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# [Math Evenings] Banach-Tarski Revisited 3: Theorem on Free Group

Now that we have proved a rudimentary version of the Banach-Tarski paradox, we will now prove one of the backbones of the actual proof of the Banach-Tarski paradox.

Theorem 2. (eq 8)

For a group F(x,y) formed by generators x and y, with x' and y' being the inverse elements of each,

holds true.

(the U and + intersected together represents disjoint unions)

Pf.

The Gothic letter F_n will be used to denote a free group formed by generators {x,y} that has length n. Also, x' and y' will denote the inverse elements of each. For instance, eq 1.

F(x,y) denotes the entire set that is formed by generators x and y. Then,

holds true. In this case, the cardinality of F(x,y) would equal the sum of 4^n when n is 0 to infinity, hence F(x,y) is a countably infinite set. Then, we define m(x) to be a set that has elements starting with x. Naturally, eq 3 would be true.

Then consider the set x'm(x) as in eq 4.

All elements in such a set would have to begin with x, y, y', or be the identity, as in eq 6.1. (In fact, it should be 5.1. My bad.) We can rigorously prove this by proving 6.2 and 6.3.

We can apply the same proof to y'm(y), which gives us eq 7.

Finally, this would give us eq 8.

Q.E.D.

Remember the Hyperwebster I talked about in my previous article in https://www.seanyoonbio.com/post/math-evenings-the-banach-tarski-paradox?

This is in fact the Hyberwebster more rigorously explained. So, say that there is a dictionary that contains all the possible combinations of the alphabet. There are 26 volumes, each only containing words that start with only one alphabet. But note here that the entire dictionary could be reduced into only one volume, since taking out the first letter "A" from volume 1 would give us the entire 26 volumes.

This is what essentially this theorem is trying to prove.

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