The Axiom of Choice
Because posting entries about my life would be terribly boring for everyone, I'll post instead what gave me yesterday a much-needed boost (particularly since the night before I got The Judgemen as my 'daily card'): I'm not totally stupid, you see and I feel incredibly proud of myself after reading about the Axiom of Choice and understanding a (dumbed down) explanation. I do love crazy mathematics: they break my brain but it's worthy.
I'm totes ready to take over the world.
It's too much of a hassle to post pics so I'll just copy the text. It's a feed, so the comments will disappear soon.
Sometimes explaining a joke makes it funnier?
theojf
2010-10-12 06:13 am (local)
The Banach-Tarski "paradox" is relatively straightforward. Consider a language with two letters and two "antiletters": call the letters a and b and the "antiletters" α and β, and what I mean by "antiletter" is that if you ever see the strings aα, αa, bβ, or βb, then you can remove them and end up with _the same_ word, or rather they are words with the same meaning. So, for example, β is the same as aαβαabβ. Oh, and any string spelled with these letters is a valid word --- the "empty word" is also valid; it's equal, for example, to bβ.
Then the right way to create a dictionary is to list only the minimal-length spellings of the words. In particular, there's the empty word, and infinitely many words whose minimal-length spellings begin with a, infinitely many that begin with α, infinitely many that begin with b, and infinitely many that begin with β. If I call those five sets E (for empty), A, Д, B, and Б, and if I call the completely language L, then we have as sets L = E ∪ A ∪ Д ∪ B ∪ Б.
Now look at the set A, say, of words that start with a in their minimal-length spellings. Then the second letter in the minimal-length spelling is definitely not α (as then the word would start aα... and you could cancel those two), but it can be any of the other three letters (or nothing at all). So consider the set αA, which is all words in A, but add an α at the beginning; the new α can cancel with the initial a, and you get minimal-length spellings of words that do not start with α, but can start with anything else: αA = E ∪ A ∪ B ∪ Б. In particular, although before we had written L as a union of four infinite sets (and one tiny set), now we have L = Д ∪ αA, the union of just two sets.
Now here's the deal. Take the hollow sphere, which is a two-dimensional surface in three dimensions, and I'll assume that it's centered at the origin in xyz space. For each word in L, I will define a way to rotate the sphere. Namely, read along the word from left to right. If you see an a, rotate the sphere 1 radian (180/π degrees) clockwise around the xДxis. If you see an α, rotate 1 radian counterclockwise around the x axis. The letters b and β are the clockwise and counterclockwise rotations by 1 radian around the y axis. Then it's clear that even if you spell a word without using its minimal-length spelling, then you get the same total rotation as if you used the minimal-length spelling (because there would just be some rotations that cancel). What's not clear, and what I'll just ask you to believe me about (although it's not too unreasonable) is that if two words are honestly different, then they lead to honestly different rotations. (Essentially, this is because π is sufficiently irrational.)
Let S denote the sphere, and s∈S some point on the sphere. Applying the rotations to the sphere, each word w∈L determines some point w(s)∈S. Now, there are a lot more points in S than there are words in L. But what I want you to do is this: pick some subset T of S so that for each s∈S, there is precisely one point t∈T and precisely one word w∈L such that s=w(t). I.e.: T is a minimal set so that to get anywhere in S, you can start somewhere in T and apply rotations that arise from words in L.
The point is that I'm not going to tell you how to find the set T. The "axiom of choice" says that such a set exists, but doesn't give a construction for it. In fact, for reasonable use of the word, it is impossible to "construct" the set T. But it exists. If you could follow infinitely-long algorithms, it would be easy to construct T, just by starting with S and removing points until you get something minimal.
(continued because it's too long)
Re: Sometimes explaining a joke makes it funnier?
2010-10-12 06:14 am (local)
(continuation)
Anyway, so we have L(T) = S, but L = E ∪ A ∪ Д ∪ B ∪ Б, so S = E(T) ∪ A(T) ∪ Д(T) ∪ B(T) ∪ Б(T). By A(T), for example, I mean "all points that you can get to by starting somewhere in T and applying the rotation corresponding to some word in A". (Note that the set E consists just of the empty word, which corresponds to no rotation, so E(T) = T.)
Ok, so I've said how to cut up S into five pieces: the pieces are E(T)=T, A(T), Д(T), B(T), and Б(T). Take the three pieces T, B(T), and Б(T) and move them out of the way, so that you're just left with A(T) and Д(T). Leave the piece Д(T) exactly where it is, but rotate the piece A(T) by the rotation corresponding to α so that you get αA(T) - this is just a rigid rotation in three-space. But L = Д ∪ αA, so Д(T) ∪ αA(T) = L(T) = S. So with just those two pieces, you can fit them back together to build a whole sphere.
Now do something similar with the B(T) and Б(T) pieces. You get a second complete sphere.
Oh, you also have that bit T=E(T) left over. If you work a bit harder, you can squeeze it into the two spheres; alternately, you can throw it out with the pumpkin seeds.
I hope that sufficiently explained the joke (and that I haven't used too much math notation),
Theo
Mathematics, UC Berkeley
And then, a couple more from Fandom Wank which I'm not pasting because those don't disappear. They explain better the thing about the ball 'pieces'.
I'm totes ready to take over the world.
It's too much of a hassle to post pics so I'll just copy the text. It's a feed, so the comments will disappear soon.
Sometimes explaining a joke makes it funnier?
theojf
2010-10-12 06:13 am (local)
The Banach-Tarski "paradox" is relatively straightforward. Consider a language with two letters and two "antiletters": call the letters a and b and the "antiletters" α and β, and what I mean by "antiletter" is that if you ever see the strings aα, αa, bβ, or βb, then you can remove them and end up with _the same_ word, or rather they are words with the same meaning. So, for example, β is the same as aαβαabβ. Oh, and any string spelled with these letters is a valid word --- the "empty word" is also valid; it's equal, for example, to bβ.
Then the right way to create a dictionary is to list only the minimal-length spellings of the words. In particular, there's the empty word, and infinitely many words whose minimal-length spellings begin with a, infinitely many that begin with α, infinitely many that begin with b, and infinitely many that begin with β. If I call those five sets E (for empty), A, Д, B, and Б, and if I call the completely language L, then we have as sets L = E ∪ A ∪ Д ∪ B ∪ Б.
Now look at the set A, say, of words that start with a in their minimal-length spellings. Then the second letter in the minimal-length spelling is definitely not α (as then the word would start aα... and you could cancel those two), but it can be any of the other three letters (or nothing at all). So consider the set αA, which is all words in A, but add an α at the beginning; the new α can cancel with the initial a, and you get minimal-length spellings of words that do not start with α, but can start with anything else: αA = E ∪ A ∪ B ∪ Б. In particular, although before we had written L as a union of four infinite sets (and one tiny set), now we have L = Д ∪ αA, the union of just two sets.
Now here's the deal. Take the hollow sphere, which is a two-dimensional surface in three dimensions, and I'll assume that it's centered at the origin in xyz space. For each word in L, I will define a way to rotate the sphere. Namely, read along the word from left to right. If you see an a, rotate the sphere 1 radian (180/π degrees) clockwise around the xДxis. If you see an α, rotate 1 radian counterclockwise around the x axis. The letters b and β are the clockwise and counterclockwise rotations by 1 radian around the y axis. Then it's clear that even if you spell a word without using its minimal-length spelling, then you get the same total rotation as if you used the minimal-length spelling (because there would just be some rotations that cancel). What's not clear, and what I'll just ask you to believe me about (although it's not too unreasonable) is that if two words are honestly different, then they lead to honestly different rotations. (Essentially, this is because π is sufficiently irrational.)
Let S denote the sphere, and s∈S some point on the sphere. Applying the rotations to the sphere, each word w∈L determines some point w(s)∈S. Now, there are a lot more points in S than there are words in L. But what I want you to do is this: pick some subset T of S so that for each s∈S, there is precisely one point t∈T and precisely one word w∈L such that s=w(t). I.e.: T is a minimal set so that to get anywhere in S, you can start somewhere in T and apply rotations that arise from words in L.
The point is that I'm not going to tell you how to find the set T. The "axiom of choice" says that such a set exists, but doesn't give a construction for it. In fact, for reasonable use of the word, it is impossible to "construct" the set T. But it exists. If you could follow infinitely-long algorithms, it would be easy to construct T, just by starting with S and removing points until you get something minimal.
(continued because it's too long)
Re: Sometimes explaining a joke makes it funnier?
2010-10-12 06:14 am (local)
(continuation)
Anyway, so we have L(T) = S, but L = E ∪ A ∪ Д ∪ B ∪ Б, so S = E(T) ∪ A(T) ∪ Д(T) ∪ B(T) ∪ Б(T). By A(T), for example, I mean "all points that you can get to by starting somewhere in T and applying the rotation corresponding to some word in A". (Note that the set E consists just of the empty word, which corresponds to no rotation, so E(T) = T.)
Ok, so I've said how to cut up S into five pieces: the pieces are E(T)=T, A(T), Д(T), B(T), and Б(T). Take the three pieces T, B(T), and Б(T) and move them out of the way, so that you're just left with A(T) and Д(T). Leave the piece Д(T) exactly where it is, but rotate the piece A(T) by the rotation corresponding to α so that you get αA(T) - this is just a rigid rotation in three-space. But L = Д ∪ αA, so Д(T) ∪ αA(T) = L(T) = S. So with just those two pieces, you can fit them back together to build a whole sphere.
Now do something similar with the B(T) and Б(T) pieces. You get a second complete sphere.
Oh, you also have that bit T=E(T) left over. If you work a bit harder, you can squeeze it into the two spheres; alternately, you can throw it out with the pumpkin seeds.
I hope that sufficiently explained the joke (and that I haven't used too much math notation),
Theo
Mathematics, UC Berkeley
And then, a couple more from Fandom Wank which I'm not pasting because those don't disappear. They explain better the thing about the ball 'pieces'.