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\newtheorem{thm}{Theorem}
\def\xx#1 {\newtheorem{#1}[thm]{#1}}
\xx Lemma
\xx Theorem
\xx Corollary
\xx Idea
\xx {First Step}
\xx Consequence
\xx Example
\xx Definition
\xx Problem
\xx{Open Questions}
\xx{Open Question}
\xx Speculation
\author{Martin Goldstern}
\title{Lattices, Interpolation, and Set Theory}
\address{Institut f\"ur Algebra\\TU Wien\\
Wiedner Hauptstr 8-10/118.2\\A-1040 Wien}
\email{Martin.Goldstern@tuwien.ac.at}
\urladdr{http://info.tuwien.ac.at/goldstern/}
\thanks{This paper is available on my homepage, and also from {\tt
arXiv.org}}
\thanks{I am grateful to the Austrian Science Foundation (FWF) for
supporting my research through grant P13325-MAT}
\setcounter{page}{23}
\begin{document}
\nocite{mfl}
\nocite{l01}
\nocite{633}
\nocite{Haviar+Ploscica:1998}
\nocite{Wille:1977}
\begin{abstract}
We review a few results concerning interpolation of monotone functions
on infinite lattices, emphasizing the role of set-theoretic
considerations.
We also discuss a few open problems.
\end{abstract}
\maketitle
\section{Introduction}
\label{section.intro}
In this article I will present a few ideas concerning the
relationship between lattice polynomials and monotone functions
on lattices. It turns out that (for infinite lattices) there are
purely algebraic questions which (apparently) cannot be solved with
purely algebraic methods; ideas from logic --- in
particular, from set theory --- have to be
used. Most prominent among these logical concepts is the notion of
distinct infinite cardinalities, but also the more subtle notion of
``cofinality'' plays a role.
Most of the results here are not new; specifically, sections
\ref{section.opc}
and \ref{section.IP}
just
highlight ideas from papers that are already published.
The reason for including those known
results here is that they serve well to show the interconnections
between algebraic and set theoretic ideas. \\
The observations in sections
\ref{section.complete}, \ref{section.sigma},
and \ref{section.ortho} are based on these
older results, but are themselves new.
I will often not give complete proofs but treat only a characteristic
case, in the hope that the information presented here is enough for
the reader to complete the proof.
For example, I
will sometimes prove (or even state) a theorem which is true for $n$-ary
functions only for unary functions. The general case is often just
an easy generalization (but sometimes needs an additional idea).
In any case, complete proofs are
available elsewhere in the literature.
To avoid some case distinctions, we
will only consider bounded lattices; in addition to the operations
$\vee $ and~$\wedge$, all lattices that we consider in this paper
come with constants $0$ and~$1$. So ``lattice'' will mean ``bounded
lattice'', all homomorphisms
have to respect 0 and 1, all sublattices contain $0$ and~$1$, etc.
\section{Order-polynomial completeness}
\label{section.opc}
The problem that got me interested in the subject was the following:
\begin{quote}
It is clear that every lattice polynomial\footnote{Recall that the set
$L[\x_1,\ldots, \x_n]$ of $n$-ary lattice polynomials [``with
coefficients from $L$''] is the free product of $L$ and the free lattice
with $n$ generators, or in other words: take
the set of well-formed expression
using the indeterminates $\x_1,\ldots,\x_n$, elements of the
lattice~$L$ and the operation symbols $\vee $ and~$\wedge$, and
divide by
the obvious equivalence relation: ``equal in all extensions of $L$''.
The $n$-ary
function induced by a lattice polynomial is called ``polynomial
function'' (or ``algebraic function'' by some authors).}
represents a monotone
function. \\For which lattices is the converse true?
\end{quote}
Let us call a lattice $L$ $k$-\opc\ ($k$-order polynomially complete) if
every monotone function $f:L^ k \to L$ is represented by a polynomial,
and call $L$ \opc\ if $L$ is $k$-\opc\ for every~$k$.
The finite \opc\ lattices were characterized by Wille (see section
\ref{section.finite}).
In this section we will show the main ideas of the proof
of the following theorem:
\begin{Theorem}
\label{opc}
There are no infinite \opc\ lattices.
\end{Theorem}
Although the problem itself
is now solved, I feel that our understanding is
not at all complete. As evidence for this lack of
understanding, see problems~\ref{A} and \ref{B}.
\subsection*{Cardinalities} If there are ``more'' monotone
functions than polynomials, then clearly not all monotone
functions are polynomials. In the finite case this does not help much,
since there are only finitely many monotone functions for every fixed
arity, while there are always infinitely many 3-ary polynomials.
{\smaller\relax [Of course, for lattices with extra properties, such as
distributive lattices, we can get a normal form for polynomials, which
means that
the set of polynomials looks rather ``small'' to us.]}
In infinite lattices the situation is reversed: The set $P$ of
polynomials on a lattice $L$ has the same
cardinality\footnote{Recall
the following elementary facts about cardinals:
\begin{itemize}
\item
We write $A \approx B$ or $|A|=|B|$ iff there is a
bijection between $A$ and~$B$. We write $ |A|$ for the
``cardinality'' of~$A$, which can informally be defined as the
equivalence class of $A$ with respect to~$\approx$.
\item We write $A \le B$ or $|A|\le|B|$ iff there is a 1-1 map from
$A$ into~$B$. Equivalently (for nonempty $A$): $|A|\le|B|$
iff $B$ can be mapped onto~$A$.
\item The notation $|A|\le|B|$ is justified by the following
theorem: If $|A|\le|B|$ and $|B|\le|A|$, then $|A| = |B|$.
\item $|A| = |A^ n | = |\bigcup_k A^ k|$ for all infinite~$A$.
\item We say that a set $A$ is countable
and write $|A| = {\aleph_0} $
iff there is a bijective map from $\N$ onto~$A$. $A$ is ``at most
countable'' [$|A|\le {\aleph_0} $] iff $A$ is countable or finite.
\end{itemize}
}
as $L$ itself,
whereas the set $L^ L$ of all functions from $L$ into $L$ has always
strictly larger cardinality.
However, not all functions are monotone, and there are even trivial
examples of lattices $L$ with only ``few'' monotone functions, i.e.,
satisfying
$$ \{f: \mbox{$f$ is monotone from $L$ to $L$}\} \approx L$$
for example the linear order $\R$ (the real numbers).
Here is the main idea that can be used for many lattices $L$ to show
that $L$ is not \opc:
\begin{enumerate}
\itm a find a ``nice'' substructure $A$ of the same cardinality as $L$
\itm b find a subset $L' \subseteq L$
such that
\begin{itemize}
\item for all ``nice'' structures $A$ of
cardinality $ \kappa$ there are $> \kappa $ many monotone maps from
$A$ to $L'$
\item the set $L'$, with the order inherited from $L$, is a complete
lattice [not necessarily a sublattice of $L$]
\end{itemize}
\itm c conclude that there are $> \kappa $ many monotone maps from
$L$ to $L$.
\end{enumerate}
Part (c) is easy: every monotone $f:A \to L'$ can be extended to $\bar
f:L \to L'$ via
$$ \bar f(x) = \sup \nolimits_{L'}\{ f(z): z\in A, z\le x\}.$$
% For (c) it is of course sufficient that $L'$ is a complete partial
% order, i.e., a partial order where every subset has a least upper and
% greatest lower bound.
\subsection*{From (anti)chains to monotone functions}
It is well-known (see, e.g., \cite{Haviar+Ploscica:1998} or
\cite{633}) that if $A \subseteq L $ is an antichain\footnote{$A
\subseteq L$ is
called an antichain if no two distinct elements of $A$ are
comparable.} then there are\footnote {If $|A| = \kappa$, we
write $2^ \kappa $ for the cardinality of~$\P(A)$. Thus, a set $B$ is
of cardinality $2 ^ {\aleph_0} $ iff $B$ can be mapped bijectively
onto $\P(\N)$ or~$\R$.}
$2^ {|A|} > |A|$ many pairwise incomparable\footnote{If
$f,g: L^ n\to L$ are
functions, we say $f\le g$ iff
$f(a_1,\ldots, a_n) \le g(a_1,\ldots, a_n)$
holds for all $(a_1, \ldots, a_n) \in L^ n$.
The relation $\le$ is then a partial order of functions.
The polynomials are (via
the functions they induce) quasiordered.
}
functions from $A$ to~$\{0,1\}$, and since $\{0,1\}$ is a complete
lattice it is trivial to extend all those functions to
monotone functions defined on all of $L$. If
$A \subseteq L$ is
well-ordered\footnote{Recall that
a linearly ordered set $(A,\le)$ is called ``well-ordered'' iff every
nonempty subset of $A$ has a least element.}, there are $2^{|A|}$
many monotone functions from $A$ to~$A$, and again they can all be
extended to total monotone functions from $L$
into the complete lattice $A \cup
\{0,1\}$.
Thus, a first approximation to the program (a)(b)(c) outlined above
is: For a given lattice~$L$, find a ``large'' antichain or
well-ordered chain in $L$ (where ``large'' means: of the same
cardinality as $L$ itself).
Unfortunately, this is not always possible but at least for countable
lattices the following observation helps:
{\em Ramsey's Theorem}\footnote{We will use the following
version of the infinitary Ramsey theorem:
Whenever the edges of the complete graph on countably
many vertices are colored with 3 colors, then there is an infinite
complete subgraph, all of whose edges have the same color.}
implies that every infinite partial order
contains either a chain isomorphic or anti-isomorphic to the natural
numbers, or an infinite antichain.
{\small Proof: We may assume that the partial order $P$ is countable:
$P = \{p_1, p_2, \ldots \}$, where all $p_i$ are distinct. Color the
edges of the complete graph on $\{1,2,\ldots\}$ with 3 colors: The
edge $\{i,j\}$ is colored according to whether the map
$i \mapsto p_i$,
$j\mapsto p_j$ is an isomorphism, anti-isomorphism, or neither. Any
infinite complete
subgraph whose edges are colored with only one color will be
the desired chain or antichain.}
Hence any infinite partial order, in particular any infinite lattice,
admits $\ge 2^ {\aleph_0} $ many
monotone functions, so an \opc\ lattice has to have size
$\ge 2^{\aleph_0} $.
\bigskip
%\showhyphens{functions}
The above method yielding many incomparable functions from an
antichain is set-theoretical. The next idea, converting a set of
incomparable functions into an antichain, is mainly algebraic. Here
it is important that we are interested in fully \opc
(and not only $1$-\opc) lattices:
\subsection*{From monotone functions to antichains}
\begin{Lemma}\label{hp}
Let $L$ be a lattice on which there are $\kappa $ many
pairwise incomparable
$k$-ary polynomials,
where\footnote {$cf(\kappa) > {\aleph_0} $
[read: $\kappa $ has uncountable cofinality]
means: $\kappa $ is the
cardinality of an infinite set $A$ with the following property:
\begin{quote} Whenever $ A = \bigcup_{n=1}^ \infty
A_n$, then for some $n$, $|A| =
|A_n|$.
\end{quote}
For example, it is true
(but not quite trivial, unless we assume the continuum hypothesis)
that the set $\R$ of real
numbers satisfies $cf(|\R|) > {\aleph_0} $.
\endgraf
The negation of this property is denoted $cf ( \kappa) = {\aleph_0}$:
$\kappa = |A| $ for some infinite set $A$ which can be written
as a countable union of sets of strictly smaller cardinality.
}
$cf(\kappa ) > {\aleph_0}$.
Then there is some $n'$ such that $L^ {n'}$ contains an antichain of
size $ \kappa $.
\end{Lemma}
\begin{proof} (This idea, and in fact the whole proof, is due to
\cite{Haviar+Ploscica:1998}.)
Assume that the polynomials $(p_i(\bar \x): i \in I)$ are pairwise
incomparable, where $|I| = \kappa $, and $\bar \x$ abbreviates
$(\x_1,\ldots, \x_k)$.
For each $i$ there is a natural
number $n_i$ such that $p_i( \bar \x)$ can be written as
$t_i(\bar \x, \bar c_i)$, where $t_i$ is a $(k+n_i)$-ary term, and $\bar
c_i \in L^{n_i}$ (the entries of the vector $\bar c_i$ are the
``coefficients'' of the polynomial $p_i$).
Since there only countably many
terms, our assumption $cf( \kappa ) > {\aleph_0}$ implies we can thin out
the set $I$ to a set $I'$ of the same cardinality such that all the
$(n_i,t_i)$ (for $i \in I'$) are equal, say $= (n', t')$. Now it is
easy to check that the ``coefficients'' $(\bar c_i: i \in I')$ form an
antichain in~$L^{n'}$.
\end{proof}
\begin{Corollary}\label{hp2}
If $L^ {n_1}$ has an antichain or well-ordered chain
of size~$\kappa$,
and $L$ is $n_1$-\opc, then there are $2^ \kappa $ many pairwise
incomparable polynomial functions $p:L^ {n_1} \to L$. So there is some
$n_2$ such that $L^ {n_2} $ has an antichain of size $2^ \kappa$.
Repeating this argument we can find some $n_3$ such that $L^ {n_3}$
has an antichain of size $2^{2^ \kappa }$, etc.
\end{Corollary}
So let $L$ be infinite and \opc{}
To
simplify some calculations, we will assume
GCH, the generalized continuum hypothesis, for the rest of this
chapter. However, all of what will be said will --- with some obvious
modifications --- remain true even without this additional
assumption.
We know that $L$ contains a chain or antichain of size~${\aleph_0}$,
so also one of size $2^ {\aleph_0} $ (since we assume GCH we can write
this as~${\aleph_1}$). Iterating corollary~\ref{hp2} we get
$$ |L| \ge {\aleph_1}, \
|L| \ge {\aleph_2}, \ |L| \ge {\aleph_3}, \ldots$$
and finally $|L| \ge {\aleph_\omega } $.
It turns out that the proof splits into three very different
cases, depending on $\kappa$, the cardinality of $L$:
\begin{description}
\item[{a}] For some $\mu < \kappa$, $2^\mu \ge \kappa$.
\item[{b}] Case (a) does not hold, and $cf(\kappa) = {\aleph_0} $.
\item[{c}] Case (a) does not hold, and $cf(\kappa) > {\aleph_0} $.
\end{description}
Case (a) is treated similarly to the cases $\kappa=\aleph_1$,
$\kappa=\aleph_2$, etc. If case (a) does not hold, i.e., if we have
$$ \forall \mu <\kappa: 2^\mu < \kappa$$
then we call $\kappa$ a ``strong limit cardinal''.
We will in this paper repeat the argument from
\cite{633} for case (b). (Case (c) was considered in \cite{opc}.)
It turns out that the case $|L| = {\aleph_\omega } $ is typical for
(b), so to again simplify the notation we will assume this equality.
Thus $L$ can be written as $L = L_0 \cup L_1 \cup \cdots\, $, where
$|L_n| = {\aleph_n} $, $L_{n+1} \approx \P(L_n)$.
Can we prove that every lattice $L$ of size ${\aleph_\omega}$
contains either an antichain or a chain (well-ordered or dually
well-ordered) of cardinality~${\aleph_\omega}$? Unfortunately this
is not true. The best we can do is to show that for each~$n$, $L$
must contain a (well-ordered or co-well-ordered) chain or antichain
$A_n$ of cardinality ${\aleph_n} $, but it is quite possible that the
union $A:= A_0 \cup A_1 \cup A_2 \cup \cdots$ is neither a chain nor
an antichain, as the following examples show:
\begin{Example}
\label{ex1} For each $n$ let $A_n$ be a well-ordered (or
co-well-ordered) set of cardinality ${\aleph_n} $, and assume that the
union
$$L = \{0\} \cup \{1\} \cup A_0 \cup A_1\cup \cdots$$
is a disjoint union. Define a partial order on $L$ by requiring $0$
and $1$ to be the least and greatest elements respectively, and by
also requiring
$$ \forall n \not= k \ \forall a \in A_n
\, \forall b \in A_k:
\qquad \mbox{$a$ and $b$ are incomparable.}$$
Thus, $L$ consists of countably many chains (of increasing sizes),
arranged side-by-side. We leave it to the reader to check that $L$
is indeed a lattice, in fact a complete lattice.
$L$ contains chains of cardinality ${\aleph_n} $ for each~$n$,
but no chain of length~${\aleph_\omega}$, and every antichain in $L$
is at most countable.
\end{Example}
\begin{Example}
\label{ex2} For each $n$ let $A_{2n+1}$ be an antichain of
cardinality~${\aleph_n}$, and let $A_{2n} = \{a_n\}$.
Again, assume that the
union
$$L = \{0\} \cup \{1\} \cup A_0 \cup A_1\cup \cdots$$
is a disjoint union. Define a partial order on $L$ by requiring $0$
and $1$ to be the least and greatest elements respectively, and by
also requiring
$$ \forall n < k \ \forall a \in A_n
\, \forall b \in A_k:
\qquad a < b .$$
Thus, $L$ consists of countably many antichains, each one on top of
the previous one. Again it is easy to check that $L$ is a complete
lattice. [If $a,b\in A_{2n+1}$ are distinct, then $a\wedge b =
a_{2n}$.]
This time
$L$ contains antichains of cardinality ${\aleph_n} $ for each~$n$,
but no antichain of length~${\aleph_\omega}$, and every chain in $L$
is at most countable.
\end{Example}
\begin{Example}
\label{ex3} For each $n$ let $A_n$ be a well-ordered set of size $\aleph_n$.
Again, assume that the
union
$$L = \{0\} \cup \{1\} \cup A_0 \cup A_1\cup \cdots$$
is a disjoint union. Define a partial order on $L$ by requiring $0$
and $1$ to be the least and greatest elements respectively, and by
also requiring
$$ \forall n < k \ \forall a \in A_n
\, \forall b \in A_k:
\qquad a > b .$$
Thus, elements of $A_0$ are above all elements of $A_1$, etc. $L$
itself is a chain of cardinality $\aleph_\omega$, but $L$ is not
well-ordered. Moreover, any well-ordered subset of $L$ has
cardinality $< \aleph_\omega$, and any co-well-ordered subset of $L$ is
countable. There are no antichains of size $>1$ in $L$.
\end{Example}
In the first two
examples we have a lattice $L$ of cardinality~${\aleph_\omega}
$, all of whose chains and antichains are of size~$<
{\aleph_\omega}$.
Still, it is easy to see that the lattices in all three examples
admit $2^ {\aleph_\omega} $ many monotone
unary functions. E.g., in example~\ref{ex2}, any map
$f:L \to L$ satisfying $f[A_n] \subseteq A_n$ for all~$n$, $f(0)=0$,
$f(1)= 1$ will be monotone, and there are
$2^ {\aleph_0} \times 2^ {\aleph_1} \times \cdots \ = 2^
{\aleph_\omega } $ many such maps.
Hence, the following theorem suffices to prove that a lattice $L$ of
cardinality ${\aleph_\omega} $ cannot be \opc:
\begin{Theorem}[GCH]
If $L$ is a partial order of cardinality~${\aleph_\omega}$, then $L$
either contains a sufficiently large antichain or
(well-ordered or co-well-ordered) chain, or
a partial order $P$ which is isomorphic or antiisomorphic to one of
those constructed in examples \ref{ex1}, \ref{ex2} and \ref{ex3}.
\end{Theorem}
This theorem can be easily deduced from the ``canonization'' theorem
of Erd\H os, Hajnal and Rado, see
\cite[28.1]{Erdos+Hajnal+Mate+Rado:1984}.
\goodbreak
The first ``open'' question is rather ill-defined:
% \begin{description}
\begin{Problem}\label{A} Find a purely algebraic proof that
every \opc\ lattice must be finite.
\end{Problem}
% \end{description}
Why do we need such a proof? First, there may be algebraists who are
uncomfortable with the notions ``cardinality'' and ``cofinality''
which were used in the proof above.
Secondly, and more to the point, a new proof of theorem~\ref{opc} may
also shed light on the following problem, which is still
open:
\begin{Problem}\label{B}
Can there be an infinite 1-\opc\ lattice? If yes,
what about 2-\opc? etc.
\end{Problem}
\begin{Speculation}
In \cite{opc} it is shown that theorem \ref{opc} cannot be proved
without some weak version of AC, the axiom of choice; this seems to
indicate that some set-theoretic sophistication is necessary for any
proof of theorem \ref{opc}. However, on closer scrutiny it turns out
that the only version of AC that was shown to be necessary for such a
proof is a statement that is strictly weaker than
\begin{quote}
``every infinite set contains a countable subset''
\end{quote}
which for most non-logicians is not even recognizable as a version of
AC, so a ``purely algebraic'' proof might still be possible.
\end{Speculation}
\section{Finite \opc\ lattices}
\label{section.finite}
For the investigation of finite lattices, set theory plays of course
no role. We give the following characterization only to contrast it
below with the situation for infinite lattices.
Let us call a function $L\to L$ ``regressive'' if
$ \forall x\in L: f(x)\le x$.
\begin{Definition}
We say that a lattice satisfies Wille's property if the only
regressive join-homomorphisms are the identity map and the constant
0.
\end{Definition}
\begin{Theorem}
A finite lattice is \opc\ iff it is simple (in the algebraic sense)
and satisfies Wille's property.
\end{Theorem}
This theorem is proved in \cite{Wille:1977}.
It is of course crucial for the proof of this theorem
that the lattice under consideration be finite, since the length of
the constructed polynomial will typically increase with the size of
the lattice.
The following easy example shows that the characterization theorem
cannot work for infinite sets (of any cardinality).
\begin{Example}
\label{example-a}
Let $A$ be any infinite set disjoint from~$\{0,1\}$. Define a
lattice $M_A = A \cup \{0,1\}$ by requiring $A$ to be an antichain and
$0\le a\le 1$ for all
$a\in A$.
Then $M_A$ is simple and
satisfies Wille's property, but is of course not \opc.
\end{Example}
% \begin{proof} Easy. \end{proof}
\section{Interpolation}
\label{section.IP}
There are several natural ways to generalize the question ``which
monotone functions are represented by polynomials'' from finite to
infinite lattices. The first approach, the property \opc,
was discussed in section \ref{section.opc}. A second approach is the
following:
\begin{Definition}
We say that a lattice $L$ has the
IP (interpolation property) iff
\begin{quote}
for every monotone function $f:L^ n\to L$ and every finite $A \subseteq
L^ n$ there is a lattice polynomial $p$ such that
$$ \forall (a_1,\ldots, a_n)\in A: \ f(a_1,\ldots,a_n) =
p(a_1,\ldots,a_n).$$
\end{quote}
\end{Definition}
In other words, if we equip $L$ with the discrete topology, and
$L^{(L^n)}$, the set of all functions from $L^n$ to~$L$, with the
product topology (= Tychonoff topology = topology of pointwise
convergence), then the IP says:
\begin{quote} For all~$n$, the set of monotone functions in
$L^{(L^n)}$ is the closure of the set of all polynomial functions.
\end{quote}
Clearly, a finite lattice has the IP iff it is \opc, but
this is not true for infinitary lattices. For example, the lattice
given in example \ref{example-a} has the IP, but is of course not
\opc{}
While there are no infinite
\opc\ lattices, the following can be shown:
\begin{Theorem}
For every lattice $L$ there is an infinite lattice $\bar L$
with the IP, where $L$ is a sublattice of~$\bar L$.
\end{Theorem}
Elementary cardinal arithmetic shows that $\bar L$ can be chosen to be
of the same cardinality as the cardinality of~$L$, as long as $L$ is
infinite. {\small (If $L$ is finite, then it is known that $L$ can be
extended to a finite \opc\ lattice.)}
This theorem can be proved with a combination of set-theoretic and
algebraic ideas. The algebraic content of the theorem is
represented by the following lemma
(which is really a
weak version of the theorem itself).
\begin{Lemma}\label{extend}
If $L$ is a lattice, $f:L\to L$ a monotone partial function,
then there is a lattice $L'$
extending~$L$, such that $f$ is the restriction of a polynomial
function with coefficients in~$L'$.
Moreover, if $L$ is complete,
then $L'$ can be chosen to be again a complete
lattice, which is moreover an ``end extension'' of~$L$, i.e.,
$$ \forall b \in L'\, \forall a \in L \setminus \{1\}: \
\bigl(b \le a \Rightarrow b \in L\bigr).$$
\end{Lemma}
(This lemma is proved in \cite{mfl} and \cite{l01}.)
It turns out that this lemma is sufficient to construct lattices with
stronger interpolation properties. For example:
\begin{Definition}
We say that a lattice $L$ has the
${\sigma}$-IP (countable interpolation property) iff
\begin{quote}
For every monotone function $L^ n\to L$ and every at most countable
$A \subseteq
L^ n$ there is a lattice polynomial such that
$$ \forall (a_1,\ldots, a_n)\in A: \ f(a_1,\ldots,a_n) =
p(a_1,\ldots,a_n).$$
\end{quote}
\end{Definition}
(There is a natural generalization of the ${\sigma}$-IP to $\kappa$-IP
for any cardinal~$\kappa$. The theorems below can easily be
generalized to these situations.
For the sake of simplicity we will not
pursue this idea here.)
\begin{Theorem}
\label{mfl}
For every lattice $L$ there is an infinite lattice $\bar L$
with the ${\sigma}$-IP, where $L$ is a sublattice of~$\bar L$.
\end{Theorem}
We will sketch a proof of this theorem,
for simplicity
only for the
case where the original lattice $L$ is of size $\le {\aleph_1}$.
The simple (or even, by today's standards, trivial)
set-theoretic idea behind this
theorem is the idea of transfinite iteration:
\begin{quote}
{\sl If at first you don't succeed --- try, try again.}
\end{quote}
This idea goes back to Cantor. The point here is, of course that
``again'' can mean here much more than ``many'' or even ``infinitely
many'' times. When we are done with infinitely many repetitions, we
start over!
As an index set for this iteration we will use the well-ordered
set\footnote{Some
authors prefer to write $\Omega$ instead of~$\omega_1$.}
$(\omega_1, <)$. Rather than giving the set-theoretic definition
of $\omega_1$ we will list some of its order-theoretic properties.
The linear order $ (\omega_1,{<})$
can be characterized (up
to isomorphism) by (1)--(4):
\begin{enumerate}
\item $({\omega_1},{<})$ is a linear order.
\item $({\omega_1},{<})$ is a well-order (i.e., every nonempty
subset has a first element).
\item ${\omega_1}$ is uncountable.
\item For every $a\in {\omega_1}$ the set of predecessors,
$\{i\in {\omega_1} : i < a\}$ is at most countable. (I.e., ``every bounded set
is countable''.)
\item Moreover: whenever $ A \subseteq {\omega_1} $ is countable,
there there is an $ a\in {\omega_1} $ such that
$ A \subseteq \{ i\in {\omega_1}: i < a \}$. (I.e., ``every
countable set is bounded.'')
\end{enumerate}
We will write ${\aleph_1} $ for $|{\omega_1} |$. We will write
$0$ for the smallest element of~$\omega_1$, and for $i\in
\omega_1 $ we will write $i+1$ for the ``successor'' of~$i$, i.e.,
$$ i+1 = \min \{j\in \omega_1: i < j \}.$$
It suffices
to construct a sequence
$(L_i:i\in
{\omega_1} )$ starting with the original lattice $L=L_0$ such that:
\begin{enumerate}
\item[(A)] $ i < j $ implies: $L_i \subseteq L_j$ (as a sublattice).
\item[(B)] Each $L_i$ is countable.
\item[(C)] For every $i \in {\omega_1}$, all natural numbers~$k$,
and for all monotone partial functions with countable domain
$f:L_i^ k \to L_i$ there is a
polynomial $p\in L_{i+1}[\x_1,\ldots, \x_k] $
such that for all $ \bar a = (a_1,\ldots,a_k) $ in the domain
of~$f$: $f(\bar a)=p(\bar a)$.
\end{enumerate}
We first explain why such a construction satisfying (A)+(B)+(C)
is possible.
A rigorous argument would have to appeal to the fundamental
theorem of Definition by Transfinite Recursion; the following
informal argument captures the essence of the proof:
Assume that this construction is not possible. Since $\omega_1$
is well-ordered this means that
there is a first index $j$ where the
construction breaks down. If $j$ has no predecessor, then $L_j:=
\bigcup_{i