Vsauce's video on the Banach-Tarski paradox

[greypilgrim]

A question to Vsauce's famous video about the Banach-Tarski paradox at 10:09:



Can you really construct the hyper-webster like that?

If you choose the order like that, you'll never get any words containing other letters than "A". Shouldn't you choose an order like A, ... , Z, AA, ..., AZ, BA, ..., BZ, ... to make sure that every word is included?

The same question arises later when he writes down the rotation sequences. Here he starts with L, LL, LLL, ... , so how can there be anything else but pure left rotations?

 

[jedishrfu]

For other viewers of this thread: Hyper-webster refers to Webster's Dictionary.

He shows two poles in his diagram so L-R are rotations about the North/South pole and U-D are rotations about the East/West pole.

 

[nuuskur]

In measure theory, the paradox arises if one assumes all subsets of $\mathbb R^3$ are measurable. As this leads to contradiction (making two out of one), the assumption must be false (and it is).

While similar, this doesn't correspond to the Banach-Tarski paradox. If the hyperwebster is to be regarded a sphere, it should contain words of length $\omega$, which it doesn't.

Seems to be based on I. Stewart's From here to infinity, but I haven't read the book. Perhaps something is lost in translation since the author of the video is not mathematician.

 

[PeroK]

In measure theory, the paradox arises if one assumes all subsets of $\mathbb R^3$ are measurable. As this leads to contradiction (making two out of one), the assumption must be false (and it is).

It's only false if you assume the axiom of choice.

 

[jedishrfu]

Vsauce is doing arguably a lot better than most or even all mathematicians with over 15 million subscribers on youtube.

https://en.wikipedia.org/wiki/Michael_Stevens_(educator)

 

[greypilgrim]

I'm sorry, but those replies don't address my (rather specific) question at all ...

 

[nuuskur]

If you choose the order like that, you'll never get any words containing other letters than "A". Shouldn't you choose an order like A, ... , Z, AA, ..., AZ, BA, ..., BZ, ... to make sure that every word is included?

As you can see, even the author of the video has trouble listing all of the words starting in Z. If $\mathcal L$ was the English alphabet, what they have in mind is
<br /> W(Z) := \left\{ Za_1a_2\ldots a_n \mid n\in\mathbb N, a_i \in \mathcal L, i=1,\ldots, n \right\}\cup\{Z\}<br />
Trickier with the rotations, since something like ..LR.. is illegal. The axiom of choice is indeed, powerful.

 

[TeethWhitener]

If you choose the order like that, you'll never get any words containing other letters than "A". Shouldn't you choose an order like A, ... , Z, AA, ..., AZ, BA, ..., BZ, ... to make sure that every word is included?

The same question arises later when he writes down the rotation sequences. Here he starts with L, LL, LLL, ... , so how can there be anything else but pure left rotations?

The ... indicate an infinite number of terms. We have (or can fairly easily formulate) a rule to pick out any single term. The same kind of notation arises in ordinal arithmetic where you have orderings of $\mathbb{N}$ like $1,3,5,\dots,2,4,6,\dots$ There is a clear order (e.g., 1<3, etc for all odd numbers, 2<4, etc for all even numbers, and all odd numbers are less than 2).

It may seem a little odd, but we deal with orderings like this on a fairly regular basis. Consider the set of the union of all rational numbers with numerator less than denominator and the natural numbers $S=\{\frac{a}{b}| a,b \in \mathbb{Z}\ \wedge a<b\} \cup \mathbb{N}$. This set could be written $\{\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \frac{2}{3},\frac{2}{4},\frac{2}{5},\dots, 1,2,3,\dots\}$, where each of the sets of dots denotes an infinite collection of numbers.

 

[jedishrfu]

Have you looked for any papers on the paradox beyond the video?

It seems this is a very specialized question.

With respect to rotations, i think you have to allow for left and right even though you may need only one to complete it.

 

[SSequence]

If the hyperwebster is to be regarded a sphere, it should contain words of length $\omega$, which it doesn't.

I am not familiar with the topic but I think so? Because if we only allow words of finite length we just have simple (algorithmic) enumeration of all the finite words.

The ... indicate an infinite number of terms. We have (or can fairly easily formulate) a rule to pick out any single term. The same kind of notation arises in ordinal arithmetic where you have orderings of $\mathbb{N}$ like $1,3,5,\dots,2,4,6,\dots$ There is a clear order (e.g., 1<3, etc for all odd numbers, 2<4, etc for all even numbers, and all odd numbers are less than 2).

Here is another way to put it. If there is a well-order of countable length/order-type [though in this case and in the linked video we are talking about well-orders of very small length] for a set, then there is also a well-order of length $\omega$. Basically the same thing can be said for dictionary with 26 alphabet and finite length words.P.S.
Not directly related to topic but a related point [which can sometimes confuse people] is when we consider all (ordered) lists of finite length whose each element is a natural number. I used to be somewhat confused with this too initially, but its a long time now though (around 8 years or so).

 

[greypilgrim]

The ... indicate an infinite number of terms. We have (or can fairly easily formulate) a rule to pick out any single term. The same kind of notation arises in ordinal arithmetic where you have orderings of $\mathbb{N}$ like $1,3,5,\dots,2,4,6,\dots$ There is a clear order (e.g., 1<3, etc for all odd numbers, 2<4, etc for all even numbers, and all odd numbers are less than 2).

It may seem a little odd, but we deal with orderings like this on a fairly regular basis. Consider the set of the union of all rational numbers with numerator less than denominator and the natural numbers $S=\{\frac{a}{b}| a,b \in \mathbb{Z}\ \wedge a<b\} \cup \mathbb{N}$. This set could be written $\{\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \frac{2}{3},\frac{2}{4},\frac{2}{5},\dots, 1,2,3,\dots\}$, where each of the sets of dots denotes an infinite collection of numbers.

Ok, if this is a common notation, I'm fine with that.

I just remember that when I first heard lectures about cardinality (only very basic ones at the beginning of a programming course) and had to solve exercises (Hilbert's Hotel and such), we had to be very careful about the construction of such lists. If I had handed in something like $1,3,5,\dots,2,4,6,\dots$, the comment probably would have been "you can't do that, at which position on the list appears 2"?
And for the rationals, we used constructions like this one, where every fraction is at a well-defined position:

 

[TeethWhitener]

Ordinality is different from cardinality. In the set with the specific order above, 2 would be at the $(\omega +1)$st position, as that particular set would have order type $\omega +\omega$.

 

[SSequence]

One can also make a list (of length $\omega$) as in OP. Actually its quite natural in this case. For example:
All words of length-0 (1 words)
All words of length-1 (26 words)
All words of length-2 (26^2 words)
All words of length-3 (26^3 words)
All words of length-4 (26^4 words)
...Also, it seems that in the video linked in the OP, the words are being listed in "dictionary order" so to speak. So let's suppose we just had two alphabet [with more alphabet only the method would be slightly more generalized and the length of well-order would change a bit]. It seems we would write something like:
A, AA, AAA, AAAA, AAAAA, ...
AB, ABA, ABAA, ABAAA, ABAAAA, ...
ABB, ABBA, ABBAA, ABBAAA, ABBAAAA, ...
ABBB, ABBBA, ABBBAA, ABBBAAA, ABBBAAAA, ...
...
B, BA, BAA, BAAA, BAAAA, BAAAAA, ...
BB, BBA, BBAA, BBAAA, BBAAAA, BBAAAAA, ...
BBB, BBBA, BBBAA, BBBAAA, BBBAAAA, BBBAAAAA, ...
BBBB, BBBBA, BBBBAA, BBBBAAA, BBBBAAAA, BBBBAAAAA, ...
...

So that seems to give a well-order of length $\omega^2 \cdot 2$.

======================================================Not sure how this is relevant to topic in OP though (seemingly it probably isn't).