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Let $\mathbb{P}$ be poset.

Let $B$ be a set. We say that a $\mathbb{P}$-name $\dot{b}$ is a nice name for member of $B$ if there is a maximal antichain $A\subseteq\mathbb{P}$ and a function $h:A\rightarrow{B}$ such that $ p\Vdash\dot{b}=h(p)$ for all $p\in{A}$. Here, we abuse of the notation and say $\dot{b}=\left<{A,h}\right>$ .

Let $D, B$ be sets. We say that a $\mathbb{P}$-name $\dot{z}$ is a nice name for a function from $D$ into $B$ if there is $\left<{A_{d},h_{d}}\right>_{d \in D}$ such that $\dot{z}(d)=\left<{A_{d},h_{d}}\right>$(as a nice name) for all $d\in D$, that is, $A_{d}$ is a maximal antichain and $h_{d}:A_{d}\to B$ such that $ p\Vdash\dot{z}(d)=h_{d}(p)$ for all $p \in A_{d}$. Here, we abuse of the notation and say $\dot{z}(d)=\left<{A_{d},h_{d}}\right>_{d}$ .

Let $p \in \mathbb{P}$ and $\tau$ be a $\mathbb{P}$-name such that $p \Vdash \tau \in \check{B}$. Then there exist a nice name $\dot{b}$ for an object in $B$ such that $p \Vdash \dot{b} = \tau.$

Also if $\sigma$ be a $\mathbb{P}$-name such that $p \Vdash \sigma:\check{D}\to \check{B}$ Then there exist a nice name $\dot{z}$ for a function from $D$ into $B$ such that $p \Vdash \dot{z} = \sigma.$

I am studying the book Kunen and I'm a little confused when defining a $\mathbb{P}$-name.

Is not a question at research but can give me a suggestion how to solve this. It is a simple fact for you ,I have not been able to resolve.

A suggestion of how to define $\dot{b}$ and $\dot{z}$. Thanks

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  • $\begingroup$ I don't think your definition of "nice name" is correct. It is satisfied by any name $\dot b$ such that $1\Vdash\dot b\in \check B$. $\endgroup$ Feb 24, 2015 at 19:27
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    $\begingroup$ This was asked many times on Math.SE. Did you check there? $\endgroup$
    – Asaf Karagila
    Feb 24, 2015 at 23:23
  • $\begingroup$ Crossposted at Math.SE: math.stackexchange.com/questions/1163238/… $\endgroup$ Feb 25, 2015 at 1:05

1 Answer 1

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I'll outline the argument for $\dot b$; the case of $\dot z$ uses the same idea. Given a name $\tau$ such that $1\Vdash\tau\in\check B$, let $D$ be the set of conditions that decide a value for $\tau$; that is, $q\in D$ iff there is some element $x\in B$ such that $q\Vdash\tau=\check x$. Verify that $D$ is dense in $\mathbb P$. Let $A$ be an antichain that is maximal among antichains $\subseteq D$, and verify that $A$ is also a maximal antichain in $\mathbb P$ (because $D$ is dense). Define $h:A\to B$ by sending any $q\in A$ to the $x\in B$ such that $q\Vdash\tau=\check x$. Then $A$ and $h$ constitute a nice name $\dot b$. Each $q\in A$ forces both $\tau$ and $\dot b$ to equal $h(q)$, and so each $q\in A$ forces $\tau=\dot b$. Because $A$ is a maximal antichain, no condition can force $\tau\neq\dot b$ (because such a condition would be incompatible with everything in $A$), and therefore $1\Vdash\tau=\dot b$.

If you only have some condition $p$ (rather than 1) forcing $\tau\in\check B$, then you'll only get a nice name $(A,H)$ in the weaker sense that $A$ is maximal among antichains of extensions of $p$.

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