# 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 $\mathrm{max}(\alpha ,\kappa )$ possible elements in the entire range of $f$, so $f$ is not surjective (or cofinal).

*Details:*

Suppose $\lambda >\kappa $ is a cardinal of $\U0001d510$ that is not a cardinal in $\U0001d510[G]$.

There is some function $f\in \U0001d510[G]$ and some cardinal $$ such that $f:\alpha \to \lambda $ is surjective. This has a name, $\widehat{f}$. For each $$, consider

$$ |

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

Notice that ${F}_{\beta}$ is definable in $\U0001d510$. Then the range of $f$ must be contained in $$. But $$. So $f$ cannot possibly be surjective, and therefore $\lambda $ is not collapsed.

Now suppose that for some $\alpha \ge \lambda >\kappa $, $\mathrm{cf}(\alpha )=\lambda $ in $\U0001d510$ and for some $$ there is a cofinal function $f:\eta \to \alpha $.

We can construct ${F}_{\beta}$ as above, and again the range of $f$ is contained in $$. But then $$. So there is some $$ such that $$ for any $$, and therefore $f$ is not cofinal in $\alpha $.

Title | proof of partial order with chain condition does not collapse cardinals |
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Canonical name | ProofOfPartialOrderWithChainConditionDoesNotCollapseCardinals |

Date of creation | 2013-03-22 12:53:43 |

Last modified on | 2013-03-22 12:53:43 |

Owner | Henry (455) |

Last modified by | Henry (455) |

Numerical id | 4 |

Author | Henry (455) |

Entry type | Proof |

Classification | msc 03E35 |