https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
He there's a set T, this set is made of all infinite sequences of binary digits, like the first element could be s1 = (0, 0, ...), the second could be s2 = (1, 1, ...), and so on. This set's assumed to be countable because you can enumerate all the elements in it, that is, write each element of the set
Then he uses a diagonal resulting from the stacking of the elements to form another infinite binary sequence, and then says this diagonal sequence doesnt exist in T, so T isnt countable
To see if this was true, I started with a set T where each sequence has a length 1. There are two possible elements. No matter how you stack these elements the diagonals are already included, or enumerated, in set T, since the diagonal needs to have the same size as the other sequences, in this case, size 1 (so you cant just put the two sequences in diagonal and say "(0,1) is a possible element that isnt contained in T, therefore T is uncountable", as that would be against what Cantor defined)
I tried a set T where each sequence has a length of 2, same deal
0, 0
0, 1
1, 0
1, 1
I tried T size 3
0, 0, 0
1, 1, 1
1, 0, 0
0, 1, 1
0, 1, 0
1, 0, 1
0, 0, 1
1, 1, 0
All of the elements in T are enumerated already, trying to form a new element of size 3 (as per the rules of Cantor) looking at diagonals will give you nothing new
The reason is extremely simple, given a sequence of binary digits of a length n and the set T that contains the elements enumerated by said sequences, being n = 3 or n = "approaching infinity", the number of elements will always be countable and equal to 2^n, this number will include all possible enumerations of elements in T, so the diagonals will already be enumerated in the stacked horizontal sequences. To prove Im wrong is very simple too, just show me at what length n this affirmation does not hold true, Im calling that it holds true even "approaching infinity" as its said in calculus
He there's a set T, this set is made of all infinite sequences of binary digits, like the first element could be s1 = (0, 0, ...), the second could be s2 = (1, 1, ...), and so on. This set's assumed to be countable because you can enumerate all the elements in it, that is, write each element of the set
Then he uses a diagonal resulting from the stacking of the elements to form another infinite binary sequence, and then says this diagonal sequence doesnt exist in T, so T isnt countable
To see if this was true, I started with a set T where each sequence has a length 1. There are two possible elements. No matter how you stack these elements the diagonals are already included, or enumerated, in set T, since the diagonal needs to have the same size as the other sequences, in this case, size 1 (so you cant just put the two sequences in diagonal and say "(0,1) is a possible element that isnt contained in T, therefore T is uncountable", as that would be against what Cantor defined)
I tried a set T where each sequence has a length of 2, same deal
0, 0
0, 1
1, 0
1, 1
I tried T size 3
0, 0, 0
1, 1, 1
1, 0, 0
0, 1, 1
0, 1, 0
1, 0, 1
0, 0, 1
1, 1, 0
All of the elements in T are enumerated already, trying to form a new element of size 3 (as per the rules of Cantor) looking at diagonals will give you nothing new
The reason is extremely simple, given a sequence of binary digits of a length n and the set T that contains the elements enumerated by said sequences, being n = 3 or n = "approaching infinity", the number of elements will always be countable and equal to 2^n, this number will include all possible enumerations of elements in T, so the diagonals will already be enumerated in the stacked horizontal sequences. To prove Im wrong is very simple too, just show me at what length n this affirmation does not hold true, Im calling that it holds true even "approaching infinity" as its said in calculus
