More specifically, if $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, does $κ$ necessarily have some form of $μ$-compactness? Is it related to strong compactness in any way?

The hypothesis is a weakening of supercompactness; it asserts that if there is any bijection from $X$ to $μ$, then at least one such bijection is in $M$. However, I doubt that it is equivalent to supercompactness. I believe it would have more relations to strong compactness (perhaps be a weakening).

On the other hand, it doesn't seem to immediately come out of the ultrapower embeddings on a measurable cardinal's ultrafilter. More specifically, in $L[U]$ (the inner model of a measurable) it holds that if $\kappa$ is measurable, then no ultrapower from an ultrafilter on a $\kappa$-sized set witnesses this large cardinal property for $\mu=\kappa^+$.

This is not to say that measurability isn't equivalent, merely that if it is equivalent to measurability, it likely would be found to be characterized by ultrafilters on sets of size above $\kappa$.

Say cardinality is $M$-downward absolute at $\mu$ when for any $X\in M$ with $|X|=\mu$, $|X|^M=\mu$.

Theorem: If $\kappa$ is measurable with $2^\kappa=\kappa^+$, and $j:V\prec M$ is an ultrapower embedding with critical point $\kappa$ from an ultrafilter on a set $S$ of size $\kappa$, then cardinality isn't $M$-downward absolute at $\kappa^+$.


  1. For every $f:S\rightarrow\kappa$, define $m(f)=[f]_U$, and it is clear that $j(\kappa)\subseteq m$$"$$\{[f]\;|\;f:S\rightarrow\kappa\}$ and so $|j(\kappa)|\leq\kappa^\kappa=\kappa^+$.
  2. $\mathcal{P}^M(\kappa)$=$\mathcal{P}(\kappa)$ so $2^\kappa\leq(2^\kappa)^M$.
  3. Clearly, $(2^\kappa)^M\leq 2^\kappa$, so $(2^\kappa)^M=2^\kappa=\kappa^+$.
  4. Assume $M$ preserves cardinality at $\kappa^+$. Then, $|j(\kappa)|^M=\kappa^+=(2^\kappa)^M$, meaning $|j(\kappa)|^M=(2^\kappa)^M$. This is a contradiction because $j(\kappa)$ is inaccessible (in fact measurable) in $M$.

The Main Questions:

  1. If $j:V\prec M$ has critical point $κ$ and for any $X\in M$ with $|X|=μ$, $|X|^M=μ$, is $κ$ necessarily related to some form of $μ$-compactness?
  2. Is there some way to characterize the existence of a $j:V\prec M$ with critical point $\kappa$ such that cardinality is $M$-downward absolute at $\mu$ as the existence of an ultrafilter on some set?
  3. Does this property relate in any way to the covering property characterization of strongly compact cardinals?
  • 1
    $\begingroup$ You may be interested in my paper "Cardinal preserving elementary embeddings", where I consider global versions of this question, see here. $\endgroup$ – Andrés E. Caicedo Oct 4 '18 at 17:03

Let me start by observing that this property is equivalent to a more standard property:

Claim: Let $M\subseteq V$ be a transitive model of $\mathrm{ZFC}$ and let $\mu \in M$ be a cardinal in $V$. The following are equivalent:

  • For every $X\in M$, $|X|^V = \mu \iff |X|^M = \mu$.
  • $(\mu^+)^V = (\mu^+)^M$.

Proof: Let us assume that $(\mu^+)^V = (\mu^+)^M$. Let $X \in M$. Let us pick a bijection $g\colon X \to \alpha$ in $M$, such that $\alpha$ is an ordinal in $M$. Since $|X|^V = \mu$, $|\alpha|^V = \mu$ and thus $\alpha < (\mu^+)^V$. By our assumption, $\alpha < (\mu^+)^M$ and therefore $|X|^M \leq \mu$. Since $\mu$ is a cardinal in $V$, $|X|^M = \mu$.

For the other direction, let us look at the ordinal $\mu < (\mu^+)^M \leq (\mu^+)^V$. If $\alpha = (\mu^+)^M < (\mu^+)^V$ then $|\alpha|^V=\mu$ and thus, by our assumption, $|\alpha|^M = \mu$ which is impossible. $\square$

As mentioned in the question, if $j\colon V \to M$ is the ultrapower embedding using a normal measure on $\kappa$, then this property fails for $\mu = \kappa^+$ (since $(\kappa^{++})^M < j(\kappa) < (\kappa^{++})^V$). Nevertheless, it holds for $\mu = (2^\kappa)^{+}$, since it and its successor are fixed points of $j$ (namely, $j(\mu) = \mu$ and $j(\mu^+) = \mu^+$). So, this property does not necessarily have consistency strength in the region of strong compactness.

| cite | improve this answer | |
  • $\begingroup$ Thanks! Why exactly does it imply that $o^K(κ)≥κ^{++}$? $\endgroup$ – Keith Millar Oct 4 '18 at 22:24
  • $\begingroup$ You're right, this is true if we further assume that $M$ is closed under $\kappa$-sequences, but without this assumption I'm not certain what is the exact consistency strength. $\endgroup$ – Yair Hayut Oct 5 '18 at 10:22
  • $\begingroup$ If you close $M$ under $\kappa$-sequences, and you make $\mu<j(\kappa)$ (like strong compactness) then it turns out to be equiconsistent to strongness; this is because every $\mu+2$-strong cardinal would have this property at $\mu$, and every such cardinal would be $\mu$-tall. Since tallness is equiconsistent to strongness, it's equiconsistent to both. $\endgroup$ – Keith Millar Oct 5 '18 at 15:15
  • $\begingroup$ Hmm... that's interesting. If we require $\mu<j(\kappa)$, does this become more 'large-cardinal-like?' More specifically, would it also hold for $\lambda<\mu$ if it held for $\mu$? $\endgroup$ – Keith Millar Oct 6 '18 at 6:33

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.