General sets Cantor's diagonal argument

illustration of generalized diagonal argument: set t = {n∈ℕ: n∉f(n)} @ bottom cannot occur anywhere in range of f:ℕ→p(ℕ). example mapping f happens correspond example enumeration s in above picture.

a generalized form of diagonal argument used cantor prove cantor s theorem: every set s, power set of s—that is, set of subsets of s (here written p(s))—has larger cardinality s itself. proof proceeds follows:

let f function s p(s). suffices prove f cannot surjective. means member t of p(s), i.e. subset of s, not in image of f. candidate consider set:

t = { s ∈ s: s ∉ f(s) }.

for every s in s, either s in t or not. if s in t, definition of t, s not in f(s), t not equal f(s). on other hand, if s not in t, definition of t, s in f(s), again t not equal f(s); cf. picture. more complete account of proof, see cantor s theorem.

consequences

this result implies notion of set of sets inconsistent notion. if s set of sets p(s) @ same time bigger s , subset of s.

russell s paradox has shown naive set theory, based on unrestricted comprehension scheme, contradictory. note there similarity between construction of t , set in russell s paradox. therefore, depending on how modify axiom scheme of comprehension in order avoid russell s paradox, arguments such non-existence of set of sets may or may not remain valid.

the diagonal argument shows set of real numbers bigger set of natural numbers (and therefore, integers , rationals well). therefore, can ask if there set cardinality between of integers , of reals. question leads famous continuum hypothesis. similarly, question of whether there exists set cardinality between |s| , |p(s)| infinite s leads generalized continuum hypothesis.

analogues of diagonal argument used in mathematics prove existence or nonexistence of objects. example, conventional proof of unsolvability of halting problem diagonal argument. also, diagonalization used show existence of arbitrarily hard complexity classes , played key role in attempts prove p not equal np.

version quine s new foundations

the above proof fails w. v. quine s new foundations set theory (nf). in nf, naive axiom scheme of comprehension modified avoid paradoxes introducing kind of local type theory. in axiom scheme,

{ s ∈ s: s ∉ f(s) }

is not set — i.e., not satisfy axiom scheme. on other hand, might try create modified diagonal argument noticing that

{ s ∈ s: s ∉ f({s}) }

is set in nf. in case, if p1(s) set of one-element subsets of s , f proposed bijection p1(s) p(s), 1 able use proof contradiction prove |p1(s)| < |p(s)|.

the proof follows fact if f indeed map onto p(s), find r in s, such f({r}) coincides modified diagonal set, above. conclude if r not in f({r}), r in f({r}) , vice versa.

it not possible put p1(s) in one-to-one relation s, 2 have different types, , function defined violate typing rules comprehension scheme.