Axioms deciding the continuum hypothesis ˇ S´ arka Stejskalov´ a Kurt G¨ odel Research Center University of Vienna logika.ff.cuni.cz/sarka Prague May 2, 2018 ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum hypothesis (CH) The continuum hypothesis says that any subset of the real numbers is at most countable or has the same size as the set of all real numbers. Or equivalently, the size of real numbers is the least possible, in a formula: 2 ℵ 0 = ℵ 1 . ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum hypothesis (CH) The continuum hypothesis says that any subset of the real numbers is at most countable or has the same size as the set of all real numbers. Or equivalently, the size of real numbers is the least possible, in a formula: 2 ℵ 0 = ℵ 1 . Formulated by G. Cantor and popularised by D. Hilbert in 1900 (problem 1 on his list of 23 problems for the new century). ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The independence CH over ZFC Suppose ZFC is consistent, then the following hold: ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The independence CH over ZFC Suppose ZFC is consistent, then the following hold: (G¨ odel) ZFC + CH is consistent. (The least inner model L , 1930’s) ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The independence CH over ZFC Suppose ZFC is consistent, then the following hold: (G¨ odel) ZFC + CH is consistent. (The least inner model L , 1930’s) (Cohen) ZFC + ¬ CH is consistent. (Forcing, 1960’s) ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
What does it mean? The independence of CH has been interpreted in many ways (not mutually exclusive): ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
What does it mean? The independence of CH has been interpreted in many ways (not mutually exclusive): A1 It means that formalising set theory is hopeless if it cannot answer such a simple question. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
What does it mean? The independence of CH has been interpreted in many ways (not mutually exclusive): A1 It means that formalising set theory is hopeless if it cannot answer such a simple question. A2 It means that ZFC is a weak theory which should be extended. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
What does it mean? The independence of CH has been interpreted in many ways (not mutually exclusive): A1 It means that formalising set theory is hopeless if it cannot answer such a simple question. A2 It means that ZFC is a weak theory which should be extended. A3 It means that CH is a wrong question. One should look for a structure behind CH which is more robust. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
What does it mean? The independence of CH has been interpreted in many ways (not mutually exclusive): A1 It means that formalising set theory is hopeless if it cannot answer such a simple question. A2 It means that ZFC is a weak theory which should be extended. A3 It means that CH is a wrong question. One should look for a structure behind CH which is more robust. Regarding A3, note that CH just postulates the existence of a bijection between 2 ℵ 0 ( R ) and ℵ 1 ; this by itself does not seem to be a structurally deep condition. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The generalized continuum hypothesis We can ”globalize” the continuum hypothesis by postulating: (GCH) The generalised continuum hypothesis: 2 ℵ α = ℵ α +1 , for all α ∈ Ord. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The generalized continuum hypothesis We can ”globalize” the continuum hypothesis by postulating: (GCH) The generalised continuum hypothesis: 2 ℵ α = ℵ α +1 , for all α ∈ Ord. Suppose ZFC is consistent, then the following hold: ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The generalized continuum hypothesis We can ”globalize” the continuum hypothesis by postulating: (GCH) The generalised continuum hypothesis: 2 ℵ α = ℵ α +1 , for all α ∈ Ord. Suppose ZFC is consistent, then the following hold: (G¨ odel) ZFC + GCH is consistent. (The least inner model L , 1930’s) ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The generalized continuum hypothesis We can ”globalize” the continuum hypothesis by postulating: (GCH) The generalised continuum hypothesis: 2 ℵ α = ℵ α +1 , for all α ∈ Ord. Suppose ZFC is consistent, then the following hold: (G¨ odel) ZFC + GCH is consistent. (The least inner model L , 1930’s) (Cohen) ZFC + ¬ GCH is consistent. (Forcing, 1960’s) ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum function Notice that the negation of GCH is not very informative: it just claims that there exists α with 2 ℵ α > ℵ α +1 . There are certainly more interesting “negations” of the GCH such as 2 ℵ α > ℵ α +1 for all α. (We will discuss later whether this strong negation is consistent). ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum function Notice that the negation of GCH is not very informative: it just claims that there exists α with 2 ℵ α > ℵ α +1 . There are certainly more interesting “negations” of the GCH such as 2 ℵ α > ℵ α +1 for all α. (We will discuss later whether this strong negation is consistent). Even more generally, let us consider the function which maps infinite cardinals κ to 2 κ . We call this function the continuum function . ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum function Notice that the negation of GCH is not very informative: it just claims that there exists α with 2 ℵ α > ℵ α +1 . There are certainly more interesting “negations” of the GCH such as 2 ℵ α > ℵ α +1 for all α. (We will discuss later whether this strong negation is consistent). Even more generally, let us consider the function which maps infinite cardinals κ to 2 κ . We call this function the continuum function . To investigate failures of GCH, we study the patterns of the continuum function which are consistent with ZFC. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum function The following is provable about the continuum function in ZFC. 1 If κ < λ , then 2 κ ≤ 2 λ , 2 cf(2 κ ) > κ (in particular 2 κ > κ ). ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
The continuum function The following is provable about the continuum function in ZFC. 1 If κ < λ , then 2 κ ≤ 2 λ , 2 cf(2 κ ) > κ (in particular 2 κ > κ ). As it turns out there is a big difference between the continuum on regular and singular cardinals. (A fact which is certainly not obvious and came as a surprise.) ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
Regular cardinals By Easton theorem (1970), ZFC has little control over the continuum function on regular cardinals. In fact, the continuum function (on regulars) only needs to satisfy the previous two conditions. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
Regular cardinals By Easton theorem (1970), ZFC has little control over the continuum function on regular cardinals. In fact, the continuum function (on regulars) only needs to satisfy the previous two conditions. In particular it is consistent that 2 ℵ α > ℵ α +1 for all regular cardinals ℵ α . ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
Singular cardinals Singular cardinals do not allow such freedom: By the following theorem, GCH cannot fail first at strong limit singular cardinal with an uncountable cofinality. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
Singular cardinals Singular cardinals do not allow such freedom: By the following theorem, GCH cannot fail first at strong limit singular cardinal with an uncountable cofinality. (Silver 1975) If κ is a singular cardinal with an uncountable cofinality and GCH holds on a stationary set below κ then 2 κ = κ + . ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
Singular cardinals Singular cardinals do not allow such freedom: By the following theorem, GCH cannot fail first at strong limit singular cardinal with an uncountable cofinality. (Silver 1975) If κ is a singular cardinal with an uncountable cofinality and GCH holds on a stationary set below κ then 2 κ = κ + . Note that as GCH holds below κ unboundedly often, κ is strong limit. ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
Singular cardinals Singular cardinals do not allow such freedom: By the following theorem, GCH cannot fail first at strong limit singular cardinal with an uncountable cofinality. (Silver 1975) If κ is a singular cardinal with an uncountable cofinality and GCH holds on a stationary set below κ then 2 κ = κ + . Note that as GCH holds below κ unboundedly often, κ is strong limit. Note that the assumption of uncountable cofinality is necessary as it is consistent that ℵ ω , the smallest singular cardinal, is the first cardinal where GCH fails. (Magidor 1977) ˇ S. Stejskalov´ a Axioms deciding the continuum hypothesis
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