# Relation between inacessible cardinals and CH

I know that the two statements “There exists an inaccessible cardinal” and “Continuum Hypothesis” are both independent of ZFC. Now, are those two statements independent of each other?

That older question of mine is related.

This is where we have to hedge quite a bit. It actually is unknown whether any large cardinal assumptions are relatively consistent with ZFC. Furthermore, we cannot prove that large cardinals are relatively consistent with ZFC without transcending ZFC itself; that is proving the relative consistency from a metatheory stronger than ZFC.

But let's press on and assume (as is the current fashion) that large cardinal assumptions are consistent.

Note that there is a theorem of Levy and Solovay that says that large cardinals are immune to "mild forcing extensions." In a bit more detail, if $\kappa$ is an inaccessible cardinal, and $P$ is a forcing notion of cardinality less than $\kappa$, then $\kappa$ remains inaccessible after forcing with $P$. (Note that "inaccessible" can be replaced everywhere in the above by "Mahlo", or "measurable", or "weakly compact", or "Ramsey", etc.).

From this it follows that after adding, say, $\aleph_2$ Cohen reals over a model of "ZFC + CH + $\exists$ inaccessible" we are left with a model of "ZFC + $\neg$ CH + $\exists$ inaccessible." Similarly, there is a mild forcing notion from which one obtains CH: essentially the family of all countable partial functions $\omega_1 \to \mathbb{R}$.

• That note about mild extensions seems awfully verbatim from Kanamori's book... :-) – Asaf Karagila Oct 15 '12 at 1:01
• @Asaf: umm... err... great minds think alike? – user642796 Oct 15 '12 at 5:50
• @arjafi: By way of clarification, does the fact that, assuming the consistency of $ZFC$ , no proof that the consistency of $ZFC$ implies the consistency of $ZFC$ + "there is an inaccessible cardinal" can be formulated in $ZFC$ (this a direct quote from the Wikipedia entry, "Inaccessible cardinal") imply that, given a model $M$ of $ZFC$ in which no inaccessibles exist, there can be no generic extension $M[G]$ of $M$ in which inaccessibles exist? – Thomas Benjamin Oct 23 '17 at 8:36

Of course. It is consistent that if an inaccessible exists that CH holds, and that it fails. Inaccessible cardinals are compatible with $V=L$, in which CH holds; and adding $\aleph_2$ reals violates CH but does not change the fact that $2^{\aleph_0}$ is below the inaccessible. One can check and see that other than the continuum no power sets were changed.

Of course it is consistent with CH that there are no large cardinals as well. If there is an inaccessible cardinal, there is a least one. In $V=L$ we can truncate the universe at the least inaccessible to have a model of ZFC+CH in which there is no large cardinals.

Do note, however, that the assumption of CH is equiconsistent with its negation, as Godel's and Cohen's work show; while this is not true for the existence of large cardinals. The assertion that there exists an inaccessible is strictly stronger than the assertion that there are none.