We say that a finite self-distributive algebra $(A,*)$ is linear if there is some $1\in A$ where $a*1=1,1*a=a$ for all $a\in A$ and where if $\preceq$ is the relation where $x\preceq y$ if and only if $x*a_{1}*...a_{n}=y$ and $x*a_{1}*...*a_{m}\neq 1$ for $0\leq m<n$, then $\preceq$ is a linear ordering on $A$.

Let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$. As always, let $*$ denote the operation on $\mathcal{E}_{\lambda}$ defined by $j*k=\bigcup_{\alpha<\lambda}j(k|_{V_{\alpha}})$. Then $(\mathcal{E}_{\lambda},*)$ satisfies the self-distributivity property $j*(k*l)=(j*k)*(j*l)$ and for each limit ordinal $\gamma<\lambda$, the relation $\equiv^{\gamma}$ defined by $j\equiv^{\gamma}k$ if and only if $j(x)\cap V_{\gamma}=k(x)\cap V_{\gamma}$ for $x\in V_{\gamma}$ is a congruence on $(\mathcal{E}_{\lambda},*)$. The algebras $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ are always locally finite.

Are there any examples of closed subalgebras models of set theory where there is some cardinal $\lambda$ and subalgebras $X\subseteq\mathcal{E}_{\lambda}$ such that $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ is linear for all ordinals $\gamma$ but where $\mathcal{E}_{\lambda}/\equiv^{\gamma}$ is not generated by a single element for sufficiently large limit ordinals $\gamma$?

  • $\begingroup$ In the last sentence, I meant to say '$X/\equiv^{\gamma}$ is linear, but $X/\equiv^{\gamma}$ is not. . .' $\endgroup$ – Joseph Van Name Feb 21 at 22:27

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