# Connected Unit Disk

• Apr 25th 2010, 11:38 PM
Drexel28
Connected Unit Disk
This is a fun problem:

Challenge: Prove in as many, and as indirect, ways that the unit disk $\displaystyle \mathbb{D}$ with the usual topology is connected and/or path connected (they are equivalent, but show either one).

Note: While any and all solutions are welcome, the challenge is to think of strange and creative ways to prove it. Use whatever subject matter you wish.
• Apr 26th 2010, 04:08 PM
Since no one is posting I'll post the first thing I though of (Note, I don't know the first thing about topology, well maybe a little metric spaces, so this is not a proof or anything... Just a though).

Consider the discrete dynamical system.

$\displaystyle x_{n+1} = x_n^k + c$

Some may recognize this as the formula to generate the Mandelbrot set when k=2. But if you increase k you get what's known as multibrot sets. As k diverges to infinity the multibrot sets converge to the unit disk.

So my thought was that since the Mandelbrot set is connected (although I'm pretty sure this proof involves the unit disk being connected lol...) perhaps that implies the multibrot sets are connected and taking k to infinity will show that the unit disk is therefore connected.

Just my thought...
• Apr 26th 2010, 04:17 PM
Drexel28
Quote:

Since no one is posting I'll post the first thing I though of (Note, I don't know the first thing about topology, well maybe a little metric spaces, so this is not a proof or anything... Just a though).

Consider the discrete dynamical system.

$\displaystyle x_{n+1} = x_n^k + c$

Some may recognize this as the formula to generate the Mandelbrot set when n=2. But if you increase n you get what's known as multibrot sets. As n diverges to infinity the multibrot sets converge to the unit disk.

So my thought was that since the Mandelbrot set is connected (although I'm pretty sure this proof involves the unit disk being connected lol...) perhaps that implies the multibrot sets are connected and taking n to infinity will show that the unit disk is therefore connected.

Just my thought...

That would certainly be "strange and creative"! That's exactly the kind of thing I wanted to see. But, I have a few problems:

A) I know so little about what you just said that I only think you are talking about fractals (I know who Benoit Mandelbrot is)

B) I think you're idea involves discussing a sequence of connected sets $\displaystyle \left\{F_n\right\}_{n\in\mathbb{N}}$ such that $\displaystyle F_n\to\mathbb{D}$. What does convergence mean for a sequence of sets? Secondly if I take a heuristic idea of what that would mean a reasonable counterexample to a sequence of connected sets converging to a connected sets would be $\displaystyle F_n=\left(-1,\tfrac{1}{n}\right)\cup\left(\tfrac{-1}{n},1\right)$ since each $\displaystyle F_n=(-1,1)$ is connected but $\displaystyle F_n\to (-1,0)\cup(0,1)$ which is not. Does that maybe capture an idea of converging of a sequence of sets?
• Apr 26th 2010, 05:06 PM
Quote:

Originally Posted by Drexel28
That would certainly be "strange and creative"! That's exactly the kind of thing I wanted to see. But, I have a few problems:

A) I know so little about what you just said that I only think you are talking about fractals (I know who Benoit Mandelbrot is)

B) I think you're idea involves discussing a sequence of connected sets $\displaystyle \left\{F_n\right\}_{n\in\mathbb{N}}$ such that $\displaystyle F_n\to\mathbb{D}$. What does convergence mean for a sequence of sets? Secondly if I take a heuristic idea of what that would mean a reasonable counterexample to a sequence of connected sets converging to a connected sets would be $\displaystyle F_n=\left(-1,\tfrac{1}{n}\right)\cup\left(\tfrac{-1}{n},1\right)$ since each $\displaystyle F_n=(-1,1)$ is connected but $\displaystyle F_n\to (-1,0)\cup(0,1)$ which is not. Does that maybe capture an idea of converging of a sequence of sets?

Ok.

A) Yeah this is fractal talk!

B blew my mind but I'll try explain...

Take that recurrence relation I posted, set $\displaystyle x_0 = 0$. Take a random point in the complex plane (this can be simplified down to points within a disk of radius 2), this is the c value.
Iterate it repeatedly. If the absolute value of any iterate is above 2 than the sequence of iterates will diverge.

The Mandelbrot set is the set of all points that do not diverge under repeated iteration.

That's it simply put. Now change k and repeat the process. This will give an entirely new set called multibrot sets (because they look like they are constructed from multiple mandelbrot sets.)

Multibrot set - Wikipedia, the free encyclopedia

Go down to the positive powers part on that link and ignore the annoying colours and focus on the black. That is the points in the set.

Now take k to infinity and the plot of the set looks like this (on a not exactly equal axis...)

http://img256.imageshack.us/i/startupw.jpg/http://img52.imageshack.us/img52/36/startupi.jpg

where black points mean in the set. (note n=10000 there)

It appear to be 'converging' (in a visual image of the set kind of way) towards a unit disk.

The connectedness part is beyond me I'm afraid.

However the dynamical system part of it is not...

The main area of each multibrot set has a attracting fixed point at the centre. I.e. points within the same 'area' converge towards it under repeated iteration. (For the original Mandelbrot set (k=2) this will be the large kidney bean shaped part)

Ignore the multibrot set thing now and just focus on the recurrence relation with $\displaystyle k \to \infty$. Any point that is contained within the OPEN unit disk will converge to the attracting fixed point at the origin under repeated iteration (no formal proof of this but it's not too hard to see). Therefore we could think of the unit disk as being the set of all points in the complex plane that do not diverge under repeated iteration from this formula...

This will be ALL points contained within the (open) unit disk and I suppose that will somehow show connectedness... (since every point in the disk will be in the set...)

I can't explain this too well right now cos it's 2am and I'm fried but was just a thought...
• Apr 26th 2010, 05:08 PM
BLAARG total mistake I see I've been saying. Take k to go to infinity. Not n (well that will go to infinity with repeated iteration anyway.)

But..!

It's the value of k we are interested in and that's what gives us the multibrot. Sorry!

EDIT: Have edited previous posts now and hope I didn't miss any.
• Apr 26th 2010, 07:39 PM
Drexel28
Quote:

Ok...

I'm sorry, the fractal stuff is beyond me as of now. I have zero previous knowledge, and without that I am not sure how you can show connectedness. Maybe another member with both knowledge of fractals and topology could help?

I'm sorry I don't have anything enlightening to say (Speechless)
• Apr 28th 2010, 04:36 PM
Drexel28
Well, there are a million, zillion proofs.

Here is the most boring.

Proof: $\displaystyle \mathbb{D}\approx\mathbb{R}^2$ which with the product topology is the product of two connected spaces and thus connected.

Yay....so fun.
• Apr 29th 2010, 06:11 PM
Drexel28
Here is a really indirect way to prove a stronger result.

Theorem: Let $\displaystyle \mathbb{B}^n,\text{ }n\geqslant 1$ be the $\displaystyle n$-th dimensional unit ball ($\displaystyle \mathbb{B}^2=\mathbb{D}$). Then, $\displaystyle \mathbb{B}^n$ is path connected.

Proof:

Lemma: Let $\displaystyle \mathbb{S}^n=\left\{\bold{x}\in\mathbb{R}^{n+1}:\| \bold{x}\|=1\right\},\text{ }n\geqslant 1$. Then, $\displaystyle \mathbb{S}^n$ is path connected.

Proof: It is fairly well known that $\displaystyle \mathbb{S}^n$ is an $\displaystyle n$-manifold and thus locally Euclidean of dimension $\displaystyle n$ and thus locally path connected. Thus, by a relatively well-known theorem we must merely prove that

$\displaystyle \mathbb{S}^n$ is connected. To see this we first remember the ,once again well-known and easy to prove (think stereographic projection), fact that $\displaystyle \mathbb{S}^n\approx\mathbb{R}_\infty^n$ where $\displaystyle \mathbb{R}_\infty^n$ is the Alexandroff compactification of

$\displaystyle \mathbb{R}^n$ (this compactification exists since $\displaystyle \mathbb{R}^n$ is Hausdorff, locally compact, and non-compact). Next, we recall that the inclusion map $\displaystyle \iota:\mathbb{R}^n\hookrightarrow\mathbb{R}_\infty ^n$ is

a topological embedding, and so $\displaystyle \mathbb{R}^n\subseteq\mathbb{R}_\infty^n$ is connected. But, since $\displaystyle \overline{\mathbb{R}^n}=\mathbb{R}_\infty^n$ and the closure of a connected set is connected it follows that $\displaystyle \mathbb{R}_\infty^n$ and thus $\displaystyle \mathbb{S}^n$ are connected and thus by previous

comment path connected. $\displaystyle \blacksquare$.

So, since every every $\displaystyle n$-th dimensional circle $\displaystyle C_\delta^n=\left\{\bold{x}\in\mathbb{R}^{n+1}:\|\b old{x}\|=\delta\right\}$ are homeomorphic to $\displaystyle \mathbb{S}^n$ it follows that they are each connected. Thus, consider the interval $\displaystyle [0,1)$, this is clearly connected

and homeomorphic to $\displaystyle [0,1)\times\underbrace{\{0\}\times\cdots\times\{0\} }_{n\text{ times}}=L^n\subseteq\mathbb{R}^{n+1}$, which is therefore connected. Note also then that since $\displaystyle L^n\cap C_\delta^n=(\delta,\cdots,0)$ that $\displaystyle L^n\cup C_\delta^n$ is path connected. Also,

$\displaystyle \left(L^n\cup C_{\delta_1}^n\right)\cap\left(L^n\cap C_{\delta_2}^n\right)\supseteq L^n,\text{ }\forall \delta_1,\delta_2\in(0,1)$ that $\displaystyle \bigcup_{\delta\in(0,1)}\left(L^n\cup C_\delta^n\right)$ is path connected. But, it is easy to verify that $\displaystyle \mathbb{B}^{n+1}=\bigcup_{\delta\in(0,1)}\left(L^n \cup C_\delta^n\right)$ from where the conclusion follows. For $\displaystyle n=1$ this is

merely the unit interval and thus trivially path connected. $\displaystyle \blacksquare$

Convolutedness factor: 8/10.
• Apr 29th 2010, 07:12 PM
Bruno J.
Quote:

Originally Posted by Drexel28
Secondly if I take a heuristic idea of what that would mean a reasonable counterexample to a sequence of connected sets converging to a connected sets would be $\displaystyle F_n=\left(-1,\tfrac{1}{n}\right)\cup\left(\tfrac{-1}{n},1\right)$ since each $\displaystyle F_n=(-1,1)$ is connected but $\displaystyle F_n\to (-1,0)\cup(0,1)$ which is not. Does that maybe capture an idea of converging of a sequence of sets?

What is your definition of the limit of the $\displaystyle F_n$'s? Because $\displaystyle 0\in F_n$ for all $\displaystyle n$, why is $\displaystyle 0 \notin \lim_{n\rightarrow \infty}F_n$?
• Apr 29th 2010, 07:16 PM
Drexel28
Quote:

Originally Posted by Bruno J.
What is your definition of the limit of the $\displaystyle F_n$'s? Because $\displaystyle 0\in F_n$ for all $\displaystyle n$, why is $\displaystyle 0 \notin \lim_{n\rightarrow \infty}F_n$?

That was my point. I have no idea what the definition of "converging sequence of sets" means. So, I just pictured the two intervals looking like $\displaystyle F_n\text{ looks like }\underbrace{(\text{ }(\text{ })\text{ })}_{\text{overlapping}}$ but as we move them apart (as n gets big) they will "approach" $\displaystyle (\text{ })(\text{ })$. I have no idea, don't ask me. I just made a guess.