# Divergence defined w.r.t infinity? Or Divergence defined w.r.t there being a finite N s.t the iteration diverges at N? Mandelbrot Set. [closed]

This question is not about the concept of divergence in general. This question is about it only specific to the the map z^2 + c in the complex plane. This question creates two sets, defined differently, and asks if both the sets are they same set?

$$\text{the question of this post is (will be elaborated on): } \\ \mathbb{M}\_{\textsf{inf}} = \mathbb{M}\_{\text{computable}} \text{?}$$

# The Question:

I am under the assumption of and I expect that there is a distinction between these two following possible definitions of the Mandelbrot set. Each of which would result in different sets, and different results. The question is not

"which one is the correct one, or which is the canonical one, which one "is" the Mandelbrot Set?",

an opinion question like that is not good for stackexchange.

My question is:

"Do the following two definitions lead to different sets, that are distinct, and not the same as each other?"

# Set $$\mathbb{M}_{\textsf{inf}}$$

"infinite operations version"

(more precise): For point $$p=a+bi \in \mathbb{C}$$. And for the mapping $$f_n=(f_{n-1})^2 + (p)$$ with $$f_0 = 0$$. Then p is a member of the set called $$\mathbb{M}_{\text{inf}}$$ under the condition that its magnitude after an infinite amount of iterations is still less than 2:

$$|\lim_{n \to \infty} f_n| < 2$$

(paraphrased): A point $$p=a+bi$$ is declared to have diverged only w.r.t its magnitude after taking infinite amount of iterations of $$z^2+c$$. If its magnitude after an infinite amount of iterations applied to it, is bigger than/eq to 2, then we declare it as diverged. You only have to iterate an infinite amount of times to know for sure (in the general case of all possible complex numbers which are uncountable). It is said to have diverged only with respect to an infinite amount of operations applied to it, in the limit, or otherwise. There may be points that do not diverge for any finite N, but do indeed diverge in the limit only at infinity.

# Set $$\mathbb{M}_{\text{computable}}$$

"finite operations version"

This is an update to the below definition. There was some confusion in the comments on the idea of divergence being defined as "the ability to find any arbitrary big-N such that a_n surpasses big-N". This is not the notion of divergence here. We only need to check if its magnitude gets bigger than 2. Not if it also gets bigger than 9,99,999.. ect. And so, it may be prudent to extend the below definition, of $$\mathbb{M}_{\text{computable}}$$. In which "divergence" is declared w.r.t there being a minimum $$f_M$$, s.t M is the smallest natural in which |$$f_M| >=2$$

(more precise): For point $$p=a+bi \in \mathbb{C}$$. And for the mapping $$f_n=(f_{n-1})^2 + (p)$$ with $$f_0 = 0$$. Then p is a member of the set called $$\mathbb{M}_{\text{computable}}$$ under the condition that: $$\nexists n \in \mathbb{N}$$ such that $$2 \leq |f_n|$$

(paraphrased): A point $$p=a+bi$$ diverges iff, that after a finite amount of iterations of $$z^2+c$$, its magnitude is bigger than/eq to 2. You only have to iterate a finite amount of times. There will be some finite N at which it diverged

As a reminder, this isn't an opinion question, but a technical one. I am asking the question:

$$\text{THE QUESTION OF THIS POST IS: } \\ \mathbb{M}\_{\textsf{inf}} = \mathbb{M}\_{\text{computable}} \text{?}$$

### Recap

To recap, the question is whether there is a nontrivial distinction between the set if you define with respect to its magnitude after only an infinite amount of operations have been applied to it. Or it with respect to the existence of some finite natural less than infinite in which it diverged at that finite natural. That, the set if you define it w.r.t an infinite amount of iterations applied to, whether that literally results in a different set.

• You really should edit this down. Math.se expects fairly well focused questions. Commented Jul 26 at 2:57
• I don't understand your question. The Mandelbrot set is, by definition, the set of complex numbers $z_0$ which have bounded orbits with respect to the recurrence $z_{n+1} = z_n^2 + z_0$. As a practical matter, it may be very difficult (or even impossible?) to know if any particular $z_0$ has a bounded orbit, so it is really only practical to approximate this set, but the underlying definition is not really up for debate... Commented Jul 26 at 2:57
• @XanderHenderson Hi, thanks for commenting, I'd be glad to clear up confusion. Pretend this question is not about the mandelbrot set, so as to avoid preconceived notions. The question creates two Sets, M_inf, and M_computable. These two sets have membership defined differently. The question is, does M_inf = M_computable. Are they the same set? I expect that they are not. Commented Jul 26 at 3:04
• The meaning of the notation $\lim_{n\to\infty} a_n$ is that for any $M > 0$, there exists an $N > 0$ such that $n > N$ implies that $a_n > M$. That is the definition of divergence to infinity. Commented Jul 26 at 3:21
• You could write this question in one paragraph - give the standard definition, give your alternative, ask if they're equal, and give at most two sentences on why you're curious and what you suspect to be the case. My guess is the down votes (mine included) are because of all the nattering about. Commented Jul 26 at 4:27

# Questions

Define Mandelbrot sequences $$f_{n\,;\,p}:=\begin{cases}0 & \text{ if }n=0\\\left(f_{n-1}\right)^{2}+p & \text{ otherwise}\end{cases}$$.

Here is my interpretation of possible intended questions:

1. Is this true?: $$p$$ is in the Mandelbrot set $$\Leftrightarrow\left(\nexists n\in\mathbb{N},\left|f_{n\,;\,p}\right|\ge2\right)$$
2. Is this true?: $$\forall p\in\mathbb{C},\left|{\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}\right|\le2\Leftrightarrow\left(\nexists n\in\mathbb{N},\left|f_{n\,;\,p}\right|>2\right)$$
3. Is this true?: $$\forall p\in\mathbb{C},\left({\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}\text{ exists}\right)\Rightarrow\left(\left|{\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}\right|\le2\Leftrightarrow\left(\nexists n\in\mathbb{N},\left|f_{n\,;\,p}\right|>2\right)\right)$$
4. Is this true?: $$\forall p\in\mathbb{C},\left({\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}\text{ exists}\right)\Rightarrow\left(\left|{\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}\right|<2\Leftrightarrow\left(\nexists n\in\mathbb{N},\left|f_{n\,;\,p}\right|\ge2\right)\right)$$
5. Is this true?: $$\forall p\in\mathbb{C},{\displaystyle \limsup_{n\to\infty}}\left|f_{n\,;\,p}\right|\le2\Leftrightarrow\left(\nexists n\in\mathbb{N},\left|f_{n\,;\,p}\right|>2\right)$$

## Question 1

Note that $$f_{1\,;\,-2}=-2$$, $$f_{2\,;\,-2}=2$$, and indeed $$f_{n\,;\,-2}=2$$ for $$n\ge2$$. (We say that $$-2$$ is in the Mandelbrot set because this sequence does not diverge - it just stays at $$2$$ after the first couple steps.)

Because of this counterexample, the answer to question 1 is no. $$\left|{\displaystyle \lim_{n\to\infty}}f_{n\,;\,-2}\right|\nless2$$ and $$\left|f_{1\,;\,-2}\right|=\left|f_{17\,;\,-2}\right|=2$$ and $$p$$ is in the Mandelbrot set because the sequence does not "diverge to infinity", for any definition of what that means.

## Question 2

The answer to Question 2 is no, because the Mandelbrot set contains points $$p$$ where the corresponding sequence $$f_{n\,;\,p}$$ is eventually periodic, so that it loops around between multiple values. (See, for instance, the Wikipedia article on Misiurewicz points.)

For such a value of $$p$$, the limit doesn't exist (so we can't say "the limit exists and is bounded by $$2$$"), despite the fact that the sequence never escapes the disk of radius $$2$$.

## Question 3

The answer to Question 3 is yes, because of how we define limits of sequences and the properties of the Mandelbrot set.

### Proof:

We will work with the contrapositive.

Suppose that for a certain $$m$$, we had $$\left|f_{m\,;\,p}\right|>2$$. By the properties of the Mandelbrot set, the sequence $$f_{n\,;\,p}$$ would diverge to infinity, and we wouldn't have $${\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}$$ existing to begin with.

Now suppose that $$\left|{\displaystyle \lim_{n\to\infty}}f_{n\,;\,p}\right|>2$$. Then there exists a limit $$L$$ with $$|L|>2$$ such that for all error tolerances $$\varepsilon>0$$, there is an $$N$$ such that $$n>N\Rightarrow\left|f_{n\,;\,p}-L\right|<\varepsilon$$. But then by choosing $$\varepsilon=|L|-2$$, and applying a version of the triangle inequality, we would have an $$m$$ such that $$\left|f_{m\,;\,p}\right|>2$$.

## Question 4

I honestly do not know the answer to Question 4, because I don't know off the top of my head if the Mandelbrot set has points $$p$$ where the sequence $$f_{n\,;\,p}$$ approaches the circle of radius $$2$$ from the inside. I don't think this was the main focus of the question, though.

## Question 5

This, using the limit superior, is the version of Question 3 that doesn't need to assume the limit exists.

The answer is still yes, and the proof is very similar to that for Question 3.

There may be points that do not diverge for any finite $$N$$, but do indeed diverge in the limit only at infinity.

I note that this isn't really true: all definitions of getting past absolute value $$2$$ in the limit require getting past $$2$$ at some finite stage.