# Proof of levy forcing and cardinal collapse

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.

• The trick as always: genericity arguments. Commented Mar 8, 2013 at 0:13

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.
• You need to add to that last part that the poset has $\lambda$-c.c. so it doesn't collapse $\lambda$ itself. Commented Mar 8, 2013 at 1:21
• 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. Commented Mar 8, 2013 at 1:26