p'.$
To get the assertions of the
final paragraph, first note that if $p=q,$ then $X''$
falls into at least three pieces on deleting $p,$ hence by
Proposition~\ref{P.glue} it cannot admit a structure of topological
lattice.
Assuming that $p\neq q$ and that we are given a topological
lattice structure on $X'',$
the desired result can be obtained by applying twice the one-whisker
result proved above: first to $X'',$ viewed as obtained by adjoining
the whisker $W'$ to the space $X',$ and then to $X',$
viewed as obtained by adjoining the whisker $W$ to~$X.$
(In the latter case, we apply our result to
the lattice structure on $X'$ that the first step
shows us it inherits from $X'',$ as the sublattice
${\uparrow}\,q$ or~${\downarrow}\,q.)$
\end{proof}
It follows from the first statement of the above corollary
that if the underlying
topological space of $\Delta(N_5)$ admits no modular
topological lattice structure having the {\em same greatest
element} as that of $\Delta(N_5)$ itself, then
the result of gluing a whisker to the top of $\Delta(N_5)$
will be a nonmodular topological lattice whose underlying space
cannot be given {\em any} structure of modular topological lattice,
and so will give our desired example.
(Here I am using the fact that if the underlying space
of $\Delta(N_5)$ admits no modular
lattice structure having the same greatest element as its standard
structure, then by symmetry, it also admits no modular
lattice structure having for least element the
greatest element of that structure.)
We can hedge our bets by noting that
the second assertion of the corollary tells us
that unless the underlying space of $\Delta(N_5)$ admits a
modular lattice structure having {\em both} the same greatest element
{\em and} the same least element as in $\Delta(N_5),$ then the
result of attaching one whisker to the top of $\Delta(N_5)$ and
another to the bottom will be an example of the desired sort.
One can also use Proposition~\ref{P.glue} to show that the
condition which Proposition~\ref{P.WT} proves is necessary
for a simplicial complex to admit a {\em distributive} lattice structure
is not sufficient, even when $X$ does admit some lattice structure.
Namely, if we glue together the topological lattices $\Delta(M_3)$
and $[0,1]^3,$ identifying the top element of one with
the bottom element of the other, the result will be a topological
lattice whose underlying space has dimension~$3$
and is clearly embeddable in $\R^3;$
but if it admitted a distributive topological lattice
structure, Proposition~\ref{P.glue} shows that its
$\Delta(M_3)$ part would also, which we have seen it does not.
\vspace{.2em}
The method of proof of Proposition~\ref{P.glue} also yields information,
though not as strong, about topological {\em semilattices}.
If $P$ is a connected $\!\vee\!$-semilattice which becomes
disconnected on deleting a point $p,$ one sees as in that proof
that any two elements ${\notin}\,{\downarrow}\,p$ lie in the same
connected component of $P-\{p\};$ in other words, all but
at most one of its connected components lie in ${\downarrow}\,p.$
Examples of such semilattices with many connected components
in ${\downarrow}\,p$ can be constructed by
starting with an arbitrary family of connected topological
$\!\vee\!$-semilattices $P_i$ $(i\in I),$ each having a greatest
element $p_i,$ and possibly one additional connected
topological $\!\vee\!$-semilattice $P_0$ $(0\notin I)$
with an arbitrary point $p_0$ chosen, identifying all the points $p_i$
(for $i\in I$ or $I\cup\{0\}$ depending on whether we have a $P_0),$
and giving the resulting set $P$ the topology and order-structure
determined in the obvious ways by those of the given structures.
(In particular, the join of elements $x_i\in P_i,$
$x_j\in P_j$ where $i\neq j$ and $i,j\in I$ is $p;$
while if there is a $P_0,$ the join of elements
$x_i\in P_i,$ $x_0\in P_0$ is $p\vee x_0\in P_0.)$
However, not all examples have this simple form.
For instance, consider the $\!\vee\!$-subsemilattice $P$ of $[0,1]^2$
consisting of the points $(x,x)$ and $(x,1)$ for all $x\in[0,1];$
and take $p=(1,1).$
Then $P-\{p\}$ has two connected components, but its
join operation does not have the form described above;
e.g., $(0,0)\vee(0,1)=(0,1),$ rather than being $p=(1,1).$
Returning to Proposition~\ref{P.glue} itself, this can be
generalized by replacing the singleton $\{p\}$ with
any convex sublattice $P\subseteq L$ (any sublattice
$P$ such that whenever
$x0,$ we
can find the desired $p$ and $q$ within distance $\varepsilon$ of $r.$
Suppose we take a small subinterval $I\subseteq [0,1],$ and
let $p$ agree with $r$ {\em except} on $I,$ where it has the value $0,$
and let $q$ likewise agree with $r$ except on $I,$ where
it has the value $1.$
Then if $I$ is small enough (for instance, of length
$<\varepsilon/(2m),$ where $m$ is the maximum of $c(0),$ $c(1),$ and the
images under $c$ of the finitely many values
of the function $r),$ then $p$ and $q$ will
each be within distance $\varepsilon$ of $r.$
Let us now consider any $s\in[p,q]$ which likewise agrees with $r$
except on $I,$ but has for value there
an arbitrary element $x\in L.$
If we take $x$ such that $c(x)$ is sufficiently
large, then $s$ will be arbitrarily far from $r;$ so
$[p,q]$ is indeed unbounded.
To prove the $\!\vee\!$-semilattice case of~(ii), we shall show
that for any $r\in\Gamma(L,c),$ $V$ contains a
finite family of points $p_i$ such that $\bigvee p_i$
is ``far'' from $r.$
The construction is similar to the preceding, so I shall
abbreviate the details.
We again start by replacing the values of $r$ on a small subinterval
$I\subseteq [0,1]$ with $0\in L,$ getting an element $p\in\Gamma(L,c).$
We then take $x\in L$ such that $c(x)$ has large value, and
let $N$ be the integer $\lceil c(x)\rceil.$
We subdivide $I$ into $N^2$ equal small
subintervals $I_i$ $(0\leq i< N^2),$
and let each $p_i$ be formed from $p$
by changing the value on $I_i$ from $0$ to $x.$
Since the length of each $I_i$ is $1/N^2$ times
that of $I,$ the distance $d(p,p_i)$ is $\leq (N+c(0))/N^2$
times the length of $I,$ hence is small if $c(x),$ and hence $N,$
is large enough; so the $p_i$ can all be made close to $p,$
which, if $I$ has been taken small, is close to $r.$
But if we form the join of these $N^2$ elements in $\Gamma(L,c),$
this has value $x$ on {\em all} of $I;$ so for fixed $I$
and large enough $c(x),$ this join is arbitrarily
far from $p,$ and hence from $r.$
For the $\!\wedge\!$-semilattice case of~(ii) we use the dual
construction.
Turning to~(iii), it is easy to see that the function $c$ defined
in the sentence containing that
statement satisfies~\eqref{d.c} and is unbounded.
To get an unbounded sublattice of $\Gamma(L,c)$ generated by $n$
elements near an element $r\in\Gamma(L,c),$ again
choose a small interval $I\subseteq [0,1],$ and this time construct
$n$ elements by replacing the values of $r$ on $I$ by each
of our $n$ generators of $L.$
It is easy to see that for $I$ small enough,
those $n$ elements all lie in $V,$ but
that no matter how small $I$ is, the unboundedness of $c$
implies that the sublattice they
generate has elements arbitrarily far from $r.$
\end{proof}
By choosing $L$ as indicated immediately before the
above lemma, we can get distributive $\Gamma(L,c)$ satisfying
conditions~(i) and~(ii), and modular $\Gamma(L,c)$ satisfying~(iii).
The construction $\Gamma(L,c)$ really uses
nothing specific to lattices, and might have other
applications in universal algebra, if it is not already known.
\section{Further notes on the order-complex construction}\label{S.further}
\subsection{Why topologists haven't looked at \texorpdfstring{$\Delta(L)\!$}{Delta(L)}}\label{S2.L-01}
In the literature I am aware of, e.g.,~\cite{MW},
the order complex construction
$\Delta(\ )$ is regularly applied, not to finite lattices $L,$ but to
their subposets $L-\{0,1\}.$
This is because $\Delta(L-\{0,1\})$ can
have nontrivial homotopy and homology, while $\Delta(L)$ cannot,
nor can $\Delta(L-\{0\})$ or $\Delta(L-\{1\}).$
Indeed,
\begin{lemma}\label{L.contractible}
If a finite partially ordered set $P$ has a least element,
a greatest element, or more generally, an element $z$ that is
comparable to all elements of $P,$ then $\Delta(P)$ is contractible.
\end{lemma}
\begin{proof}
Given $z$ comparable to all elements of $P,$ let us
define for each $s\in[0,1]$ a map $H_s:\Delta(P)\to\Delta(P);$
namely, for $f\in\Delta(P),$ let
$H_s(f) = (1-s)\cdot f + s\cdot({\downarrow}\,z).$
To see that $H_s(f)$ lies in $\Delta(P),$
note that $f$ is a convex linear
combination of characteristic functions of principal ideals
corresponding to
a chain of elements of $P,$ and since $z$ is comparable with
all elements of $L,$ throwing it in still leaves a chain.
Since $H_0$ is the identity map of $\Delta(P)$
and $H_1$ is a constant map (i.e., $H_1(f)={\downarrow}\,z$ is
independent of $f),$ $\Delta(P)$ is contractible.
\end{proof}
Examining the above proof, one sees that the simplicial
complex $\Delta(P)$ is a cone on $\Delta(P-\{z\}).$
In particular, if $L$ is a finite lattice with more than one element,
then $\Delta(L)$ is a {\em cone on a cone} on $\Delta(L-\{0,1\}).$
For instance, if $L=M_3,$ then $L-\{0,1\}$ is a $\!3\!$-element
antichain, hence $\Delta(L-\{0,1\})$ is a $\!3\!$-point discrete space.
The simplicial complex $\Delta(L-\{0\})$ or $\Delta(L-\{1\})$
is a cone on that set, and so looks like
%
\raisebox{2pt}[12pt][8pt]{ % oY
\begin{picture}(20,20)
\qbezier(6,7)(6,7)(6,0)
\qbezier(0,-4)(0,-4)(6,0)
\qbezier(12,-4)(12,-4)(6,0)
\end{picture}}
%
(with the center point as greatest or least element, respectively);
while $\Delta(L),$ a cone on the latter space, has, as we have
seen, the form of three triangles connected along a common edge.
Another triviality result, specific to lattices, is
\begin{lemma}\label{L.retract}
If $L$ is a finite lattice, then $\Delta(L)$ is a retract
of $[0,1]^L$ by a piecewise linear order-preserving map.
More generally, if $\r{cl}$ a closure operator on a finite
set $X,$ then $\Delta(X,\r{cl})$ is a retract
of $[0,1]^X$ by such a map.
\end{lemma}
\begin{proof}[Sketch of proof]
To get a retraction $r$ in the more general situation,
map each $f\in[0,1]^X$ to the function $r(f)$ such that for
$t\in\ho,$
$r(f)_t=\r{cl}(f_t);$ equivalently, such that
for $x\in L,$ $r(f)(x)$ is the largest $t$ such that $x\in\r{cl}(f_t).$
\end{proof}
For another such result, see \cite[Theorem~6.2]{WT77}.
Neither of the above lemmas holds for the class of lattices of the
form $\Delta(L,S)$ described in Proposition~\ref{P.pairs}.
For example, if $L=\{0,1\}$ and $S=\{(1,1)\},$ then
$\Delta(L,S)$ is a $\!2\!$-point disconnected lattice.
However, if we exclude pairs of the form $(x,x)$ from $S,$ then
the result of Lemma~\ref{L.contractible} does go over, though
it takes more work to prove:
\begin{lemma}\label{L.*D(L,S)_cnctd}
Let $L$ be a nontrivial finite lattice, and $S\subseteq L\times L$
a family of pairs $(x,y)$ all of which satisfy $x