1 The Continuum Hypothesis 1398 G del showed Con ZFC Con ZFC CH 1962 Cohen showed Con ZFC Con ZFC CH Theorem Hartogs s Theorem For every set X there is a least cardinal such that there is no injection from to X We denote this by Hartogs s aleph of X X Theorem For every set X there is no injection from P X into X We denote this by 2 X Using Axiom of Choice well order P X and get ordinal 2 X We have X 2 X Notation Define 0 1 0 1 2 Clearly Continuum Hypothesis CH 1 1 Generalised Continuum Hypothesis GCH Why continuum Lemma CH if and only if X X is uncountable X means there is a bijection Proof Obvious since 1 Suppose 1 Well order the reals in order type 1 r Consider X r w 1 This is a subset of the reals of cardinality 1 hence is an uncountable subset which is not in bijection with it Reminder G del showed Con ZFC Con ZFC CH Cohen showed Con ZFC Con ZFC CH Relative consistency proofs By Completeness Theorem this means If there is M ZFC then I can construct N ZFC CH Analogy from algebra L fields 0 1 1 Axioms of fields Fields Let 2 x x x 1 1 Note 2 Idea Start with and extend to get F Fields 2 Construct by algebraic closure not in the usual sense here we just mean adding in 2 and then adding everything else that this requires us to add Obtain 2 Fields 2 This is easy because everything that matters Fields and 2 is determined by equations all formulas we need to check are atomic Definition 1 1 absolute If M N and M N are L structures and an L formula then we say is absolute between M and N if for all x 1 x m M M x 1 x n N x 1 x n If the direction holds then we say upwards absolute and if the direction holds then we say downwards absolute Theorem Substructure Lemma All atomic formulas are absolute between substructures WHat if we have models of ZFC Have L No function symbols nor constant symbols So almost nothing is atomic M N if and only if M is an L substructure of N And the formulas that we care about are definitely not atomic but instead very complex Try to imagine a proof of If M ZFC then there is N M such that N ZFC CH Without loss of generality M CH else we are done For simplicity let s consider the case 1 2 What can we do to get rid of 1 Maybe a surjection f 1 Maybe we can form M f M to get a smallest model M f ZFC Clearly in M f 1 M is not a cardinal anymore Does that show CH All sorts of things can happen Assuming it is actually possible to form this smallest model M f there are lots of ways that this might not end up being enough to deduce CH For example Maybe M f M Maybe 2 M is not a cardinal either A fundamental problem of non absoluteness Consider x z z x which means x is empty Consider M ZFC Therefore there are e e such that M e and M z z e z e Consider N M e N is an L substructure of M But N e even though M e So is not absolute between substructures Instead of substructures we will restrict out attention to transitive substructures in particular to M transitive x x M x M or equivalently x M y x y M such that M ZFC Next time theorems about absoluteness for transitive substructures Definition Absolute formula We say is absolute for M if for all x 1 x n M we have M x 1 x n x 1 x n is true Clearly if is absolute for M and absolute for N then it s absolute between M and N Cohen s proof becomes If M is a countable transitive set such that M ZFC then there is a a countable transitive set N M such that N ZFC CH