In the end of the last section, we failed at finding a bijection from the numbers between 0 and 1 (including 0 but not 1) and the naturals, but we knew from an earlier example that sometimes these bijections take a lot of cleverness to come up with, and so maybe we just weren't clever enough to find the right bijection. But what if there weren't one out there to be found? How could we be sure of this? And what would that statement, that there is no bijection between the naturals and the numbers between 0 and 1, mean?
Let's address the second question, what the fact that there is no bijection between the naturals and the numbers between 0 and 1 would mean, first, and then try to show that it is true that there is no bijection. So, assume that there isn't a bijection, no matter how clever we are, we can't go out and find one. At the very beginning of this set of lectures, we said that two sets have the same cardinality exactly when there is a bijection between them, so this must mean that these two sets have different cardinality. Also, as we saw in our first attempt at a bijection, we can always find a buddy for every natural in the numbers between 0 and 1, but we can't find every number between 0 and 1 a buddy in the naturals. This should suggest to us that there are somehow more numbers between 0 and 1 than there are naturals, even though both of these sets are infinite. In other words, if we try to count up the number of things in them we will never finish. In fact, these two sizes of infinity come up often enough that they have names. Sets in bijection with the naturals are called countable and sets that are infinite but not in bijection with the naturals are called uncountable. Sets that are actually in bijection with the numbers between 0 and 1 are said to have cardinality of the continuum
Now let's try to make sure that the statement "There is no bijection between the naturals and the numbers between 0 and 1" is true. We are going to do this by assuming that there is a bijection from the naturals to the numbers between 0 and 1 and then showing that something must be wrong, so our original assumption must have been false.
So let's write down the list of buddy pairs, writing each number between 0 and 1 as a (possibly infinite, non-repeating) decimal. We are going to write the kth digit in the decimal expansion (k places away from the decimal point, starting with 0 as the tenths place) of the number whose buddy is n as a_(n,k), so if 0.14159265357... is buddied with 62, a_(62,8) = 3.
Now that list will look like:
0 - 0.a_(0,0)a_(0,1)a_(0,2)a_(0,3)a_(0,4)... 1 - 0.a_(1,0)a_(1,1)a_(1,2)a_(1,3)a_(1,4)... 2 - 0.a(2,0)a_(2,1)a_2(2,2)a_(2,3)a_(2,4)... 3 - 0.a_(3,0)a_(3,1)a_(3,2)a(_3,3)a_(3,4)... 4 - 0.a_(4,0)a_(4,1)a_(4,2)a_(4,3)a_(4,4)... 5 - 0.a_(5,0)a_(5,1)a_(5,2)a_(5,3)a_(5,4)... . . .
Let's think about the number 0.a_(0,0)a_(1,1)a_(2,2)a_(3,3)a_(4,4)... . Now let us change this number a little bit, by adding one to every digit in the number, unless the digit is a nine, in which case we will replace it with a zero (this is called arithmetic mod 10, and is another topic in this lecture series). Let's call the new digit we created from a_(k,k) by changing every digit b_k. The number 0.b_0b_1b_2b_3b_4... is now a number between 0 and 1, and so must have a buddy somewhere in the naturals, lets call that buddy n. Then, we can look on our list from before and find n's buddy in the numbers between 0 and 1. Of course, not knowing what n is, we can only write down that this buddy is 0.a_(n,o)a_(n,1)a_(n,2)...a_(n,n-1)a_(n,n)a_(n,n+1).... But wait, something is amiss here! Wasn't this buddy supposed to be 0.b_0b_1b_2...b_n-1b_nb_n+1...? And, specifically, we chose that number so that the digit b_n wasn't a_(n,n). Since this was true for every n, there can't possibly be an n that is buddies with out new number 0,b_1b_2b_3.... This tells us that the bijection was bad, but since we didn't assume anything about the bijection, other than the fact that it was one, it must not have been one.
The argument used in the above proof is due to Georg Cantor, and is commonly called Cantor's diagonalization argument.