# Cantor's diagonal argument  Main Article Discussion Related Articles  [?] Bibliography  [?] External Links  [?] Citable Version  [?] This editable Main Article is under development and subject to a disclaimer. [edit intro]

Cantor's diagonal argument provides a convenient proof that the set $2^{\mathbb {N} }$ 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.

## 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, $s_{i}\in \{0,1\}$ for all i. This is done as follows. Take $s_{1}$ 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 $s_{1}$ is defined to be 1. Take $s_{2}$ to be different from the second digit of the second member on the list (in our example $s_{2}=0$ ). 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 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.

## Formal argument

To prove that the family of all subsets of $\mathbb {N}$ is not countable, we associate to any set $S\subset \mathbb {N}$ its characteristic function $\phi :\mathbb {N} \rightarrow \{0,1\}$ by setting $\phi (n)=1$ if $n\in S$ and $\phi (n)=0$ , 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 $F:\mathbb {N} \rightarrow 2^{\mathbb {N} }$ , that allows us to assign an index $i=F^{-1}(S)$ to every subset S. In other words, all the functions $\phi :\mathbb {N} \rightarrow \{0,1\}$ can be enumerated as $\{\phi _{1},\phi _{2},\phi _{3},\ldots \}$ . Assuming this has been done, we proceed to construct a function $\psi :\mathbb {N} \rightarrow \{0,1\}$ that is not in this list. Consequently, the corresponding set, $T$ cannot be in the range of $F$ .

For each $i$ , either $\phi _{i}(i)=0$ or $\phi _{i}(i)=1$ , and so we define $\psi (i)=1-\phi _{i}(i)$ . Clearly, $\psi (i)\in \{0,1\}$ and $\psi (i)\not =\phi _{i}(i)$ .

It follows that $\psi \not =\phi _{i}$ for any $i$ , and it must therefore correspond to a set not in the range of $F$ . This contradiction shows that $2^{\mathbb {N} }$ cannot be countable.

## 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

$C=\{x\in X:x\not \in f(x)\}.\,$ 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

$c\in C\Leftrightarrow c\in \{x\in X:x\not \in f(x)\}\Leftrightarrow c\not \in f(c)\Leftrightarrow c\not \in C,\,$ which is a contradiction. Hence the supposition that f is a surjection is untenable. In particular, then, f cannot be a bijection.