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I'm looking at lemma IV.$3.10$ from Kunen's Set Theory ($M$ denotes a model of $\sf{ZFC}$:

If $\mathbb{P}\in M$ and $\tau,\mu\in M^\mathbb{P}$ (the set of names in $M$), then there is a nice name $\vartheta\in M^\mathbb{P}$ for a subset of $\tau$ such that $\mathbb{1}\Vdash (\mu\subset\tau\rightarrow\mu=\vartheta)$.

The proof starts by saying:

Work entirely in $M$, let $\vartheta=\bigcup\{\{\sigma\}\times A_\sigma:\sigma\in\text{dom}(\tau)\}$, where each $A_\sigma$ is chosen so that:

  1. $A_\sigma$ is an antichain in $\mathbb{P}$.
  2. For all $p\in A_\sigma$, $p\Vdash \sigma\in\mu$.
  3. $A_\sigma$ is maximal with respect to (1)(2).

This can be done in $M$ by using $\sf{AC}$ in $M$ and the Definability Lemma.

I'm lost as to how exactly he's forming the $A_\sigma$. I can see that, by the Definability Lemma, we can form the set $\{p\in M:p\Vdash \sigma\in\mu\}$, and this set must belong to $M$, but how to we get an antichain?

I'm guessing condition $(3)$ follows by an application of Zorn's lemma (which is valid since $\sf{AC}$ holds in $M$), although I haven't sat down to check the details since condition $(1)$ seems much more mysterious. Is this the correct path, assuming we can form an $A_\sigma$ satisfying $(1)$ and $(2)$?

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Let $B_{\sigma} = \{p \in M : p \Vdash \sigma \in \mu\}$. Select $x_0 \in B_{\sigma}$. For each $\alpha > 0$, let $x_{\alpha}$ be any element of $B_{\sigma}$ that is incomparable with every $x_{\beta}$ for $\beta < \alpha$, if it exists. At some point, we cannot choose another $x_{\alpha}$; for this $\alpha$, the class $\{x_i \mid i < \alpha\}$ is a maximal antichain. By the Axiom of Choice, this class is a set (we need Choice to pick each new $x_i$, of course).

Alternatively: antichains exist because a single-element set is an antichain - pick any single element from $B_{\sigma}$. As you noted, Zorn's Lemma implies the existence of maximal antichains, provided antichains exist at all - so starting from a single-element antichain, a maximal antichain exists.

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  • $\begingroup$ Ah, got it know, thank you very much! $\endgroup$ – Reveillark Feb 11 '17 at 3:52

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