Apologies if the question is somewhat naïve - I'm not a set theorist, just an enthusiast.

One possible intuition we could have about the generalized continuum hypothesis is that power sets are so difficult to well-order that they must be very large on the ordinal scale (so, at the very least, they should be weakly Mahlo). Questions about this intuition have appeared on mathoverflow before, e.g. here, here, and here.

For the continuum, we have the nice result that if the cardinality of the continuum is real-valued measurable, then it is automatically greatly Mahlo, and there is the pleasing result due to Prikry that in this case for all $\aleph_0 \le \kappa < 2^{\aleph_0}$, we have $2^{\kappa} = 2^{\aleph_0}$ (I found these in sections 4 and 5 of Fremlin's Real-valued-measurable cardinals). There is the additional advantage that real-valued measurability is a concept which is easy to explain to a non-specialist, such as myself.

However, these results about real-valued measurability are very specific to the continuum: no larger powerset can be real-valued measurable because measurable cardinals are necessarily strongly inaccessible.

I want to know if there is some analogue of this for larger powersets, such as $\kappa = 2^{2^{\aleph_0}}$. A first guess might be to ask for something like a $\kappa$-additive "surreal-valued" measure, but I don't know of any way to make this idea coherent.

In an attempt to be more specific, what I wish for is some property $X$ such that:

  • The assumption "Every infinite powerset has property $X$" isn't obviously inconsistent with ZFC,
  • There is some convincing analogy between property $X$ and real-valued measurability,
  • Any cardinal $\kappa$ with property $X$ is automatically weakly Mahlo, and
  • If $2^\kappa$ has property $X$ and $\kappa \le \lambda < 2^\kappa$, then $2^\lambda = 2^\kappa$ (or some variation of this).

Edit: After sleeping on this, I noticed that the first bullet point may be too strong, since it contradicts the Singular Cardinal Hypothesis - perhaps it should be weakened to "For every regular $\kappa$, $2^\kappa$ has property $X$". Also, the second bullet point could be made slightly less vague, by asking for the existence of a cardinal with property $X$ to have equivalent consistency strength to the existence of a measurable cardinal.

| cite | improve this question | | | | |
  • $\begingroup$ Do you want something directly analogous to a measure? Some kind of field-valued additive function on sets? $\endgroup$ – Monroe Eskew 2 days ago
  • $\begingroup$ That would be nice, but I'm willing to settle for a less direct analogy: perhaps an abstraction of the ideal of null sets of such a measure, if the existence of such a thing is sufficiently strong. $\endgroup$ – zeb 2 days ago

Your Answer

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

Browse other questions tagged or ask your own question.