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If $G$ is a subset of some forcing poset $P$ and for $x \in V$ (where I think $V$ is some model of ZFC but I'm not clear what it means) the canonical $\mathbb{P}$-name is defined as $$ \underset{\dot{\hphantom{x}}}{x}:=\{\langle\underset{\dot{\hphantom{y}}}{y},0\rangle\mid y\in x\} $$

then how do I see the second equality:

$$ \underset{\dot{\hphantom{x}}}{x}[G] = \{ \underset{\widetilde{\hphantom{y}}}{y}[G] \mid \exists p \in G: \langle \underset{\widetilde{\hphantom{y}}}{y}, p \rangle \in \underset{\dot{\hphantom{x}}}{x} \} = \{ \underset{\dot{\hphantom{y}}}{y}[G] : y \in x \}$$

assuming $0 \in G$ where $0$ is the smallest element in $P$? Thanks for your help!

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Whoever voted to close... are you familiar with the terms in the question at all? – Asaf Karagila Nov 27 '11 at 22:53
Don't overthink it, it's just an immediate consequence of the definitions and the assumption $0 \in G$. For the simple matter of understanding that second equality, you don't actually have to know anything about forcing. You probably just need to think about it more (or perhaps less), but maybe it would be less confusing if we replaced $\tilde{y}$ with some other symbol, since the $\tilde{y}$ and $\dot{y}$ don't really have anything to do with each other. Moreover, $\dot{y}$ is something made from $y$, whereas $\tilde{y}$ is just supposed to be some random thing. – Amit Kumar Gupta Nov 28 '11 at 11:22
$\dot{x}[G] = \{\tau[G]\ |\ \exists p\in G:\langle\tau,p\rangle\in\dot{x}\} = \{\dot{y}[G]\ |\ y\in x\}$ – Amit Kumar Gupta Nov 28 '11 at 11:23
@AmitKumarGupta: Thanks! – Rudy the Reindeer Nov 30 '11 at 18:45
@AsafKaragila: So if I want to compute $ \underset{\dot{\hphantom{p}}}{p}[\{ p \}]$, how do I do that? – Rudy the Reindeer Nov 30 '11 at 18:48
up vote 3 down vote accepted

In forcing we start with a model of ZFC, of course we cannot prove (from ZFC) that such model exists, but it would be quite strange to assume the theory we want to work with is inconsistent.

So we have a class, $V$ and a relation $E$ defined on $V$ that we understand as $xEy$ to mean $x\in y$. Now we can use all the machinery of forcing.

First we define $\mathbb P$-names, these are defined by induction as follows:

  • $V^\mathbb P_0 = \varnothing$,
  • $V^\mathbb P_{\alpha+1} = \{\langle x,p\rangle\mid x\in V^\mathbb P_\alpha, p\in\mathbb P\}$
  • If $\beta$ is a limit ordinal then $V^\mathbb P_\beta=\bigcup_{\alpha<\beta} V^\mathbb P_\alpha$

Later when we have fixed a generic filter $G$, by identifying the $p\in G$ as "true" and $q\notin G$ as false, we can interpret the names into actual sets, and the truth values of the resulting model will be given by the set which are interpreted nicely.

We would like it if the ground model $V$ will be a subclass of $V[G]$, and if it will be canonically defined. Luckily, this is just what the canonical $\mathbb P$-names are for. We define those, again, by induction

  • $\check\varnothing=\varnothing$ (I have taken the notation of Jech here),
  • $\check x = \{\langle\check y,0\rangle\mid y\in x\}$.

This tells us that $0$ will force that $y\in x$, where as $0$ is the least element of the forcing poset, so if $0$ will force $y\in x$ every condition will have to force that as well.

Lastly, if $\sigma$ is a $\mathbb P$-name, we define the interpretation of $\sigma$ by $G$ as:

$$\sigma[G] = \{y[G] \mid \exists p\in G:\langle y,p\rangle\in\sigma\}$$

This exactly means that some $p\in G$ was forcing that $\dot y\in\dot\sigma$, or using the $\Vdash$ relation: $p\Vdash\dot y\in\dot\sigma$. If the generic filter $G$ was chosen, the result will be that indeed $y\in\sigma$.

We now want to check that the canonical names are interpreted exactly as the original elements of $V$. For this we need only to verify that indeed $G$ interprets $\check x$ as $x$ again.

$$\check x[G] = \{\check y[G]\mid \langle \check y,0\rangle\in\check x\}$$

Inductively assume that every rank below the rank of $\check x$ was interpreted correctly, then $\check x[G] = \{y[G]\mid y\in x\} = \{y\mid y\in x\}$ as wanted.

Note: Since $G$ is a generic filter, every two elements are compatible. Namely, if $p\Vdash\varphi$ and $q\in G$ then $q$ cannot force $\lnot\varphi$ (it may not decide, but it cannot force the negation). Since $0\in G$ for every $G$, we have that if $0\Vdash\varphi$ then for every $p\in G: p\Vdash\varphi$.

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Thanks for your nice answer. When you write $V$, does it stand for the von Neumann universe? I read the Wikipedia article on forcing and it says: "...forcing consists of expanding the set theoretical universe $V$ to a larger universe $V^\ast$..." so I was wondering if it's "the universe" because it's always the von Neumann universe. – Rudy the Reindeer Nov 28 '11 at 19:48
@Matt: Every universe of ZF can be constructed in the "von Neumann" method, it is in internal construction. – Asaf Karagila Nov 28 '11 at 19:54

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