This is a follow-up to an older question of mine:
Suppose $\mathfrak{A}=(A;…)$ is an algebra (in the sense of universal algebra) and $E$ is an equivalence relation – not necessarily a congruence – on $A$. For each choice-of-representatives function $f:A\rightarrow A$ (that is, $f$ maps each $E$-class to an element of that class), let $\mathfrak{A}_f$ be the algebra with underlying set $im(f)$ and basic operations given by $(f(a_1),…,f(a_n))\mapsto f(g(f(a_1),…,f(a_n)))$ for each $n$-ary basic operation $g$ of $\mathfrak{A}$.
In general, the equational theory of $\mathfrak{A}_f$ may fall well short of that of $\mathfrak{A}$. Each such pair $(\mathfrak{A},E)$ as above gives rise to a partial order $\mathbb{P}_\mathfrak{A}(E)$ of "realizable" equational theories, namely the $Th_{Eq}(\mathfrak{A}_f)$s for $f$ appropriate. Sometimes this poset is trivial (e.g. if $E$ is a congruence), but other times it is more interesting. This question is a first stab at trying to understand it better, at least in a natural special case:
Let $\mathfrak{R}=(\mathbb{R};0,1,+,\times)$. Is there an equivalence relation $E$ on $\mathbb{R}$ such that $\mathbb{P}_\mathfrak{R}(E)$ is not upwards-directed?
I would be especially interested in an example where $E$ is Borel.
Best Answer
Let $\mathbb{R}=(\{\textrm{real numbers}\};0,1,+,\times)$. Is there an equivalence relation $E$ on $\mathbb{R}$ such that $\mathbb{P}_\mathbb{R}(E)$ is not upwards-directed?
I will assume that the order on $\mathbb{P}_\mathbb{R}(E)$ is the inclusion order. If that is what is meant, then the answer to the question is Yes, there is such an equivalence relation.
Let $E$ be the equivalence relation on $\mathbb R$ whose classes are $A=(-\infty,0]$ and $B=(0,\infty)$. Let $f\colon \mathbb R\to \mathbb R$ be the $E$-transversal for which $f(A)=\{0\}$ and $f(B)=\{1\}$. $\mathbb R_f$ is a $2$-element algebra and the operation tables for ${\mathbb R}_f$ are $$ 0^{{\mathbb R}_f} = 0, \quad 1^{{\mathbb R}_f} = 1, \quad \begin{array}{|c||c|c|} \hline + & 0 & 1 \\ \hline \hline 0 & 0 & 1\\ \hline 1 & 1 & 1\\ \hline \end{array}, \quad \begin{array}{|c||c|c|} \hline \cdot & 0 & 1 \\ \hline \hline 0 & 0 & 0\\ \hline 1 & 0 & 1\\ \hline \end{array}. $$
Now let $g\colon \mathbb R\to \mathbb R$ be the $E$-transversal for which $g(A)=\{-1\}$ and $g(B)=\{1\}$. $\mathbb R_g$ is a $2$-element algebra and the operation tables for ${\mathbb R}_g$ are $$ 0^{{\mathbb R}_g} = -1, \quad 1^{{\mathbb R}_g} = \hphantom{-}1, \quad \begin{array}{|c||c|c|} \hline + & -1 & \hphantom{-}1 \\ \hline \hline -1 & -1 & -1\\ \hline \hphantom{-}1 & -1 & \hphantom{-}1\\ \hline \end{array}, \quad \begin{array}{|c||c|c|} \hline \cdot & -1 & \hphantom{-}1 \\ \hline \hline -1 & \hphantom{-}1 & -1\\ \hline \hphantom{-}1 & -1 & \hphantom{-}1\\ \hline \end{array}. $$
If $\mathbb{P}_\mathbb{R}(E)$ were up-directed by inclusion, then there would exist an $E$-transversal $h\colon \mathbb R\to \mathbb R$ such that $\textit{Th}_{\textrm{Eq}}(\mathbb{R}_f)\cup \textit{Th}_{\textrm{Eq}}(\mathbb{R}_g)\subseteq \textit{Th}_{\textrm{Eq}}(\mathbb{R}_h)$. There are many ways to see that there no such $h$. The easiest might be this: $0^{{\mathbb R}_f}$ is a neutral element for $+^{{\mathbb R}_f}$ and $0^{{\mathbb R}_g}$ is an absorbing element for $+^{{\mathbb R}_g}$. If $\textit{Th}_{\textrm{Eq}}(\mathbb{R}_f)\cup \textit{Th}_{\textrm{Eq}}(\mathbb{R}_g)\subseteq \textit{Th}_{\textrm{Eq}}(\mathbb{R}_h)$ then $0^{{\mathbb R}_h}$ would be both a neutral element and an absorbing element for $+^{{\mathbb R}_h}$. This leads to a deduction $$ x\stackrel{neut.}{\approx} (x+0) \stackrel{abs.}{\approx} (y+0) \stackrel{neut.}{\approx} y, $$ or $x\approx y$ in $\textit{Th}_{\textrm{Eq}}(\mathbb{R}_h)$. But this is impossible, since $\mathbb{R}_h$ must be a $2$-element algebra if $h$ is a transversal for an equivalence relation with $2$ classes.