*

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