# proof of partial order with chain condition does not collapse cardinals

Outline:

Given any function $f$ purporting to violate the theorem by being surjective  (or cofinal) on $\lambda$, we show that there are fewer than $\kappa$ possible values of $f(\alpha)$, and therefore only $\max(\alpha,\kappa)$ possible elements in the entire range of $f$, so $f$ is not surjective (or cofinal).

Details:

Suppose $\lambda>\kappa$ is a cardinal of $\mathfrak{M}$ that is not a cardinal in $\mathfrak{M}[G]$.

There is some function $f\in\mathfrak{M}[G]$ and some cardinal $\alpha<\lambda$ such that $f:\alpha\rightarrow\lambda$ is surjective. This has a name, $\hat{f}$. For each $\beta<\alpha$, consider

 $F_{\beta}=\{\gamma<\lambda\mid p\Vdash\hat{f}(\beta)=\gamma\}\text{ for some }% p\in P$

$|F_{\beta}|<\kappa$, since any two $p\in P$ which force different values for $\hat{f}(\beta)$ are incompatible and $P$ has no sets of incompatible elements of size $\kappa$.

Notice that $F_{\beta}$ is definable in $\mathfrak{M}$. Then the range of $f$ must be contained in $F=\bigcup_{i<\alpha}F_{i}$. But $|F|\leq\alpha\cdot\kappa=\max(\alpha,\kappa)<\lambda$. So $f$ cannot possibly be surjective, and therefore $\lambda$ is not collapsed.

Now suppose that for some $\alpha\geq\lambda>\kappa$, $\operatorname{cf}(\alpha)=\lambda$ in $\mathfrak{M}$ and for some $\eta<\lambda$ there is a cofinal function $f:\eta\rightarrow\alpha$.

We can construct $F_{\beta}$ as above, and again the range of $f$ is contained in $F=\bigcup_{i<\eta}F_{i}$. But then $|\operatorname{range}(f)|\leq|F|\leq\eta\cdot\kappa<\lambda$. So there is some $\gamma<\alpha$ such that $f(\beta)<\gamma$ for any $\beta<\eta$, and therefore $f$ is not cofinal in $\alpha$.

Title proof of partial order with chain condition does not collapse cardinals ProofOfPartialOrderWithChainConditionDoesNotCollapseCardinals 2013-03-22 12:53:43 2013-03-22 12:53:43 Henry (455) Henry (455) 4 Henry (455) Proof msc 03E35