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We know that the Mandelbrot fractal contains a countable number of copies of itself.

See :

Does the Mandelbrot fractal contain countably or uncountably many copies of itself?

Where that is explained.

Notice that polynomials have a finite amount of zero's and entire functions have a countable amount of zero's.

So I started to wonder :

Is there a Julia fractal that contains uncountable many copies of itself ?

And if so, can they be iterations of entire functions ?

What are typical examples ?

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    $\begingroup$ Think about how e.g. you can't pack uncountably many disjoint "X-shapes" (for pretty much any interpretation of that phrase) in the plane. The same basic argument will apply to any set that's much more complicated than a line segment. $\endgroup$
    – Noah Schweber
    Nov 20 at 19:51
  • $\begingroup$ @NoahSchweber If the size shape has a positive area on a surface or a positive volume in a space, you are probably correct.... But what if it does not ? Like some kind of sponge or so ? $\endgroup$
    – mick
    Nov 20 at 20:52
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    $\begingroup$ If packing the shapes obligates that the "holes" between them have positive area (which I believe is the case NoahScheweber is asking you to consider) then from your comment I think you already agree with him, no? $\endgroup$
    – Ron Kaminsky
    Nov 20 at 20:57
  • $\begingroup$ @RonKaminsky well it makes alot of sense. But I would not call it a proof yet ? $\endgroup$
    – mick
    Nov 20 at 21:33
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    $\begingroup$ The Julia set of $z \to z^2 - 2$ is a line segment, which contains uncountably many line segments; however only countably-many are related to the dynamics, so this is more of a coincidence than anything structural. (Thanks for @NoahSchweber's hint.) $\endgroup$
    – Claude
    Nov 21 at 18:56

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