Cantor's diagonal argument: Difference between revisions

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==Consequences==

===Ordering of cardinals===

With equality defined as the existence of a bijection between their underlying sets, Cantor also defines a [[preorder]] of cardinalities <math>|S|</math> and <math>|T|</math> in terms of the [[Cardinality#Comparing_sets|existence of injections]] between <math>S</math> and <math>T</math>, here written "<math>\le</math>". One can embed the naturals into the binary sequences, thus proving various ''injection existence'' statements explicitly, so that in this sense <math>|{\mathbb N}|\le|2^{\mathbb N}|</math>, where <math>2^{\mathbb N}</math> denotes the function space <math>{\mathbb N}\to\{0,1\}</math>. But following from the argument in the previous sections, there is ''no surjection'' and so also no bijection, and in this sense <math>|{\mathbb N}|<|2^{\mathbb N}|</math>, i.e. the set is uncountable. Also <math>|S|<|{\mathcal P}(S)|</math>, as has been shown, and at the same time it is the case that <math>\neg(|{\mathcal P}(S)|\le|S|)</math>, for all sets <math>S</math>.

Assuming the [[law of excluded middle]], [[characteristic functions]] surject onto powersets, and impliesthen <math>|2^S|=|{\mathcal P}(S)|</math>,. and soSo <math>2^{\mathbb N}</math> is also not enumerable. Hereand it can also be mapped onto <math>{\mathbb N}</math>. [[classical mathematics|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 <math>{\mathbb N}</math> is then in bijection with <math>{\mathbb N}</math> itself, and every [[subcountable]] set (a property in terms of surjections) is then already countable, i.e. in the surjective image of <math>{\mathbb N}</math>. In this context the possibilities are then exhausted, making "<math>\le</math>" a [[partial order|non-strict partial order]], or even a [[total order]] when assuming [[axiom of choice|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 <math>S</math> were the set of all sets, then <math>{\mathcal P}(S)</math> would at the same time be bigger than <math>S</math> and a subset of <math>S</math>.

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| title = One hundred years of Russell's paradox

| volume = 6

| year = 2004}}</ref><ref>Rathjen, M. "[http://www1.maths.leeds.ac.uk/~rathjen/acend.pdf Choice principles in constructive and classical set theories]", Proceedings of the Logic Colloquium, 2002</ref> The elaborate collection of subsets of a set is constructively not exchangeable with the collection of its [[characteristic function]]sfunctions and also the existence of injections from the uncountable <math>2^{\mathbb N}</math> or <math>{\mathbb N}^{\mathbb N}</math> into <math>{\mathbb N}</math> is here possible.<ref>Bauer, A. "[http://math.andrej.com/wp-content/uploads/2011/06/injection.pdf An injection from N^N to N]", 2011</ref> So the cardinal relation fails to be [[Antisymmetric relation|antisymmetric]]. Consequently, also in the presence of function space sets that are even classically uncountable, [[intuitionist]]s do not accept this relation to constitute a hierarchy of transfinite sizes.<ref>{{cite book |title=Mathematics and Logic in History and in Contemporary Thought |author=Ettore Carruccio |publisher=Transaction Publishers |year=2006 |page=354 |isbn=978-0-202-30850-0}}</ref>

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 [[Axiom schema of predicative separation|Predicative Separation]].