Cantor's diagonal argument

Cantor's diagonal argument provides a convenient proof that the set of subsets of the natural numbers (also known as its power set) is not countable. More generally, it is a recurring theme in computability theory, where perhaps its most well known application is the negative solution to the halting problem.

[edit] Informal description

The original Cantor's idea was to show that the family of 0-1 infinite sequences is not countable. This is done by contradiction. If this family is countable then its members can be enumerated or enlisted. Such a list gives a table of digits, like in the following arbitrarily chosen example:

0, 1, 0, 1, 0, ...

1, 1, 1, 1, 0, ...

1, 0, 1, 0, 0, ...

Now, we construct a sequence s = (s₁, s₂, s₃, ....), which is not on the list while still, for all i. This is done as follows. Take s₁ to be different from the first digit of the first member on the list. In our example the digit is 0 (in boldface) and so s₁ is defined to be 1. Take s₂ to be different from the second digit of the second member on the list (in our example s₂ = 0). Generally, define s_n as different from the n-th digit of the n-th entry on the list. In other words, the sequence s = (s₁, s₂, s₃, ....) contains "the complement in {0,1}" of the diagonal of our table. It follows that that the sequence s itself is not on the list, since it is different from every member by the definition. The list was supposed to contain all the 0-1 sequences. The contradiction shows that such sequences can not be enumerated (or they are not countable).

The role of the diagonal clearly explains the name of the argument.

[edit] Formal argument

To prove that the family of all subsets of is not countable, we associate to any set its characteristic function by setting if and , otherwise. Conversely, every such function defines a subset. Observe also that every such function corresponds to a 0-1 sequence and vice-versa.

If power set is countable, there is a bijective map , that allows us to assign an index i = F ^{− 1}(S) to every subset S. In other words, all the functions can be enumerated as . Assuming this has been done, we proceed to construct a function that is not in this list. Consequently, the corresponding set, T cannot be in the range of F.

For each i, either or , and so we define . Clearly, and .

It follows that for any i, and it must therefore correspond to a set not in the range of F. This contradiction shows that cannot be countable.

[edit] Application to general sets

More generally, the argument shows that a set cannot be put into one-to-one correspondence with its power set: equivalently, the cardinality of a set is strictly less than that of its power set. The argument proceeds by showing that there cannot be a surjection from a set X to the set PX of subsets of X. Suppose if possible that there were such a surjection f. Form the set

Since C is a subset of X then from the assumption that f is surjective there is an element c of X such that f(c) = C. We now ask whether c is in C. We find that

which is a contradiction. Hence the supposition that f is a surjection is untenable. In particular, then, f cannot be a bijection.

Some content on this page may previously have appeared on Citizendium.