1,$ since that lattice contains an embedded copy
of $\Conv(\R^{n-1},\,\cp)$ (cf.\ proof of Lemma~\ref{X.13}),
to which the above observation applies.
While on the topic of join- and meet-semidistributivity, I will
note some questions and examples which are easier to state
now that we have named several sublattices of $\Conv(\R^n).$
Kira Adaricheva
(personal communication) has posed several questions of the following
form: If we take one of the sublattices
of $\Conv(\R^n)$ that we know to be join- or
meet-semidistributive, and extend it by adjoining,
within $\Conv(\R^n),$ a single ``nice'' outside element,
is the property in question already lost, and if so, does the resulting
lattice in fact contain copies of $M_3?$
In many cases this does happen.
For instance, the sublattice $L \subseteq\Conv(\R^2)$ generated by the
join-semidistributive sublattice $\Conv(\R^2,\,\cp),$ and the open disk,
which we shall denote $B,$ contains a copy of $M_3.$
To describe it, let $p_0,$ $p_1,$ $p_2$ be any three distinct points
on the unit circle, and for $i=0,1,2$ define the
point $q_i=1/5\,p_i + 2/5\,p_{i+1} + 2/5\,p_{i+2}$ (subscripts
evaluated\ \ mod~$3;$ I suggest making a sketch).
Let $x_i=(\ch(q_i,p_{i+1})\wedge B) \vee (\ch(q_i,p_{i+2})\wedge B).$
Then $x_1,\ x_2,\ x_3$ can be seen to generate a copy
of $M_3$ in $\Conv(\R^{2}),$ resembling,
though not identical to, the $n\,{=}\,2$ case
of the first example described in Lemma~\ref{X.60}.
An analogous construction gives a family of elements
of $L$ resembling the example
described immediately following that lemma, showing that $L$ is
not \!$m$\!-join semidistributive for any $m.$
As another example let $L$ be the sublattice
of $\Conv(\R^3)_{\geq\{0\}}$ generated
by $\Conv(\R^3,\,\ob)_{\geq\{0\}}$ and the cube $C=[-1,+1]^3.$
Letting $B$ denote the interior of $C,$ I
claim that for any triangle $T$ drawn on a
face $F$ of $C,$ $L$ contains the union
of $B$ with the interior of $T$ relative to $F.$
Indeed, we can find in $\Conv(\R^3,\,\ob)_{\geq\{0\}}$ an
open pyramid $P$ (with apex near $0)$
meeting no face of $C$ except $F,$ and
meeting $F$ in precisely the relative interior of $T.$
Then $(P\wedge C)\vee B$ will be the desired set.
If we construct three sets of this sort using as our $T$'s three
triangles in a common face $F $ of $C,$ whose relative
interiors form a copy of the $n\,{=}\,2$ case of the first example of
Lemma~\ref{X.60}, then the sublattice generated
by the three resulting sets $(P\wedge C)\vee B$ will
have the form $M_3$ (with $B$ as its least element;
cf.\ the last sentence of Corollary~\ref{X.69}).
The same trick can again be used to get examples like those of
the paragraph following Lemma~\ref{X.60}, so this $L,$ too, is not
\!$m$\!-join semidistributive for any~$m.$
On the other hand, I do not know what can be said about the lattices
gotten by
adjoining to $\Conv(\R^n,\,\ob)_{\geq\{0\}}$ a
closed ball $D$ about $0,$ respectively an open ball $B$ not
containing $0,$ except that the second of these lattices is not
meet-semidistributive, since we can find three ``fingers''
in $\Conv(\R^n,\,\ob)_{\geq\{0\}}$ which
meet $B$ in disjoint subsets $x,$ $y_1,$ $y_2,$ such
that $x\wedge(y_1\vee y_2)$ is nonempty.
\section{Relatively convex sets}\label{NHRel}
Let $S$ be a subset of $\R^n,$ in general non-convex.
We shall call a subset $x \subseteq S$ convex {\em relative
to} $S$ if $x=\ch(x)\cap S;$ equivalently, if $x$ is the
intersection of $S$ with some convex subset of $\R^n.$
Such sets $x$ will form a
lattice $\RelConv(S),$ with meet operation given, as
in (\ref{x.04}), by intersection, but join now given by
%
\[\label{x.83}
x\vee y\ =\ \ch(x\cup y)\,\cap\,S.\]
%
Note that the map $x \mapsto \ch(x)$ gives
a bijection between the relatively convex subsets of $S,$ and
the subsets of $\R^n$ which are convex hulls of subsets
of $S,$ the inverse map being given by $- \cap S.$
Convex sets of the latter sort form a lattice with
join as in~(\ref{x.04}), but with meet operation
%
\[\label{x.85}
x\wedge y\ =\ \ch(x\cap y\cap S).\]
%
We observe that Carath\'eodory's Theorem (with the
``refinement'' given in the final sentence), regarded as a statement
about the closure operator $\ch(-)$ on $\R^n,$ entails
the same properties for the closure operator $\ch(-)\cap S$ on $S.$
Hence the proofs of Lemma~\ref{X.23} and of the positive assertions
of Theorem~\ref{X.25} immediately yield
%
\begin{proposition}[{cf.\ Huhn \cite[Lemma~3.2]{AH.n-d}}]\label{X.88}
If $n$ is a natural number and $S$ a subset of $\R^n,$ then
$\RelConv(S)$ satisfies the
identity $((D_n)\vee z)\wedge y_1\wedge y_2,$ and
if $p$ is a point of $S,$ $\RelConv(S)_{\geq\{p\}}$ satisfies the
identity $D_n.$\qed
\end{proposition}
However, the next lemma shows that the corresponding results fail
badly for the identities involving $D_n\op$ obtained
in Theorem~\ref{X.40}.
In this lemma, the second assertion embraces the first (plus two obvious
intermediate results not stated); however I include the first assertion
because both its statement and the example proving
it are more transparent than for the second.
%
\begin{lemma}\label{X.90}
For $S$ a subset of $\R^2,$ $\RelConv(S)$ need
not satisfy the identity $D_n\op$ for any positive integer $n.$
In fact, for $p$ an element of such
an $S,$ $\RelConv(S)_{\geq\{p\}}$ need not satisfy the
identity $(((D_n\op)\wedge z')\vee z)\wedge(\bigwedge_i\,y_i).$
\end{lemma}
\begin{proof}
To get the first assertion, let $S$ be the set consisting
of the unit circle and its center, $0.$
Given $n,$ let $q_1,\dots,q_{2n+2}$ be the successive
vertices of a regular \!$(2n{+}2)$\!-gon on that circle;
for $i=1,\dots,n{+}1,$ let $y_i=\{q_1,\dots,q_{n+1}\}-\{q_i\},$ and
let $x=\{q_{n+2},\dots,q_{2n+2}\}.$
Observe that each intersection of $n$ of the $y$'s consists
of one of the points $q_i,$ and that $x$ contains
the antipodal point; hence the join of $x$ with each intersection
of $n$ $y$'s contains~$0.$
On the other hand, the intersection of all the $y$'s is
empty, hence its join with $x$ is $x,$ which does not contain~$0.$
So $D_n\op$ fails for this choice of arguments.
To get an example where all the given sets have a common
element $p,$ and where, moreover, the indicated weaker
identity fails, let us
take for $S$ the same set as in the preceding example, with the
addition of one arbitrary point $r$ outside the circle,
at distance $\geq 2$ from $0.$
% to get cvxty of z'; cd be much weakened
This time, let $q_1,\dots,q_{2n+6}$ be the successive
vertices of a regular \!$(2n{+}6)$\!-gon on $S,$ placed so
that $q_{n+2}$ lies on
the line connecting $0$ with the external point $r.$
For $i=1,\dots,n{+}1,$ let $y_i=\{q_1,\dots,q_{n+3}\}-\{q_i\},$
and let $x=\{q_{n+3},\dots,q_{2n+4}\}.$
We note that the set of subscripts of the $q$'s occurring in each
of these sets lies in an interval of length $