# Is the limit of classical Laver tables connected anywhere?

Let $A_{n}=(\{1,\dots,2^{n}\},*_{n})$ be the $n$-th classical Laver table. Then $*_{n}$ is the unique operation on $\{1,\dots,2^{n}\}$ where $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ and $x*_{n}1=x+1\mod 2^{n}$ whenever $x,y,z\in A_{n}$.

Let $L_{n}:\{0,\dots,2^{n}-1\}\rightarrow\{0,\dots,2^{n}-1\}$ be the mapping that reverses the ordering of the digits in the binary expansion of the number in $\{0,\dots,2^{n}-1\}$. More explicitly, $$L_{n}(\sum_{k=0}^{n-1}2^{k}a_{k})=\sum_{k=0}^{n-1}2^{n-1-k}a_{k}$$ whenever $a_{0},\dots,a_{n-1}\in\{0,1\}$.

Let $L_{n}^{\sharp}:\{1,\dots,2^{n}\}\rightarrow\{1,\dots,2^{n}\}$ be the mapping where $L_{n}^{\sharp}(x)=L_{n}(x-1)+1$. Define a binary operation $\#_{n}$ on $\{1,...,2^{n}\}$ by $$x\#_{n}y=L_{n}^{\sharp}(L_{n}^{\sharp}(x)*_{n}L_{n}^{\sharp}(y)).$$

Let $$C_{n}=\{(2^{-n}\cdot x,2^{-n}\cdot(x\#_{n}y))\mid x,y\in\{1,\dots,2^{n}\}\}.$$ Then $C_{n}\subseteq[0,1]\times[0,1]$ for all $n$. Furthermore, if we give the hyperspace $H([0,1]^{2})$ (where $H(X)$ is the collection of all closed subsets of $X$) the Hausdorff distance metric, then $C_{n}\rightarrow C$ for some compact space $C$. The following 512x512 image is an image of the compact space $C$.

While some portions of the space $C$ appear to be just vertical or diagonal lines, if there exists a rank-into-rank cardinal, then every open subset of $C$ looks like a fractal (if you look very closely, then you will see some tiny branches coming out of the vertical and diagonal lines). Does there exist a connected (path-connected) open subset $U$ of $C$? Does there exist a component (path-component) of $C$ with a non-empty interior? Does there exist a single element component (path-component) of $C$?

I suspect that the answer to these questions is no, but if the answer to these questions is no, then I also suspect that one will need large cardinal hypotheses to form a proof.