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Let $M$ be the Mandelbrot set, and $\partial M$ its boundary. So $\partial M$ is the set of those points $z\in M$ such that every neighborhood of $z$ contains a point of $\mathbb R^2\setminus M$.

Let $U$ be the component of $\infty$ in $\mathbb R^2\setminus M$.

Clearly $\partial U\subset \partial M$.

Is $\partial M=\partial U$?

It is known that equality is achieved if we replace $M$ with the Julia set $J$ of any complex polynomial. Maybe it is also known for $M$ and I am just unaware.

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Yes, and the proof of this fact is elementary. Mandelbrot set $M$ consists of those $c$ for which the trajectory of $0$ is bounded. So if $c_0\in\partial M$, then there are $c$ close to $c_0$ such that the trajectory of $0$ under $f_c$ is unbounded. But every unbounded trajectory tends to $\infty$, so such $c$ belong to $D_\infty$, and therefore $\partial M\subset \partial D_\infty$. The converse is evident.

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  • $\begingroup$ Why do all the terms of a trajectory need to be in the same component? $\endgroup$
    – Andrej Bauer
    Commented Dec 14, 2023 at 14:36
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    $\begingroup$ Because $D_\infty$, the set of all $c$ such that $f_c(0)\to\infty$ is connected. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 14, 2023 at 14:39

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