Is there a forcing notion that collapses $\aleph_{\omega_1}$ to $\aleph_\omega$ while preserving every cardinal below $\aleph_\omega$?


1 Answer 1


The existence of such a forcing is independent of ZFC.

First, let us see that that it is possible, we will need some large cardinals to make it happen. The example will be an instance of forcing with the "stationary tower forcing". Suppose $\delta$ is a Woodin cardinal and $G$ is generic for the (full) stationary tower forcing $\mathbb{P}_{<\delta}$ at $\delta$. Then in $V[G]$ there is an elementary embedding $$j\colon V\rightarrow M$$ with critical point $<\delta$, $M$ is transitive and $V[G]\models M^{<\delta}\subseteq M$. In fact, the critical point can be any regular uncountable cardinal of $V$ strictly below $\delta$, in particular it can be $\aleph_{\omega+1}$. In this case, $\aleph_{\omega+1}$ has to be collapsed as $\aleph_{\omega}^M=\aleph_{\omega}^V$ and $M\models(\aleph_{\omega+1})^V<\aleph_{\omega+1}=j((\aleph_{\omega+1})^V)$. Also all cardinals below $\aleph_{\omega}$ are preserved as $\mathcal P(\aleph_\omega^V)^{V[G]}\subseteq M$ (by the closure of $M$ in $V[G]$) and $M$ and $V$ agree about which ordinals below $\mathrm{crit}(j)$ are cardinals by elementarity.

Before using the stationary tower, we may first collapse $\aleph_{\omega_1}$ to $\aleph_{\omega+1}$ by forcing with $\mathrm{Col}(\aleph_{\omega+1},\aleph_{\omega_1})$. This preserves all cardinals $\leq\aleph_{\omega+1}$ and also preserves $\delta$ as a Woodin cardinal as the forcing is small. Now we can follow this up by forcing with the stationary tower at $\delta$ as described above.

(In fact, the first collapse forcing is not necessary as we can arrange with the stationary tower that $\mathrm{crit}(j)=(\aleph_{\omega+1})^V$ and $j((\aleph_{\omega+1})^V)>(\aleph_{\omega_1})^V$.)

All of the facts about the stationary tower are due to Woodin and can be found in Paul Larson's very nice book "The stationary tower".

On the flip side, a Woodin cardinal is in some sense neccessary for this. If there is such a forcing you are asking about then there is an inner model of ZFC with a Woodin cardinal. This is a consequence of the "Covering lemma below a Woodin cardinal". Don't be fooled by the term "lemma", this is actually a very deep theorem. It roughly states the following:

If there is no inner model with a Woodin cardinal, then "the core model" $K$ exists and it has the following properties:

  • $K$ computes successors of singular cardinals correctly, that is if $\kappa$ is a singular cardinal in $V$ then $(\kappa^+)^V=(\kappa^+)^K$ (but $\kappa$ is not neccessarily singular in $K$).
  • $K$ is immune to forcing, that is if $V[G]$ is a forcing extension of $V$ then $K^V=K^{V[G]}$.

In this form, this result follows from the papers

Mitchell, W. J.; Schimmerling, E., Weak covering without countable closure, Math. Res. Lett. 2, No. 5, 595-609 (1995). ZBL0847.03024.


Jensen, Ronald; Steel, John, (K) without the measurable, J. Symb. Log. 78, No. 3, 708-734 (2013). ZBL1348.03049.

Now suppose toward a contradiction that $\mathbb P$ is a forcing which collapses $\aleph_{\omega_1}$, or even just $\aleph_{\omega+1}$, and preserves all cardinals below $\aleph_\omega$, but there is no inner model with a Woodin cardinal. Let $V[G]$ be an extension of $V$ by $\mathbb P$. Then there is no inner model of $V[G]$ with a Woodin cardinal (this is not obvious!). Now applying the covering lemma above in $V$ and in $V[G]$, we find that $(\kappa^+)^V=(\kappa^+)^{K^V}$ and $(\kappa^+)^{V[G]}=(\kappa^+)^{K^{V[G]}}$ by the first bulletpoint where $\kappa=(\aleph_\omega)^V$. But by the second bulletpoint, $K^V=K^{V[G]}$, contradiction.

It follows for example that there is no such forcing in the constructible universe $L$: The only inner model of $L$ is $L$ itself and there are no measurable cardinals in $L$, much less Woodin cardinals. For $L$ this can also be seen from Jensen's covering lemma instead of using the covering lemma below a Woodin cardinal.

  • $\begingroup$ Wonderful answer. It seems to me that the same argument gives that the large cardinal strength of "Collapsing $\aleph_{\omega+1}$ to $\aleph_\omega$ while preserving all cardinals below $\aleph_\omega$" is also that of Woodin cardinal. Am I correct? $\endgroup$
    – Hanul Jeon
    Oct 10, 2023 at 22:59
  • 1
    $\begingroup$ Yes, thats exactly rigth! Additionally, one can replace $\alpeh_\omega$ by any other singular cardinal here. $\endgroup$ Oct 11, 2023 at 8:36

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