Conjecture

We have proved in Sec. 3 that $\mathcal{M}_{2}$ is a connected set. It is natural to ask whether the Tetrabrot is also connected. Until now, the exploration of the Tetrabrot seems to confirm that the Tetrabrot is connected. However, if we enlarge Fig. 7 in the centre of the Tetrtabrot above the zone B, we notice (see Fig. 24.A) that there is a piece which seems to be disconnected from the main figure (Fig. 25 focuses on this piece). Because we work with divergence layers and a computational approximation, we are far from knowing if the piece is really unconnected or if there is point inside the piece which is also inside the Tetrabrot. However this is enough to formulate a conjecture:

Conjecture 1   The Tetrabrot is unconnected.

It is possible to translate the conjecture into a question about the geometry of the Mandelbrot set. For this, we need to prove the following lemma which is itself of interest:

Lemma 3   The Tetrabrot can be characterized as follows:

$$\mathcal{T}=\bigcup\limits_{y\in[-m,m]}\{[(\mathcal{M}_{1}-yi_1)\cap(\mathcal{M}_{1}+yi_1)]+yi_2\}$$

where $m:=sup\{q\in\mathbb {R}:\exists p\in\mathbb {R}\mbox{ such that }p+qi_1\in\mathcal{M}_{1}\}$.

Proof. By definition,

$$\mathcal{T}=\{c=c_1+c_2i_2\in\mathbb {C}_{2}:Im(c_{2})=0 \mbo... ...P_{c}^{\circ n}(0)\mbox{ is bounded }\forall n\in\mathbb {N}\}.$$

Let $c=(c_1-c_2i_1)e_1+(c_1+c_2i_1)e_2$ with $c_1=c_{11}+c_{12}i_1$ and $c_2=c_{21}+c_{22}i_1$ where $c_{11},c_{12},c_{21},c_{22}\in\mathbb {R}$. Now, if $Im(c_2)=0$, we have $c_2=c_{21}+0i_1$ and therefore, $c=(c_1-c_{21})e_1+(c_1+c_{21}i_1)e_2$ whenever $Im(c_2)=0$. Hence $\mathcal{T}=\{(c_1-c_{21}i_1)e_1+(c_1+c_{21}i_1)e_2: P_{c}^{\circ n}(0)\mbox{ i... ... }P_{c_1+c_{21}i_1}^{\circ n}(0) \mbox{ are bounded }\forall n\in\mathbb {N}\}.$To continue the proof, we need to remark the following fact: $\forall z\in\mathbb {C}_{1},$

$$\{c\in\mathbb {C}_{1}:P_{c+z}^{\circ n}(0) \mbox{ is bounded }\forall n\in\mathbb {N}\}=\mathcal{M}_{1}-z.$$

By definition, $P_{c_1-c_{21}i_1}^{\circ n}(0)$ and $P_{c_1+c_{21}i_1}^{\circ n}(0)$ are bounded $\forall n\in\mathbb {N}$ if and only if $c_1-c_{21}i_1$, $c_1+c_{21}i_1\in\mathcal{M}_{1}$, and by the remark, it is also if and only if $c_1\in(\mathcal{M}_{1}-c_{21}i_1)\cap(\mathcal{M}_{1}+c_{21}i_1)$. Hence, if we express $(c_1-c_{21}i_1)e_{1}+(c_1+c_{21})e_{2}=c_1+c_{21}i_2=c_{11}+c_{12}i_1+c_{21}i_2$, the Tetrabrot can be characterized as follows:

$$\mathcal{T}=\{c_{11}+c_{12}i_1+c_{21}i_2:c_{11}+c_{12}i_1\in (\mathcal{M}_{1}-c_{21}i_1)\cap(\mathcal{M}_{1}+c_{21}i_1)\}$$

$$=\bigcup\limits_{y\in\mathbb {R}}\{[(\mathcal{M}_{1}-yi_1)\cap(\mathcal{M}_{1}+yi_1)]+yi_2\}.$$

It is possible to be more precise with the last expression. In fact,

$$=\bigcup\limits_{y\in[-m,m]}\{[(\mathcal{M}_{1}-yi_1)\cap(\mathcal{M}_{1}+yi_1)]+yi_2\}$$

because $(\mathcal{M}_{1}-yi_1)\cap(\mathcal{M}_{1}+yi_1)=\emptyset$ whenever $y\in[-m,m]^{c}$. This comes from the fact that $\mathcal{M}_{1}\subset\{z\in\mathbb {C}_{1}:\vert Im(z)\vert\leq m\}$.

Moreover, $(\mathcal{M}_{1}-yi_1)\cap(\mathcal{M}_{1}+yi_1)\not=\emptyset$ $\forall y\in[-m,m]$. To see this, we just have to prove that $E_y:=\{c=c_{11}+0i_1+yi_2:P_{c}^{\circ n}(0)\mbox{ is bounded }\forall n\in\mathbb {N}\}$ is nonempty $\forall y\in[-m,m]$ because $E_y\subset\{c=c_{11}+c_{12}i_1+c_{21}i_2:P_{c}^{\circ n}(0)\ \mbox{ is bounded ... ..._{11}+c_{12}i_1\in (\mathcal{M}_{1}-c_{21}i_1)\cap(\mathcal{M}_{1}+c_{21}i_1)\}$. In fact, the set $E_y$ is the algorithm for the Mandelbrot set iterates, with the imaginary part in ``$i_2$" fixed at $y$. By the compactness and the symmetry of the Mandelbrot set $\mathcal{M}_{1}$, there must exist $x_m$ such that $x_m-mi_2$, $x_m+mi_2\in E_m$. Therefore, because $\mathcal{M}_{1}$ is connected, we must have $E_y\neq\emptyset$ $\forall y\in[-m,m]$.

Theorem 6   Let

$$R_1:=R(-1.3939+0.0848i_1; -1.3893+ 0.0803i_1),$$

$$L:=R(-1.3939+0.0803i_1; -1.3939+0.0728i_1),$$

$$R_2:=R(-1.3939+0.0728i_1; -1.3893+0.0683i_1),$$

$$L_3:=R(-1.3939+0.1304i_1; -1.3893+0.1304i_1),$$

$$L_4:=R(-1.3939+0.1259i_1; -1.3893+0.1259i_1),$$

$$L_5:=R(-1.3893+0.1304i_1; -1.3893+0.1259i_1),$$

$$L_6:=R(-1.3939+0.1304i_1; -1.3939+0.1259i_1),$$

where

$$R(a+bi_1,c+di_1):=\{\alpha_{1}(a+bi_1)+\alpha_{2}(c+bi_1)+\al... ...1) \vert\sum_{\i=1}^{4}\alpha_{i}=1,\mbox{ }\alpha_{i}\geq 1\}.$$

If

$$F_1:=R_1\cup L\cup R_2$$

and

$$F_2:=L_3\cup L_4\cup L_5$$

are disjoint from the Mandelbrot set and,

$$z_{1}^{*}:=-1.391816306+0.129472959i_1$$

and

$$z_{2}^{*}:=-1.392873019+0.077172405i_1$$

are inside the Mandelbrot set, then the Tetrabrot is unconnected.

Proof. The goal of the proof is to construct, from the hypothesis about the Mandelbrot set, a box where the algorithm for the Tetrabrot diverges with, inside the box, a point where the algorithm converges. The box that we want to construct is a box around the piece of Fig. 25. Also, to understand better with which zones we work, Fig. 26.A gives an indication where the zones of the hypothesis are on the Mandelbrot set (Figs. 27, 28 and 29 are enlargements of Fig. 26.A where the specific sets $F_1$, $F_2$, $z_{1}^{*}$ and $z_{2}^{*}$ are illustrated).

The ``box of divergence" is constructed as follows: let $y_1:=0.0228$ and $y_2:=0.0288$,

$$B_i:=R_i+y_ii_1-y_ii_2 \mbox{ for i=1,2,}$$

$$B_i:=\bigcup\limits_{y\in[y_1,y_2]}(L_i-yi_1-yi_2)\mbox{ for i=3,...,6}.$$

Each ``$B_i$" is a side of the box; then the ``box of divergence" is $B:=\bigcup\limits_{i=1}^{6}B_{i}$.

First, we have to confirm that each ``$B_i$" is a set where the algorithm for the Tetrabrot diverges. This is possible with Lemma 3 and the assumption that $F_1\cup F_2$ is not in the Mandelbrot set.

For $B_1$, by Lemma 3, we just need to prove that:

$$B_1\cap ([(\mathcal{M}_{1}-y_1i_1)\cap (\mathcal{M}_{1}+y_1i_1)]-y_1i_2)=\emptyset.$$

This is clear because if $B_1\cap ([(\mathcal{M}_{1}-y_1i_1)\cap (\mathcal{M}_{1}+y_1i_1)]-y_1i_2)\not= \emptyset$, we obtain that there exists $z_1\in R_1$ such that $z_1+y_1i_1\in[(\mathcal{M}_{1}-y_1i_1)\cap (\mathcal{M}_{1}+y_1i_1)]$. However, this is impossible because if $z_1+y_1i_1\in\mathcal{M}_{1}+y_1i_1$ we obtain that $z_1\in\mathcal{M}_{1}$ and this contradicts the hypothesis.

A similar proof is possible for $B_2$. The cases of $B_3$, $B_4$ and $B_5$ are also along the same lines. For example:

$$B_3\cap ([(\mathcal{M}_{1}-yi_1)\cap (\mathcal{M}_{1}+yi_1)]-yi_2) =\emptyset\mbox{ }\forall y\in [y_1,y_2],$$

because if it is not true, there must exist $y\in [y_1,y_2]$ and $z_3\in L_3$ such that $z_3-yi_1\in[(\mathcal{M}_{1}-yi_1)\cap (\mathcal{M}_{1}+yi_1)]$. However, this is impossible because if $z_3-yi_1\in\mathcal{M}_{1}-yi_1$ we obtain that $z_3\in\mathcal{M}_{1}$ and this contradicts the hypothesis. For $B_6$, the same argument is possible if we remark that $L_6-yi_1\subset F_1+yi_1$ $\forall y\in [y_1,y_2]$.

Now, we have to confirm that each $B_i$ is on the same box. For this, we will just remark the following fact:

$$L_i-y_ki_1-y_ki_2\subset B_k=R_k+y_ki_1-y_ki_2\mbox{ }\forall i=3,...,6\mbox{ and }\forall k=1,2$$

because $L_i-y_ki_1\subset R_k+y_ki_1$ $\forall i=3,...6$ and $\forall k=1,2$. Then each $B_i$, for $i=3,...6$, touches $B_1$ and $B_2$ at their extremity. To be more specific and to confirm that the $B_i$ form a box, we have to check directely with the exact coordinates given in the hypothesis of the theorem.

Finally, we have to prove, with the assumptions of the theorem, that there is a point inside the box $B$ for which the algorithm for the Tetrabrot converges. For this, we remark that $z_{2}^{*}$ is between $R_1$ and $R_2$. Moreover, $Re(z_{1}^{*})>Re(z_{2}^{*})$; then we must have in the set $A:=\{x+yi_1\in\mathbb {C}_{1} :x=Re(z_{1}^{*})\mbox{ and } 0.0728<y<0.0803\}$ a point $z^{*}\in\mathcal{M}_{1}$. If not, by hypothesis, the Mandelbrot set would not be connected because $z_{2}^{*}$ would be separated from $z_{1}^{*}$ by $F_1\cup A$ since $(F_1\cup A)\cap\mathcal{M}_{1}=\emptyset$.

Now, let

$$y^{*}:=\frac{Im(z_{1}^{*})-Im(z^{*})}{2}$$

where $z_{1}^{*},z^{*}\in\mathcal{M}_{1}$. We note that $z_{1}^{*}-y^{*}i_1=z^{*}+y^{*}i_1$ and $y^{*}\in[-1,1]\subset[-m,m]$. Then by Lemma 3, $z_{1}^{*}-y^{*}i_1-y^{*}i_2\in\mathcal{T}$ because $z_{1}^{*}-y^{*}i_1-y^{*}i_2=z^{*}+y^{*}i_1-y^{*}i_2\in [(\mathcal{M}_{1}-y^{*}i_1)\cap (\mathcal{M}_{1}+y^{*}i_1)]-y^{*}i_2\subset\mathcal{T}$. Moreover, $z_{1}^{*}-y^{*}i_1-y^{*}i_2$ is inside the ``box of divergence" because $y_1<y^{*}<y_2$ and $z_{1}^{*}$ is inside the rectangle formed by $L_3\cup L_4\cup L_5\cup L_6$.$\Box$