What is the logical complexity of the Mandelbrot Local Connectivity conjecture? (Is it equivalent to a statement of arithmetic?)

Question

Denote by MLC the statement “the Mandelbrot set is locally connected” and MHC the statement “hyperbolic components are dense in the Mandelbrot set” (it is known that MLC implies MHC, and whether either of these is a theorem of ZFC is, at the moment, unknown). Formally, these are statements of third-order arithmetic, i.e., they talk about real numbers (≈ sets of integers) and sets of real numbers, and I think at least MHC can be easily brought down to second-order. But it is often the case that open problems of this sort can be shown to be equivalent to arithmetical statement (e.g., the Riemann hypothesis is equivalent to a $\Pi_1$ statement of first-order arithmetic, see here; similarly, the Hodge conjecture is equivalent to a first-order arithmetical statement, although its exact level in the arithmetic hierarchy is not so clear). My question is: is this the case for MLC or MHC? In other words:

• Is there a statement of first-order arithmetic which we know how to prove (in ZFC) to be equivalent to MLC, or perhaps MHC? If so, what is its level in the arithmetical hierarchy? If not, how far down the analytical hierarchy can we get down? Are there, at least, some arithmetical implications of MHC that are not known from ZFC alone?

• Variant question, with a less logical flavor: I understand (please correct me if this is incorrect) that there is a “combinatorial” or “abstract” Mandelbrot set, and that statements like MLC or MHC tell us, in a vague sense, that this combinatorial Mandelbrot set correctly describes the (complex) Mandelbrot set; but do they also have implications on the combinatorial object in question? (Such implications would probably be somewhere in the arithmetical hierarchy and thus answer at least part of the previous item's questions.)

Answer 1

Here is an argument that MLC is equivalent to an arithmetic statement. Let $M$ denote the Mandelbrot set, and let $\mathbb{C}_{\mathbb{Q}}$ denote the set of rational complex numbers. First note that by a fairly standard argument, the set of pairs $\{(z,\varepsilon)\in \mathbb{C}_{\mathbb{Q}}\times \mathbb{Q}: B_{\leq \varepsilon}(z)\cap M = \varnothing\}$, where $B_{\leq \varepsilon}(z)$ is the closed ball of radius $\varepsilon$ centered at $z$, is computably enumerable. This means that there is a uniform procedure for computing an element of $B_{< 2\varepsilon}(z)\cap M$ (for $z \in \mathbb{C}_{\mathbb{Q}}$ and $\varepsilon \in \mathbb{Q}$ with $B_{\leq \varepsilon} \cap M \neq \varnothing$) from $0'$, and therefore $0'$ uniformly computes a countable dense sequence $(m_i)_{i<\omega}$ of elements of $M$. (It is highly likely that there is a uniformly computable dense sequence in $M$, but I do not see an argument.) Let $M_0$ denote the set $\{m_i:i<\omega\}$.

Say that a set $F$ in a metric space $X$ is $\delta$-connected if for any $x,y \in F$, there is a finite chain $z_0,z_1,\dots,z_{n}$ with $z_0 = x$, $z_n = y$, and $d(z_i,z_{i+1}) < \delta$ for each $i<n$. Note that in a compact metric space, a closed set is connected if and only if it is $\delta$-connected for every $\delta > 0$. (Furthermore, recall that $M$ is a compact metric space.)

I claim that MLC is equivalent to the following arithmetic statement:

For every rational $\varepsilon > 0$, there is an $n < \omega$ such that for every rational $\delta >0$, there are finite sets $F_0,F_1,\dots,F_{n-1} \subset M_0$ such that

• each $F_i$ has diameter less than $\varepsilon$,
• each $F_i$ is $\delta$-connected, and
• $M$ is covered by $\bigcup_{i<n} \bigcup_{x \in F_i} B_{<\delta}(x)$.

First note that this is indeed arithmetic. Each of the bulleted conditions is $0'$-semi-decidable and therefore $0''$-decidable. (The last one relies on the fact that the complement of $M$ in $\mathbb{C}$ is computably enumerable.) This means that ostensibly the bullet points each have complexity $\Sigma^0_2$. Since the quantifiers out front are $\forall \exists \forall \exists$, this implies that the complexity of the entire statement is $\Pi^0_5$. If we knew there was a uniformly computable dense sequence in $M$, I think this shows that it is $\Pi^0_4$. Also note that there's nothing particularly special about the Mandelbrot set in all of this. All we're really using is that it's a compact co-c.e. subset of a computable metric space (where I mean co-c.e. in the sense of computable analysis).

Let's see that MLC implies the statement. Assume that $M$ is locally connected. Fix a rational $\varepsilon > 0$. Find a finite open cover $U_0,\dots,U_{n-1}$ of $M$ such that each $U_i$ is connected and has diameter less than $\frac{1}{2}\varepsilon$. Let $G_i$ be the closure of $U_i$ for each $i<n$. Note that the diameter for $G_i$ is at most $\frac{1}{2}\varepsilon < \varepsilon$. Now fix a rational $\delta > 0$. Since each $G_i$ is also connected, we can find a finite subset $H_i$ of $G_i$ such that $G_i \subseteq \bigcup_{x \in H_i} B_{<\frac{1}{2}\delta}(x)$. Since $G_i$ is connected, it is $\frac{1}{2}\delta$-connected and we can find a finite set $E_i \subseteq G_i$ with $H_i \subseteq E_i$ such that $E_i$ is $\frac{1}{2}\delta$-connected. (Just add a chain between each pair of elements of $H_i$.) $E_i$ still covers $G_i$ in the same way, so $M \subseteq \bigcup_{i<n} \bigcup_{x \in E_i} B_{<\frac{1}{2}\delta}(x)$. Finally, since $E_i \subseteq G_i$, the diameter of $E_i$ is less than $\frac{1}{2}\varepsilon$. Now finally, for each $x \in E_i$, find $m_x \in M_0$ such that $d(x,m_x) < \min\left(\frac{1}{3}\varepsilon, \frac{1}{5}\delta\right)$, and let $F_i = \{m_x : x \in E_i\}$. It is immediate that the diameter of $F_i$ is less than $\frac{1}{2}\varepsilon + \frac{1}{3}\varepsilon < \varepsilon$. Furthermore, $G_i \subseteq \bigcup_{x \in F_i} B_{< \delta}(x)$ and each $F_i$ is $\delta$-connected by the triangle inequality (since $\frac{1}{5}\delta + \frac{1}{2}\delta + \frac{1}{5}\delta < \delta$).

Now let's see that the arithmetic statement implies MLC. We'll need a small lemma.

Lemma. A compact metric space $X$ is locally connected if and only if for every rational $\varepsilon > 0$, there is a finite closed covering $G_0,G_1,\dots,G_{n-1}$ of $X$ such that each $G_i$ is connected and has diameter at most $\varepsilon > 0$.

Proof. The forward direction is obvious. For the reverse direction, suppose that the second statement is true. Fix $\varepsilon > 0$ and find a finite covering $G_0,G_1,\dots,G_{n-1}$ of $X$ such that each $G_i$ is connected and has diameter at most $\frac{1}{2}\varepsilon > 0$. For any $x \in X$, let $G_x = \bigcup\{G_i:i<n,~x \in G_i\}$. Note that since the $G_i$'s form a closed cover, we have that $x$ is in the interior of $G_x$. Furthermore, $G_x$ is connected (since it is a union of connected sets that contain a common point) and has diameter at most $\varepsilon$. Since we can do this for any $x \in X$ and any rational $\varepsilon > 0$, we have that $X$ is connected im kleinen at every point, which implies local connectedness (in any topological space). $\square_{\text{Lemma}}$

Fix a rational $\varepsilon > 0$. By assumption, there is an $n$ such that for every rational $\delta >0$, there are finite sets $F_0,F_1,\dots,F_{n-1} \subset M_0$ satisfying the stated properties above. For each $k < \omega$, let $F^k_0,F^k_1,\dots,F^k_{n-1}$ be a particular choice of such sets for $\delta = 2^{-k}$. Since the space of closed subsets of $M$ is compact under the Hausdorff metric, we can find an increasing sequence $(k(j))_{j<\omega}$ of natural numbers such that $(F^{k(j)}_i)_{j<\omega}$ converges in the Hausdorff metric for each $i<n$. Let $G_i = \lim_{j \to \infty} F^{k(j)}_i$. Fairly easy arguments show that each $G_i$ has diameter at most $\varepsilon$, each $G_i$ is $\delta$-connected for every $\delta > 0$ and therefore is connected, and $M$ is covered by $\bigcup_{i <n} \bigcup_{x \in G_i} B_{<\delta}(G_i)$ for every $\delta > 0$ and so $M = \bigcup_{i<n} G_i$. Therefore, by the lemma, $M$ is locally connected.