Collapsing a cardinal to $\omega$: $P$ is the set of all finite sequences of ordinals less than a given cardinal $\lambda$. If $\lambda$ is uncountable then forcing with this poset collapses $\lambda$ to $\omega$.

Collapsing a cardinal to another: $P$ is the set of all functions from a subset of $\kappa$ of cardinality less than $\kappa$ to $\lambda$ (for fixed cardinals $\kappa$ and $\lambda$). Forcing with this poset collapses $\lambda$ down to $\kappa$.

Levy collapsing: If $\kappa$ is regular and $\lambda$ is inaccessible, then $P$ is the set of functions $p$ on subsets of $\lambda \times \kappa$ with domain of size less than $\kappa$ and $p(α,ξ)<α$ for every $(α,ξ)$ in the domain of $p$. This poset collapses all cardinals less than $\lambda$ onto $\kappa$, but keeps $\lambda$ as the successor to $\kappa$.

What would be the proof of these? I learned how to use countable chain condition to prevent cardinal collapse, but I am not sure why this kind of cardinal collapse would hold.

  • 1
    $\begingroup$ The trick as always: genericity arguments. $\endgroup$
    – Asaf Karagila
    Mar 8, 2013 at 0:13

1 Answer 1


In these two cases showing that certain cardinals are collapsed is easier than showing that certain other cardinals aren't collapsed:

If $G \subset \text{Col}(\kappa,\lambda)$ is a $V$-generic filter then its union $f = \bigcup G$ is a function whose domain is a subset of $\kappa$ and whose range is a subset of $\lambda$. For every ordinal $\beta < \lambda$ the set $D_\beta \subset \text{Col}(\kappa,\lambda)$ of conditions whose range contains $\beta$ (and which therefore force the range of $f$ to contain $\beta$) is dense. The filter $G$ is $V$-generic and these dense sets $D_\beta$ are all in $V$, so $G$ intersects each of these dense sets, and therefore $f$ is a surjection from a subset of $\kappa$ onto $\lambda$. (In fact a similar argument shows that the domain of $f$ is $\kappa$, but we do not need this to see that $\lambda$ is collapsed.)

The argument that forcing with $\text{Col}(\kappa,\mathord{<}\lambda)$ collapses everything less than $\lambda$ is very similar.

  • $\begingroup$ You need to add to that last part that the poset has $\lambda$-c.c. so it doesn't collapse $\lambda$ itself. $\endgroup$
    – Asaf Karagila
    Mar 8, 2013 at 1:21
  • $\begingroup$ @Asaf Except that (rather surprisingly) the OP said he or she understands that part but not this part. $\endgroup$ Mar 8, 2013 at 1:23
  • $\begingroup$ Read again, the OP wrote only for ccc. Of course this has the same properties, but one has to argue as for why this poset is $\lambda$-c.c. as well. $\endgroup$
    – Asaf Karagila
    Mar 8, 2013 at 1:26
  • $\begingroup$ @Asaf Oh, you're right. I'm too lazy to prove the chain condition for the Levy collapse right now though, so you should feel free to write a competing answer. $\endgroup$ Mar 8, 2013 at 1:29
  • 1
    $\begingroup$ I wrote an answer and got to the same conclusion. I was hoping that you won't be as lazy! :-) $\endgroup$
    – Asaf Karagila
    Mar 8, 2013 at 1:32

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