April 6th, 2017, 09:03 PM  #31 
Senior Member Joined: Aug 2012 Posts: 2,343 Thanks: 732  
April 16th, 2017, 08:45 PM  #32 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 Math Focus: Number theory 
Porosity is referenced on one of the aforementioned websites. I was curious whether it had anything to do with density. Take a ball of radius R. Slice it into alternating spherical shells of infinitesimal thickness. Reconstruct the odd or even shells into one ball each of radius R. 
April 17th, 2017, 10:22 AM  #33  
Senior Member Joined: Aug 2012 Posts: 2,343 Thanks: 732  Quote:
What they are trying to say is that one of the pieces must be nonmeasurable. I might use words like "weird" or "hard to visualize" for nonmeasurable sets, but porous wouldn't be a word I'd use. A nonmeasurable set is porous in the sense that it contains no contiguous intervals; but you could say the same for the rationals or the irrationals, both of which are porous in exactly the same way, yet perfectly measurable. I'd ignore the word porosity. It's a Wikipedia error. Quote:
But there's no problem generating paradoxes with cardinality. We can decompose the counting numbers 1, 2, 3, ... into the evens and the odds; and each of those can be bijected to the original set of natural numbers. We could do the same thing with a ball in 3space, using a cardinality argument to make two balls out of one. What's amazing about the BanachTarski theorem is that it shows that we can do this not with a mere cardinality argument  after all, cardinality is a very weak and counterintuitive measure of size  but with rigid motions, or isometries. That is, we can decompose the ball in 3space into as few as five pieces, move the pieces around space via rigid, distancepreserving motions, and end up with two balls, each the same size as the original. Last edited by Maschke; April 17th, 2017 at 10:24 AM.  
April 17th, 2017, 03:23 PM  #34 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 Math Focus: Number theory 
What is the fourth (lower righthand), unlabled set on the Cayley graph at https://en.wikipedia.org/wiki/Banach...sition_F_2.svg Is the Cayley graph of 2, 3 or other dimensions? Can Banach–Tarski be understood by means of fractals? Are isometries restricted to the Cayley graph in a nonmeasurable but continuous way? 
April 17th, 2017, 03:52 PM  #35  
Senior Member Joined: Aug 2012 Posts: 2,343 Thanks: 732  Quote:
$ab^2a^{5}$ or $a^{12}b^{47}$ and so forth, if we hit them all on the left with $a^{1}$, we end up with every possible word except the ones that begin with a single 'a'. To get to this point I need to do some more exposition. I have a post on the free group on ONE letter, which is a helpful prereq to the free group on two letters. I'll post that soon. Perhaps if I get time in the next few days I'll answer your direct question in more detail. The point is that the set of "words" made up of two letters and their inverses is paradoxical, it can be split into subsets that can be rotated back to form two copies of the original set. This is the fundamental idea that we can then lift into 3space in the next part of the proof. The picture shown is the graph of $F_2$, the free group on two letters. It's all we need for BanachTarski. There are of course free groups with 3 generators, 4, 5, and even infinitely many generators, but we don't need them for BT. You've motivated me to continue with my exposition so I'll start talking about free groups soon. Heck if I know. I don't think it can be "understood" at all. The best we can do is try to follow the logic. All proofs involving the axiom of choice are like that. I've never seen anything describing any connection to fractals but that doesn't mean there isn't some connection somewhere. I don't personally think this is a productive line of inquiry else someone would have a web page about it. Quote:
The Cayley graph doesn't have anything to do with isometries. In this step of the proof we're only showing that $F_2$ is paradoxical. Then in the next step, we show that the isometry group of 3space contains a copy of $F_2$ and use that fact to get the BT paradox.  
April 17th, 2017, 05:15 PM  #36 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 Math Focus: Number theory 
Is AplanisTophet an anagram?

April 18th, 2017, 12:28 AM  #37  
Senior Member Joined: Aug 2012 Posts: 2,343 Thanks: 732  Quote:
To begin with, the picture of the Cayley graph on the Wiki page is MISLABELED. I will fix this for you here. In what follows I'm using this picture of the same graph but drawn a little differently on this page: https://en.wikipedia.org/wiki/Free_group. It's the same picture but a little simpler. So now first, here is the picture with the properly labeled quadrants. Say we start in the center at the identity word $e$. (All undefined terminology is defined in the Wiki pages and will be defined in the next two articles on free groups, as I say I'm a little ahead of myself here). Now the first letter of our word might be $a$, $b$, $a^{1}$, or $b^{1}$. This corresponds to branching to the right, up, left, or down subgraphs, respectively. Now say our word starts with $a$. What are the possible choices for the second letter? They can be $a$, which would result in $a^2$; or $b$, resulting in $ab$; or $b^{1}$, resulting in $ab^{1}$. But what CAN'T be the second letter is $a^{1}$; because that would reduce to $aa^{1} = e$. That's why once you make your initial move into one of the four quadrant subgraphs, from then on there are only three possible next choices. [Did I say that clearly?] So now all our nonidentity words fall into four classes, representing the east, north, west, and south subgraphs. These are: * $S(a)$, the set of words starting with $a$; * $S(b)$, the set of words starting with $b$; * $S(a^{1})$, the set of words starting with $a^{1}$; * $S(b^{1})$, the set of words starting with $b^{1}$. Reasonable, yes? So we can write: $F_2 = \{e\} \cup S(a) \cup S(b) \cup S(a^{1}) \cup S(b^{1})$. Now, the magic happens. This is the mathematical version of the tower of Babel. Suppose we take every word in $S(a^{1})$ and hit it on the left by $a$. In other words we're "translating" it by $a$, which simply means applying $a$ to every word in $S(a^{1})$. What happens? * Any word that began $a^{1}bx$ where $x$ is any continuation, now becomes $bx$. In other words $S(b) \subset aS(a^{1})$. * Likewise any word that began $a^{1}b^{1}x$ now becomes $b^{1}x$. So $S(b^{1}) \subset aS(a^{1})$. * The word $a^{1} \in S(a^{1})$, so $aa^{1} = e\in aS(a^{1})$. * If a word was already in $S(a^{1})$, let's say $a^{5}b$ for example, hitting it on the left with $a$ simply reduces the exponent to $a^{4}b$. So all the words starting with $a^2$ and higher exponents end up in $aS(a^{1}$. And the words starting with $a^{1}b$ or $a^{1}b^{1}$ we already covered, they end up in $aS(a^{1})$ as noted. So what did we just prove? That $F_2 = aS(a^{1}) \cup S(a)$. We took one of the four quadrants, translated it, and it ate two of the other quadrants! That's the heart of the proof right there. Now I didn't draw a picture, but in the exact same way we can see that $F_2 = bS(b^{1}) \cup S(b)$. So we can start with the decomposition $F_2 = \{e\} \cup S(a) \cup S(b) \cup S(a^{1}) \cup S(b^{1})$, translate two of the pieces, and end up with TWO copies of $F_2$. Now all that remains is to translate this magic to Euclidean 3space, which we'll do later by showing that the isometry group of 3space contains a copy of $F_2$. This is the heart of the proof. I'd just add that none of this is completely obvious the first time through, you have to work at this a bit to see what's going on with all the notation and such, and play around with some examples to see how $aS(a^{1})$ works with the translating. Now that I've gone through this in detail I see your point. This has something to do with the fractal nature of the quadrants. Tower of Babel again, each sublibrary contains a copy of all the other sublibraries. But this isn't used directly in the proof, it's sort of lurking in the background. Last edited by Maschke; April 18th, 2017 at 12:41 AM.  
April 23rd, 2017, 08:38 PM  #38  
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 Math Focus: Number theory 
Maschke, Quote:
 
April 27th, 2017, 09:31 PM  #39 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 Math Focus: Number theory 
Isn't "doubling the cube," proved impossible in 1837, a form of Banach–Tarski?

April 28th, 2017, 05:45 AM  #40  
Senior Member Joined: Aug 2012 Posts: 2,343 Thanks: 732  Quote:
https://en.wikipedia.org/wiki/Doubling_the_cube Last edited by Maschke; April 28th, 2017 at 05:51 AM.  

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