116c- Lecture 20

June 6, 2008

We showed that \mathsf{AD} implies a weak version of choice, \mathsf{AC}_\omega({}^\omega\omega), namely, every countable family of non-empty sets of reals admits a choice function. This implies that \omega_1 is regular and suffices to develop classical analysis in a straightforward fashion (in particular, to construct Lebesgue measure and to prove its basic properties).

Coupled with the fact that all sets of reals have the perfect set property, this implies that \omega_1>\omega_1^{L[x]} for any real x and therefore \omega_1^V is strongly inaccessible in L[x] for any real x.

We closed the course by showing that, in fact, \omega_1 is a measurable cardinal. We proved this result of Solovay by showing Martin’s result that the “cone measure” is indeed a non-atomic measure on the structure {\mathcal D} of the Turing degrees and then “pulling back” this measure to \omega_1.

Finally, given any measurable cardinal \kappa, let \mu be a (\kappa-complete, non-principal) measure on \kappa. Then L[\mu] is a model of choice in which \kappa is measurable. In particular, {\rm Con}({\sf ZF}+{\sf AD})\Rightarrow{\rm Con}({\sf ZFC}+\mbox{There is a measurable cardinal}).

Since, under choice, any measurable cardinal is strongly inaccessible and the limit of strongly inaccessible cardinals, this shows that {\sf AD} has significant consistency strength.


116c- Lecture 19

June 4, 2008

As discussed during Lecture 13, for the theories one encounters when studying set theory, no absolute consistency results are possible, and we rather look for relative consistency statements.  For example, the theories A=\mathsf{ZFC}+“There is a weakly inaccessible cardinal” and B=\mathsf{ZFC}+“There is a strongly inaccessible cardinal” are equiconsistent. This means that a weak theory (much less than \mathsf{PA} suffices) can prove \mathrm{Con}(A)\Leftrightarrow\mathrm{Con}(B). Namely: A is a subtheory of B, so its inconsistency implies the inconsistency of B. Assume B is inconsistent and fix a (say, Hilbert-style) proof \phi_0,\dots,\phi_n of an inconsistency from B. Then a proof \psi_0,\dots,\psi_m of an inconsistency from A can be found by showing (by induction on i\le n) that each \phi_i^L is a theorem of A, and this argument can be carried out in a theory (such as \mathsf{PA}) where the syntactic manipulations of formulas that this involves are possible.

It is a remarkable empirical fact that the combinatorial statements studied by set theorists can be measured against a linear scale of consistency, calibrated by the so-called large cardinal axioms, of which strongly inaccessible cardinals are perhaps the first natural example. Hypotheses as unrelated as the saturation of the nonstationary ideal or determinacy have been shown equiconsistent with extensions of \mathsf{ZFC} by large cardinals. One direction (that models with large cardinals generate models of the hypothesis under study) typically involves the method of forcing and will not be further discussed here. The other direction, just as in the very simple example of weak vs strong inaccessibility, typically requires showing that certain transitive classes (such as L) must have large cardinals of the desired sort. We will illustrate these ideas by obtaining large cardinals from determinacy in the last lecture of the course.

We defined the axiom of determinacy \mathsf{AD}. It contradicts choice but it relativizes to the model L(\mathbb{R}). This is actually the natural model to study \mathsf{AD} and, in fact, from large cardinals one can prove that L(\mathbb{R})\models\mathsf{AD}.

We illustrated basic consequences of \mathsf{AD} for the theory of the reals by showing that it implies that every set of reals has the perfect set property (and therefore a version of \mathsf{CH} is true under \mathsf{AD}). Similar arguments give that \mathsf{AD} implies that all sets of reals have the Baire property and are Lebesgue measurable. In the last lecture of the course we will use the perfect set property of sets of reals to show that the consistency of \mathsf{AD} implies the consistency of strongly inaccessible cardinals. (Though this is beyond the scope of this course, by using more sophisticated ideas, one can prove the optimal stronger result that the consistency of \mathsf{AD} implies the consistency of \mathsf{ZFC}+ “there exist infinitely many Woodin cardinals”.)

Relativizations of P vs NP

June 1, 2008

In Relativizations of the {\mathcal P}=?{\mathcal N}{\mathcal P} question by Theodore Baker, John Gill and Robert Solovay, SIAM J. Comput. 4 (1975), no. 4, 431-442 (available through JSTOR) it is shown that the question of whether P equals NP cannot be solved with the kind of arguments typical in computability theory, since these arguments relativize to Turing machines with oracles. Among the results shown there, two oracles f and g are found such that {\sf P}^f={\sf NP}^f and {\sf P}^g\ne{\sf NP}^g. I discussed these results during 117c, the course on decidability in computability theory.

There has been some recent attempts (not very successful, and not too serious in my opinion) to show that the P vs NP question is independent of {\sf PA} or even stronger systems. Apparently, part of the motivation for trying to show independence comes from the results in the Baker-Gill-Solovay paper.

I reproduce below a posting by Timothy Chow to the Foundations of Mathematics list where this motivation is shown lacking.

[FOM] Amusing observation about independence of complexity results
Thursday, May 22, 2008 8:33 PM
Timothy Y. Chow

Here is an amusing observation regarding the idea that the existence of contradictory relativizations of assertions such as {\sf P} = {\sf NP} is evidence that said assertions are independent of some strong theory.  I doubt this observation is new, but I haven’t seen it explicitly before.

We can write down (thanks to Levin, I think) an explicit machine M with the property that, if {\sf P} = {\sf NP}, then M solves {\sf SAT} in polynomial time. (Essentially M multitasks over all polytime algorithms until it jackpots.)

Suppose now that a statement such as

\varphi:=M correctly solves {\sf SAT} in at most n^{100} + 100 steps on length-n inputs”

is independent of {\sf PA} (for example).  The statement \varphi is stronger than the statement that {\sf P} = {\sf NP}, but we might imagine that if {\sf P} = {\sf NP} is independent of {\sf PA} then something like \varphi will also be independent of {\sf PA}.

Now \varphi is \Pi^0_1, so if it is independent of {\sf PA} then it is true, and therefore {\sf P} = {\sf NP}.  It follows that {\sf NP}\ne{\sf EXP}, by the time hierarchy theorem for example.

This line of reasoning can itself be formalized, and this shows that in some system call it {\sf S}! that is slightly stronger than {\sf PA}, we can prove “if \varphi is independent of {\sf PA}, then {\sf NP}\ne{\sf EXP}.” This in turn means that if we can prove “\varphi is independent of {\sf PA}” in {\sf S}, then we certainly can’t prove “{\sf NP} = {\sf EXP} is independent of {\sf S}.”

Informally speaking, the upshot is that since we know that {\sf P}\ne{\sf EXP}, it is probably too much to expect that both{\sf P} = {\sf NP}” and “{\sf NP} = {\sf EXP}” are provably independent of strong systems.  On the other hand, both “{\sf P} = {\sf NP}” and “{\sf NP} = {\sf EXP}” admit contradictory relativizations.  So it seems we should be wary of drawing too tight a connection between contradictory relativizations and logical independence.