Is $2^{\aleph_0}$ a weakly Mahlo cardinal? Can it be? That is, are there conditions (such as the negation of the continuum hypothesis or something) under which it is, and other conditions under which it isn't? When it is not weakly Mahlo, is there a clear example of a club set which it does not intersect? Thanks.


1 Answer 1


First of all, the existence of a weakly Mahlo cardinal is equiconsistent with the existence of a strongly Mahlo cardinal, both of which are far beyond the reaches of what $\sf ZFC$ can prove.

So since we can't prove that there exists a weakly Mahlo cardinal to begin with, we can't prove that $2^{\aleph_0}$ is weakly Mahlo.

But even if there is a weakly Mahlo cardinal, it is still consistent that $2^{\aleph_0}=\aleph_1$ or $\aleph_2$ or some other value far below that cardinal. So even if there exists a weakly Mahlo cardinal, we still can't necessarily prove that the continuum is weakly Mahlo.

But it is consistent that if there is a weakly Mahlo cardinal, then it is the continuum. This can be done by starting with a model that has a strongly Mahlo cardinal $\kappa$, and adding $\kappa$ Cohen reals.

So we cannot disprove that $2^{\aleph_0}$ is weakly Mahlo either.

This makes the rest of the question quite difficult to answer. We don't know the exact value of the continuum, and we know even less on the cardinal structure below the continuum, so it's hard to come up with a particular example.

We can say the following, weakly Mahlo cardinals are weakly inaccessible (regular limit cardinals). So if $2^{\aleph_0}$ is a successor cardinal, or a singular cardinal, then it is not weakly Mahlo.


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