Here's a disproof of cantor's diagonal argument.
let be an arbitrarily long sequence of binary numbers. Then is the set of all possible sequences with that length. It has a cardinality of 2^length. For example, sequences two digits long have a cardinality of 4.
Here is the proof by mathematical induction.
There exists a bijection between the first two natural numbers and the set of sequences with length 1. and .
Assume there is a set of sequences that has a bijection to a subset of the natural numbers. Then the sequences are to . Then the set of sequences with length= has a bijection to a subset of the natural numbers. They are labeled to , indicating a bijection to the subset of the first natural numbers.
By mathematical induction, there exists a bijection between sequences of any length and a subset of the natural numbers. By analytical continuation, that, includes those of infinite length.
let be an arbitrarily long sequence of binary numbers. Then is the set of all possible sequences with that length. It has a cardinality of 2^length. For example, sequences two digits long have a cardinality of 4.
Here is the proof by mathematical induction.
There exists a bijection between the first two natural numbers and the set of sequences with length 1. and .
Assume there is a set of sequences that has a bijection to a subset of the natural numbers. Then the sequences are to . Then the set of sequences with length= has a bijection to a subset of the natural numbers. They are labeled to , indicating a bijection to the subset of the first natural numbers.
By mathematical induction, there exists a bijection between sequences of any length and a subset of the natural numbers. By analytical continuation, that, includes those of infinite length.
