I wonder if any can sketch for me in very broad lines the proof of the fact that the existence of inaccessible cardinals implies the consistency of ZFC? I don´t know much about set theory, but I find it extremely interesting that this should be the case.

  • $\begingroup$ You can verify the axioms for the sets smaller than the given cardinal, so that gives a model of ZFC. $\endgroup$ – Berci Nov 25 '13 at 19:50
  • $\begingroup$ Equivalently, you can verify that $V_\kappa$ is a model of ZFC, where $\kappa$ is an inaccessible cardinal. $\endgroup$ – Zhen Lin Nov 25 '13 at 19:51

In $\sf ZFC$ one can define a "rank" function on sets. This means, in rough words, how many times we need to iterate the power set function before we can generate a set (taking union at limit stages).

If $\kappa$ is inaccessible then the sets whose rank is smaller than $\kappa$ form a model of $\sf ZFC$. Therefore by the completeness theorem one has the $\sf ZFC$ is consistent, if there exists an inaccessible cardinal.

I'm not getting into details of what exactly are the sets of rank less than $\kappa$, or how to show that all the axioms of $\sf ZFC$ hold in that set. But if one is familiar with these basics definitions then one can easily show that it is the case.

Finally, even if we assume that $\sf ZFC$ is consistent, and therefore has a model, it is still far from implying that there is an inaccessible cardinals. There is a long and curious hierarchy of stronger and stronger assertions regarding the consistency of $\sf ZFC$ (we can require just the consistency of $\sf ZFC$, or the consistency of the theory $\sf ZFC+\rm Con(\sf ZFC)$, and so on; we can require the there are "nice" models of $\sf ZFC$; and more and more).

  • $\begingroup$ But if k isn´t part of the model of ZFC, isn´t the model incomplete? $\endgroup$ – Adam Nov 25 '13 at 19:56
  • $\begingroup$ Why would it be incomplete? Does the fact that $\Bbb Q$ is a proper subfield of $\Bbb R$ makes it any less of a field? $\endgroup$ – Asaf Karagila Nov 25 '13 at 19:57
  • $\begingroup$ Well, ZFC are axioms from which we can derive all usual mathematics. What then is a "model" of ZFC? Is it like the "space" where all those axioms hold - ie. the whole world of mathematics? $\endgroup$ – Adam Nov 25 '13 at 20:05
  • 1
    $\begingroup$ Adam, a model of $\sf ZFC$, like any model of any first-order theory, is a set with a structure, and this structure satisfies some axioms. The point is that we can define internally to a model of $\sf ZFC$ a lot of modern mathematics, that is, we can write formulas in the language of set theory defining and describing the things we do in mathematics. Whereas in most theories we sometimes need to argue "outside" the model (i.e. talk about the subsets or substructures of the model, or superstructure of the model), in $\sf ZFC$ we don't need to... well, for developing mathematics anyway. $\endgroup$ – Asaf Karagila Nov 25 '13 at 20:11
  • $\begingroup$ I see. (some useless characters) $\endgroup$ – Adam Nov 25 '13 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.