## The Mandelbrot Set: Compact?

Several weeks ago, I read something on Wikipedia that shocked me: “The Mandelbrot set is a compact set.

At first I didn’t believe it. How could the Mandelbrot set, in its infinite complexity, be compact?

### Attempt 1: Prove False — Failed

My first attempt to disprove this was very naive: I’d prove it was Lindelöf by putting an open set around each one of the seahorses in seahorse valley, thereby constructing an open cover with only countable subcover.

That something is wrong is immediately clear if we carry this to its natural conclusion: we could put an open cover around each `sprout’ on each seahorse, and recurse indefinitely, creating an open cover with a $\aleph_0^{\aleph_0}$ subcover. The Mandelbrot set is a subset of the complex plane so it is clear that this is nonsense.

What is wrong? We have to put an open set around the main component of all of these. It will contain some of the off shoots. We can have an arbitrarily large finite number outside it, but that is very different from infinity…. So that argument doesn’t work.

### Attempt 2: Prove False — Failed

The second attempt arose from consider the definition of the Mandelbrot set: all points $x$ that don’t diverge under $\lim_{n\to\infty} (z\to z^2+x)^n$

If the Mandelbrot set, $M$ is compact, it’s image in $(z\to z^2+x)^n$ is compact for any finite $n$ (by the continuity of finite compositions of a continuous function). But this can’t be compact, since the set contains all numbers that don’t diverge to infinity…

This is the mistake, right here. It essentially arises from me misunderstanding the Mandelbrot set. At this point, if you had asked me to define the Mandelbrot set, I would have written, and have written several times on this site (sorry!) $\{x | x\in\mathbb{C}; \lim_{n\to\infty}\left|(z\to z^2+x)^n(x)\right|\neq\infty\}$. But this isn’t correct.

It wasn’t even what I was thinking. In my head, I was thinking $\{x | x\in\mathbb{C}; \left|(z\to z^2+x)^{\aleph_0}(x)\right|\neq\infty\}$ which is also wrong, but at least intelligible. In the previous one, the set would have been all of real numbers because applying the function a finite number of times (even a very large, approaches infinity number of times) won’t reach infinity, just get arbitrarily close. When we iterate a countable infinity number of times, we can at least reach an infinite magnitude and thus have something like the Mandelbrot Set.

Attempt 3: Prove True — Success Failed

So what is the correct definition of the Mandelbrot Set? $\{x | x\in\mathbb{C}; \lim_{n\to\infty} |(z\to z^2+x)^n(x)| \leq 2 \}$

Why? Because once the magnitude exceeds $2$, it must escape. Even in the worst case $z$ and $x$ are in opposite directions, $2$ is the turning point because $2^2-2=2$. (This is why the Mandelbrot Set is bound in the $2$ unit circle.)

Using this definition, our proof works. The Mandelbrot Set is compact iff it’s image in $x\to\lim_{n\to\infty}(z\to z^2+x)^n(x)$ is compact. By definition, we select that set as $\{x | x\in\mathbb{C}; |x|\leq 2\}$ which is compact since it is closed and bounded (Bolzano–Weierstrass theorem). QED.

Update: Or not. There are a number of problems with this proof that I didn’t see when I made it. I blame the fact that I was very new to topology at the time.

The first one is that the implication for compactness goes the wrong way, ie. the image of a compact set must be compact, but the reverse is not necessarily true. If this was the only problem, however, we could save our proof by using the fact that the inverse image of a closed set under a continuous function is closed along with the fact that we know the Mandelbrot set is bounded.

The second is that the composition of a non-finite number number of continuous functions is not necessarily continuous (trivial example, think about what would happened if squared over and over again: the limit would exist but it would not be continuous). In order to get continuity we would need uniform convergence, which we don’t have. In fact, the limit of the function does not even exist in this case (because of stable cycles), nor even necessarily a convergent subsequence.

The real proof is very much non-trivial.

### Corollary: The Mandelbrot Set is Closed

Again, Bolzano–Weierstrass theorem.

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