# Cantor's diagonal argument

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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 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 sn 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.

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