Connected Unit Disk
This is a fun problem:
Challenge: Prove in as many, and as indirect, ways that the unit disk 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.
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.
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...
Originally Posted by Drexel28
A) Yeah this is fractal talk!
B blew my mind but I'll try explain...
Take that recurrence relation I posted, set . 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...)
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 . 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...
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.)
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.
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?
Originally Posted by Deadstar
I'm sorry I don't have anything enlightening to say (Speechless)
Well, there are a million, zillion proofs.
Here is the most boring.
Proof: which with the product topology is the product of two connected spaces and thus connected.
Here is a really indirect way to prove a stronger result.
Theorem: Let be the -th dimensional unit ball ( ). Then, is path connected.
Lemma: Let . Then, is path connected.
Proof: It is fairly well known that is an -manifold and thus locally Euclidean of dimension and thus locally path connected. Thus, by a relatively well-known theorem we must merely prove that
is connected. To see this we first remember the ,once again well-known and easy to prove (think stereographic projection), fact that where is the Alexandroff compactification of
(this compactification exists since is Hausdorff, locally compact, and non-compact). Next, we recall that the inclusion map is
a topological embedding, and so is connected. But, since and the closure of a connected set is connected it follows that and thus are connected and thus by previous
comment path connected. .
So, since every every -th dimensional circle are homeomorphic to it follows that they are each connected. Thus, consider the interval , this is clearly connected
and homeomorphic to , which is therefore connected. Note also then that since that is path connected. Also,
that is path connected. But, it is easy to verify that from where the conclusion follows. For this is
merely the unit interval and thus trivially path connected.
Convolutedness factor: 8/10.
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 but as we move them apart (as n gets big) they will "approach" . I have no idea, don't ask me. I just made a guess.
Originally Posted by Bruno J.