\section{$\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3})$ is not hereditarily simple}

Let $\mathbf{3}$ be the structure with universe $\set{0,1,2}$ in which all three elements are named by constant symbols. Then \(\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3}) = \DSuit_1(\set{0,1,2})\), which has 55 elements. Part of the lattice structure of $\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3})$ is shown in Figure \ref{CsIFG(3)}. For simplicity, we only show the interval above $\Omega$. Furthermore, we omit the falsity coordinate and denote each truth coordinate by listing the maximal winning teams. For example, the vertex labeled $\set{0,1}, \set{2}$ denotes the element \(\tuple{\powerset(\set{0,1}) \cup \powerset(\set{2}), \set{\emptyset}}\), and the vertex labeled $\emptyset$ denotes \(\tuple{\set{\emptyset}, \set{\emptyset}} = \Omega\). Readers familiar with the cover of \cite{Balbes:1974} will recognize that Figure \ref{CsIFG(3)} is isomorphic to the free distributive 1-lattice on three generators. To obtain the full lattice structure of \(\Cyls_{\mathrm{IFG}_{1}}(\mathbf 3)\) it is necessary to flip the figure upside-down to get the interval below $\Omega$, then fill in the sides with every possible double suit incomparable to $\Omega$.


\input{figure-CsIFG3}

The goal of this section is to show that \(\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3})\) is simple, but not hereditarily simple. In fact, every IFG$_N$-algebra whose universe is the collection of all double suits over a set $A$ is simple.


\begin{prop} 
\label{finite constant structures have simple cylindric set algebras}
\(\DSuit(\^NA)\) is simple.
\end{prop}

\begin{proof}
Suppose $X$ and $Y$ are distinct elements of 
\(\DSuit(\^NA)\) such that \(X \cong Y\). Without loss of generality we may assume that there exists a \(V \in Y^+ \setminus X^+\). Let \(Z = \tuple{\powerset(\^NA \setminus V), \powerset(V)}\). Since \(V \notin X^+\) we know that for every \(U \in X^+\) there is an \(\vec a \in V \setminus U\). Hence \(U \cup (\^NA\setminus V) \not= \^NA\). Thus \(1 \not= X +_\emptyset Z \cong Y +_\emptyset Z = 1\). Therefore $\cong$ is the total congruence.
\end{proof}


Recall that in the proof that $\Cyls_{\mathrm{IFG}_{1}}(\mathbf{2})$ is hereditarily simple, we used the fact that \(A +_\emptyset A = 1\) but \(B +_\emptyset B \not= 1\). For any element $X$ of an IFG-algebra, let $nX$ be a abbreviation for \( \underbrace{X +_\emptyset \cdots +_\emptyset X}_{n} \).


\begin{defn} 
The \emph{order}\index{order of an element|mainidx} of an element $X$ is the least positive integer $n$ such that \(nX = 1\). If no such positive integer exists then the order of $X$ is infinite.
\end{defn}


\begin{lem} 
\label{congruent different order}
Let $\cong$ be a congruence on a double-suited IFG$_N$-algebra. If any two elements of different order are congruent, then $\cong$ is the total congruence.
\end{lem}

\begin{proof}
Let \(X \cong Y\). If the order of $X$ is less than the order of $Y$, then for some positive integer $n$ we have \(1 = nX \cong nY \not= 1\).
\end{proof}


We know by \propref{all double suits} and \propref{finite constant structures have simple cylindric set algebras} that $\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3})$ is simple, but we can verify this directly by using the lemmas and the technique of perspective edges. For example, if \(\set{0},\set{1} \cong \set{0,1}\), then \(\set{0}, \set{1}, \set{2} \cong \set{0,1},\set{2}\). But \(\set{0}, \set{1}, \set{2}\) has order 2, while \(\set{0,1},\set{2}\) has order 1, so by \lemref{congruent different order} we have that $\cong$ is the total congruence.


\begin{thm} 
\label{CsIFG_1(3) is not hereditarily simple}
\(\Cyls_{\mathrm{IFG}_{1}}(\mathbf{3})\) is not hereditarily simple.
\end{thm}

\begin{proof}
Let \(A = \tuple{\powerset(\set{0,1}),\ \set{\emptyset}}\) and \(B = \tuple{\powerset(\set{0}) \cup \powerset(\set{1}),\ \set{\emptyset}}\). The subalgebra \(\gen{B} = \set{0, \n{A}, \n{B}, \Omega, B, A, 1}\) is closed under 
$+_\emptyset$ and $+_{\set 0}$ because\begin{alignat*}{2}
A +_\emptyset A &= A, 		&\qquad	 	A +_{\set{0}} A &= A, \\
A +_\emptyset B &= A,		&\qquad		A +_{\set{0}} B &= A, \\
B +_\emptyset B &= A,		&\qquad		B +_{\set{0}} B &= B, \\
A +_\emptyset \n{A} &= A,		&\qquad		A +_{\set{0}} \n{A} &= A, \\
A +_\emptyset \n{B} &= A,		&\qquad		A +_{\set{0}} \n{B} &= A, \\
B +_\emptyset \n{A} &= B,		&\qquad		B +_{\set{0}} \n{A} &= B, \\
B +_\emptyset \n{B} &= B,		&\qquad		B +_{\set{0}} \n{B} &= B, \\
\n{A} +_\emptyset \n{A} &= \n{A},	&\qquad		\n{A} +_{\set{0}} \n{A} &= \n{A}, \\
\n{A} +_\emptyset \n{B} &= \n{B},	&\qquad		\n{A} +_{\set{0}} \n{B} &= \n{B}, \\
\n{B} +_\emptyset \n{B} &= \n{B},	&\qquad		\n{B} +_{\set{0}} \n{B} &= \n{B}. 
\end{alignat*}
All of the $+_{\set 0}$ calculations are easy to check by looking at the lattice. The $+_\emptyset$ calculations require some computation. First, \(A +_\emptyset A = A\) because $A$ is flat. Second, \(A +_\emptyset B = A\) because \(\set{0, 1} = \set{0} \cup_\emptyset \set{1}\), where \(\set{0} \in A^+\) and \(\set{1} \in B^+\), while \((A +_\emptyset B)^- = A^- \cap B^- = A^-\). Similarly, \(B +_\emptyset B = A\). The remaining $+_\emptyset$ calculations all follow from \propref{X < Omega,Y} or \propref{flat absorption}. Finally, the set is closed under $C_{0,J}$ by \propref{C_0,J...C_N-1,J double suit}.

Let $\cong$ denote the equivalence relation that makes \(A \cong B\) and \(\n{A} \cong \n{B}\), but makes no other pair of distinct elements equivalent. To verify that $\cong$ is a congruence, observe that $\cong$ is preserved under $\n{\,}$ because \(\n{A} \cong \n{B}\) and \(\n{(\n{A})} = A \cong B = \n{(\n{B})}\). It is preserved under $C_{0,J}$ because \(C_{0,J}(A) = 1 = C_{0,J}(B)\) and \(C_{0,J}(\n{A}) = \Omega = C_{0,J}(\n{B})\). Finally, the calculations above show that $\cong$ is preserved under $+_\emptyset$ and $+_{\set{0}}$.
Thus $\cong$ is a nontrivial, non-total congruence. Therefore $\gen{B}$ is not simple.
\end{proof}
{"config": "arxiv", "file": "0806.4785/7-CsIFG3.tex"}