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 $\\alpha$ .