# Is the Mandelbrot set Suslinian?

The Mandelbrot set is known to be (path-)connected and compact. A non-degenerate space with these properties is called a continuum.

A continuum $$X$$ is Suslinian if every collection of non-degenerate pairwise disjoint subcontinua of $$X$$ is countable.

My question is: Is the (boundary of the) Mandelbrot set Suslinian?

If not, it would in fact contain a continuous collection of continua with a Cantor set as its quotient; something like the product $$\text{(Cantor set})\times [0,1]$$.

This could be closely related to the MLC "Mandelbrot Locally Connected" conjecture.

If $$\partial M$$ is Suslinian, then is it locally connected?

If $$\partial M$$ is locally connected, then is it Suslinian?

I suspect that the latter will be easier to answer, possibly using the pinched disk model of $$M$$.

• I'm confused. Doesn't the Mandelbrot set contain an open subset? Wouldn't this give you an uncountable collection of pairwise disjoint homeomorphic copies of $[0,1]$? Commented Nov 18, 2023 at 1:10
• @JamesHanson I think the Mandelbrot set here refers to the boundary of what you're calling the Mandelbrot set. (I think there's some inconsistency in usage.) Commented Nov 18, 2023 at 1:34
• That is correct, I am only thinking about the boundary. Commented Nov 18, 2023 at 1:56

There are different versions of the Suslinian property for continua. One is called finitely Suslinian, a term I think I learned about from the OP - if I recall correctly, it asks that, for any eps>0, there is no infinite collection of pairwise disjoint subcontinua all having diameter at least eps.

For boundaries of planar simply-connected domains, being finitely Suslinian is equivalent to local connectivity. The boundary of the Mandelbrot is such a set (the complement of the Mandelbrot set in the sphere is a simply-connected domain). So local connectivity certainly implies Suslinian. In particular, IF the Mandelbrot set is locally connected (as is widely believed), it is Suslinian.

The opposite implication is false for general boundaries of simply-connected domains; consider for example the Warsaw circle. On the other hand, the Knaster buckethandle is an example of a non-Suslinian continuum that is also the boundary of a simply-connected domain.

It is plausible that, if the Mandelbrot set turned out to be not locally connected, it would also not be Suslinian. (Not only because I think it is very likely that the Mandelbrot set is locally connected.) Indeed, if local connectivity fails, then we know this would have to be at some "infinitely renormalisable" parameters; i.e., there would be some nested sequence of little Mandelbrot copies that did not shrink to a point. But then one would surely expect to have several choices of little Mandelbrot copies at each stage, resulting in uncountably many different choices of such continua.

Of course, this is a heuristic argument, not a proof. It is not clear to me whether it would be possible to give a proof without also having sufficient control to settle the question of local connectivity of the Mandelbrot set. :) But perhaps someone else can comment.

I remark that in all cases of NOT locally connected quadratic Julia sets that we understand sufficiently to answer the question, the Julia set is NOT Suslinian.

• Thanks Lasse! Just for a little background, I originally asked this question due to a recent result (Theorem A in arxiv.org/pdf/2401.10206.pdf) that I've been working on with several coauthors. The result implies that every plane continuum which contain Erdos space (or the endpoints of the Julia set of $e^z-1$) must be non-Suslinian. So my idea was that if I could embed that space into the Mandelbrot set, then the Mandelbrot set would be non-Suslinian and therefore probably not locally connected... Commented Feb 6 at 22:54
• Your thoughts on "$\partial M$ Suslinian$\Rightarrow$MLC'' are also interesting. That would be a very nice theorem, but like you say, maybe it is just as difficult as MLC. By the way, this question is also open for Julia sets of rational maps. Commented Feb 6 at 23:04
• @D.S.Lipham Thank you for explaining. I think that (from what you write) it does not seem plausible that one could embed Erdos space into the boundary of the Mandelbrot set without already knowing that the Mandelbrot set is not locally connected. :) Commented Feb 11 at 22:17