The question is in the title. For Easton's theorem see Wikipedia. Loosely speaking we can use forcing to manipulate the powerset function on regular cardinals as much as we like given we satisfy the following restrictions: (a) $\kappa<\lambda$ implies $2^\kappa\le 2^\lambda$ and (b) $\kappa<cf(2^\kappa)$.

In addition, Easton's forcing preserves cardinals.

Question: Does Easton forcing preserve measurable cardinals? In particular, if $\mu$ is the first measurable, does it remain the first measurable in the extension (measurable here means "there is an $\aleph_1$-complete non principal ultrafilter on $\mu$). If the answer is "no" does the restriction of Easton's theorem below the first measurable $\mu$ preserve the measurability of $\mu$?

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    Presumably you don't intend to let Easton blow up the continuum (or power sets of other cardinals below $\mu$) past $\mu$, because that would certainly destroy its measurability. – Andreas Blass Jun 14 at 0:09
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    A slightly less trivial comment: GCH cannot first fail at a measurable cardinal. So if $\mu$ is the first cardinal whose power set's cardinality you increase, you'll destroy its measurability. – Andreas Blass Jun 14 at 0:11
  • @AndreasBlass. Yes, I don't want to blow any cardinals below $\mu$ to larger than $\mu$. The second comment is indeed worth knowing. – Ioannis Souldatos Jun 14 at 7:21
up vote 9 down vote accepted

There are a variety of positive and negative results on this topic, depending on the Easton function and the set-theoretic background.

The Kunen-Paris theorem, for example, provides a large number of positive instances.

Theorem. If $\kappa$ is a measurable cardinal and $2^\kappa=\kappa^+$, then for any Easton function $E:A\to\kappa$ defined on a set $A\subset\kappa$ of regular cardinals below $\kappa$, with $A\notin\mu$ for some normal measure $\mu$ on $\kappa$, then in the Easton forcing extension $V[G]$, the cardinal $\kappa$ remains measurable.

Proof. Let $\newcommand\P{\mathbb{P}}\P$ be the Easton forcing for $E$, and suppose that $G\subset\P$ is $V$-generic. Let $j:V\to M$ be the ultrapower by $\mu$. If you consider $j(\P)$, you can factor it at $\kappa$ and see that $j(\P)=\P\times\P^*$, where $\P^*$ is the forcing beyond stage $\kappa$. There is no forcing at coordinate $\kappa$ precisely because $\kappa\notin j(A)$. It follows that $\P^*$ is $\leq\kappa$-closed in $M$. Furthermore, the size of $\P^*$ is $|j(\kappa)|^M$, and so the number of subsets of it in $M$ is $|j(2^\kappa)|^M=|j(\kappa^+)|^M$, and this is $\kappa^+$ in $V$, since $(\kappa^+)^\kappa=\kappa^+$. So in $V$ we may line up the dense subsets of $\P^*$ in $M$ into a $\kappa^+$ sequence. Since $M^\kappa\subset M$, we may therefore perform a diagonalization to build in $V$ an $M$-generic filter $G^*\subset\P^*$, simply by meeting the dense sets one-by-one. Since $G$ is $V$-generic, it is also $M[G^*]$-generic, and so $G\times G^*$ is $M$-generic for $j(\P)$. This filter fulfills the lifting criterion, and so we may lift the embedding to $j:V[G]\to M[j(G)]$, with $j(G)=G\times G^*$. Thus, $\kappa$ remains measurable in $V[G]$, as desired. $\Box$

Note that the assumption $2^\kappa=\kappa^+$ can be easily forced without adding subsets to $\kappa$ and thefore while preserving the measurability of $\kappa$. So the argument is applicable to any measurable cardinal, whether or not the GCH holds at $\kappa$, provided one also forces $2^\kappa=\kappa^+$.

Since the construction of $G^*$ has wide flexibility, there are many such $G^*$ and consequently it follows that in $V[G]$ there are $2^{2^\kappa}$ many normal measures on $\kappa$. This is how Kunen and Paris proved the consistency of having many normal measures on a measurable cardinal.

There are many similar further such lifting arguments in many of my papers, but let me point you to:

Using results in that paper, one can show that $\kappa$ is not necessarily the least measurable cardinal in $V[G]$, since it could be that the measurability of some smaller cardinal is turned on by the forcing.

The general case of $A\in\mu$ is not always possible, since if you control some pattern of the continuum function on a set $A$ of measure one, this will have consequences for the continuum function at $\kappa$ itself in $M$. And so the general rule is that the function $E$ must exhibit certain kinds of self-reference in order to be able to preserve the large cardinal property at $\kappa$.

Another limitation or negative result is that the consistency strength of $\kappa$ is measurable plus $2^\kappa>\kappa^+$ is strictly stronger than a measurable cardinal. One needs $\kappa$ to be $\kappa^{++}$-tall. (See my paper, Hamkins, Joel D., Tall cardinals, Doi:10.1002/malq.200710084, Math. Log. Q. 55, No. 1, 68-86 (2009). ZBL1165.03044, blog post.) So it will be necessary to start with a larger large cardinal situation if this is desired.

Meanwhile, one can achieve a large number of positive instances when one begins with a supercompact cardinal. In many of these cases it is often easier to use the Easton-support iteration rather than the Easton product forcing.

There is a whole literature on combining Easton's theorem with large cardinals, and perhaps people can post further references. Let me mention a few papers of Brent Cody, who first looked at this topic in his disseration (under my supervision).

  • Every time I ask here on MO, a new window opens and the view is always a lot nicer and a lot wider than I thought. Thank you for your answer. – Ioannis Souldatos Jun 14 at 21:19
  • Your welcome! It was a pleasure, and this is a very nice topic, which I find to be very rich with mathematical ideas. – Joel David Hamkins Jun 14 at 21:28

Suppose $GCH$ holds, $U$ is a normal measure on $\kappa$ and $j: V \to M$ is an ultrapower embedding.

Let $F$ be an Easton function on regular cardinals and $P_F$ be the corresponding Easton forcing. Also let $G_F$ be $P_F$-generic filter.

First note that we can assume that $dom(F)$ contains regular cardinals $\leq \kappa$ (as the rest is $\kappa^+$-closed and does not have any effects on measurability of $\kappa$).

If $\kappa \notin dom(j(F)),$ then $\kappa$ remains measurable in $V[G_F].$ This is due to Kunen-Paris: Boolean extensions and measurable cardinals, Annals of Mathematical Logic 2 Issue 4 (1971) pp 359-377.

Also if we force with $Add(\lambda, 1)$ for all regular $\lambda < \kappa$, then $\kappa$ fails to be measurable in $V[G_F].$ See pages 375-376 of the above paper of Kunen-Paris (the argument works if we also force at all regular $\lambda \leq \kappa$).

  • Than you for the reference to Kunen-Paris. You answer the question too. I marked Joel's answer, because it came first and it is very detailed. – Ioannis Souldatos Jun 14 at 21:23
  • Actually, Ioannis, Mohammad's answer came slightly before mine. I was finishing my writing when he posted his answer. – Joel David Hamkins Jun 14 at 21:27
  • @JoelDavidHamkins Strange! When I click the "oldest" tab, your answer appears first. – Ioannis Souldatos Jun 15 at 7:38

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