Cantor's diagonal argument

Consequences

Ordering of cardinals

With equality defined as the existence of a bijection between their underlying sets, Cantor also defines binary predicate of cardinalities ${\displaystyle |S|}$ and ${\displaystyle |T|}$ in terms of the existence of injections between ${\displaystyle S}$ and ${\displaystyle T}$. It has the properties of a preorder and is here written "${\displaystyle \le }$". One can embed the naturals into the binary sequences, thus proving various injection existence statements explicitly, so that in this sense ${\displaystyle |\mathbb{N}|\le |2^{\mathbb{N}}|}$, where ${\displaystyle 2^{\mathbb{N}}}$ denotes the function space ${\displaystyle \mathbb{N}\to \{0,1\}}$. But following from the argument in the previous sections, there is no surjection and so also no bijection, i.e. the set is uncountable. For this one may write ${\displaystyle |\mathbb{N}|<|2^{\mathbb{N}}|}$, where "${\displaystyle <}$" is understood to mean the existence of an injection together with the proven absence of a bijection (as opposed to alternatives such as the negation of Cantor's preorder, or a definition in terms of assigned ordinals). Also ${\displaystyle |S|<|\mathcal{P}(S)|}$ in this sense, as has been shown, and at the same time it is the case that ${\displaystyle \mathrm{\neg }(|\mathcal{P}(S)|\le |S|)}$, for all sets ${\displaystyle S}$.

Assuming the law of excluded middle, characteristic functions surject onto powersets, and then ${\displaystyle |2^S|=|\mathcal{P}(S)|}$. So the uncountable ${\displaystyle 2^{\mathbb{N}}}$ is also not enumerable and it can also be mapped onto ${\displaystyle \mathbb{N}}$. Classically, the Schröder–Bernstein theorem is valid and says that any two sets which are in the injective image of one another are in bijection as well. Here, every unbounded subset of ${\displaystyle \mathbb{N}}$ is then in bijection with ${\displaystyle \mathbb{N}}$ itself, and every subcountable set (a property in terms of surjections) is then already countable, i.e. in the surjective image of ${\displaystyle \mathbb{N}}$. In this context the possibilities are then exhausted, making "${\displaystyle \le }$" a non-strict partial order, or even a total order when assuming choice. The diagonal argument thus establishes that, although both sets under consideration are infinite, there are actually more infinite sequences of ones and zeros than there are natural numbers. Cantor's result then also implies that the notion of the set of all sets is inconsistent: If ${\displaystyle S}$ were the set of all sets, then ${\displaystyle \mathcal{P}(S)}$ would at the same time be bigger than ${\displaystyle S}$ and a subset of ${\displaystyle S}$.

In the absence of excluded middle

Also in constructive mathematics, there is no surjection from the full domain ${\displaystyle \mathbb{N}}$ onto the space of functions ${\displaystyle {\mathbb{N}}^{\mathbb{N}}}$ or onto the collection of subsets ${\displaystyle \mathcal{P}(\mathbb{N})}$, which is to say these two collections are uncountable. Again using "${\displaystyle <}$" for proven injection existence in conjunction with bijection absence, one has ${\displaystyle \mathbb{N}<2^{\mathbb{N}}}$ and ${\displaystyle S<\mathcal{P}(S)}$. Further, ${\displaystyle \mathrm{\neg }(\mathcal{P}(S)\le S)}$, as previously noted. Likewise, ${\displaystyle 2^{\mathbb{N}}\le {\mathbb{N}}^{\mathbb{N}}}$, ${\displaystyle 2^S\le \mathcal{P}(S)}$ and of course ${\displaystyle S\le S}$, also in constructive set theory.

It is however harder or impossible to order ordinals and also cardinals, constructively. For example, the Schröder–Bernstein theorem requires the law of excluded middle. In fact, the standard ordering on the reals, extending the ordering of the rational numbers, is not necessarily decidable either. Neither are most properties of interesting classes of functions decidable, by Rice's theorem, i.e. the set of counting numbers for the subcountable sets may not be recursive and can thus fail to be countable. The elaborate collection of subsets of a set is constructively not exchangeable with the collection of its characteristic functions. In an otherwise constructive context (in which the law of excluded middle is not taken as axiom), it is consistent to adopt non-classical axioms that contradict consequences of the law of excluded middle. Uncountable sets such as ${\displaystyle 2^{\mathbb{N}}}$ or ${\displaystyle {\mathbb{N}}^{\mathbb{N}}}$ may be asserted to be subcountable. This is a notion of size that is redundant in the classical context, but otherwise need not imply countability. The existence of injections from the uncountable ${\displaystyle 2^{\mathbb{N}}}$ or ${\displaystyle {\mathbb{N}}^{\mathbb{N}}}$ into ${\displaystyle \mathbb{N}}$ is here possible as well. So the cardinal relation fails to be antisymmetric. Consequently, also in the presence of function space sets that are even classically uncountable, intuitionists do not accept this relation to constitute a hierarchy of transfinite sizes. When the axiom of powerset is not adopted, in a constructive framework even the subcountability of all sets is then consistent. That all said, in common set theories, the non-existence of a set of all sets also already follows from Predicative Separation.

In a set theory, theories of mathematics are modeled. Weaker logical axioms mean fewer constraints and so allow for a richer class of models. A set may be identified as a model of the field of real numbers when it fulfills some axioms of real numbers or a constructive rephrasing thereof. Various models have been studied, such as the Cauchy reals or the Dedekind reals, among others. The former relate to quotients of sequences while the later are well-behaved cuts taken from a powerset, if they exist. In the presence of excluded middle, those are all isomorphic and uncountable. Otherwise, variants of the Dedekind reals can be countable or inject into the naturals, but not jointly. When assuming countable choice, constructive Cauchy reals even without an explicit modulus of convergence are then Cauchy-complete and Dedekind reals simplify so as to become isomorphic to them. Indeed, here choice also aids diagonal constructions and when assuming it, Cauchy-complete models of the reals are uncountable.

Diagonalization in broader context

Russell's paradox has shown that set theory that includes an unrestricted comprehension scheme is contradictory. Note that there is a similarity between the construction of T and the set in Russell's paradox. Therefore, depending on how we modify the axiom scheme of comprehension in order to avoid Russell's paradox, arguments such as the non-existence of a set of all sets may or may not remain valid.

Analogues of the diagonal argument are widely used in mathematics to prove the existence or nonexistence of certain objects. For example, the conventional proof of the unsolvability of the halting problem is essentially a diagonal argument. Also, diagonalization was originally used to show the existence of arbitrarily hard complexity classes and played a key role in early attempts to prove P does not equal NP.

Version for Quine's New Foundations

The above proof fails for W. V. Quine's "New Foundations" set theory (NF). In NF, the naive axiom scheme of comprehension is modified to avoid the paradoxes by introducing a kind of "local" type theory. In this axiom scheme,

{ s ∈ S: s ∉ f(s) }

is not a set — i.e., does not satisfy the axiom scheme. On the other hand, a modified diagonal argument would begin by noticing that

{ s ∈ S: s ∉ f({s}) }

is a set in NF. In which case, if P₁(S) is the set of one-element subsets of S and f is a proposed bijection from P₁(S) to P(S), one is able to use proof by contradiction to prove that |P₁(S)| < |P(S)|.

The proof follows by the fact that if f were indeed a map onto P(S), there would be an r in S, such that f({r}) coincides with the modified diagonal set, above. This would show that if r is not in f({r}), then r is in f({r}) and vice versa.

It is not possible to put P₁(S) in a one-to-one relation with S, as the two have different types, and so any function so defined would violate the typing rules for the comprehension scheme.