I’m taking the GRE tomorrow, and had a question. Based on the answer key, this practice test states that the set of all functions from N to {0, 1} is not countable.
Can’t you map the natural numbers to these functions, as follows?
i 1 2 3 4 5 6 7 8 ... f0 = 0 0 0 0 0 0 0 0 ... f1 = 1 0 0 0 0 0 0 0 ... f2 = 0 1 0 0 0 0 0 0 ... f3 = 1 1 0 0 0 0 0 0 ... f4 = 0 0 1 0 0 0 0 0 ... That is, f4(1)=0, f4(2)=0, f4(3)=1, and f4(anything else)=0. Won’t this eventually cover all possible kinds of these functions? And we can definitely map the natural numbers to this set.
All entries in your list will contain a finite number of ones. Where in your list would the function that returns 0 for all evens but 1 for all odds appear or the function which always returns 1? A diagonalization argument can show that no other numbering scheme can work either. To do this, consider a function which returns 1-(fi(i)) at position i. Then this function differs from each entry in the list in at least one place, so it’s not in the list.