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Will you look at my answer to the following homework and tell me where I made a mistake? Thanks for your help!

Let $\mathbb{P} = (P, \leq)$ be a forcing poset and let $p,q \in P$ denote two different conditions. We define the following $\mathbb{P}$-names:

$$\begin{align} &\underset{\widetilde{\hphantom{x}}}{x} := \{ \langle \emptyset , p \rangle\} \\ &\underset{\widetilde{\hphantom{y}}}{y} := \{ \langle \emptyset , q \rangle, \langle \underset{\widetilde{\hphantom{x}}}{x}, q \rangle\}\\ &\underset{\widetilde{\hphantom{z}}}{z} := \{ \langle \underset{\widetilde{\hphantom{y}}}{y} , p \rangle, \langle \underset{\widetilde{\hphantom{x}}}{x} , q \rangle, \langle \underset{\widetilde{\hphantom{x}}}{x} , p \rangle\}\\ &\underset{\widetilde{\hphantom{u}}}{u} := \{ \langle \underset{\widetilde{\hphantom{z}}}{z} , p\rangle, \langle \underset{\dot{\hphantom{q}}}{q} , p\rangle, \langle \underset{\dot{\hphantom{p}}}{p} , q\rangle \}\\ &\underset{\widetilde{\hphantom{v}}}{v} := \{  \langle \underset{\widetilde{\hphantom{u}}}{u} , q\rangle, \langle \underset{\dot{\hphantom{p}}}{p} , p\rangle, \langle \underset{\widetilde{\hphantom{z}}}{z} , p \rangle \} \end{align} $$

Determine $ \underset{\widetilde{\hphantom{v}}}{v}[\{ p \}], \underset{\widetilde{\hphantom{v}}}{v}[\{ q \}], \underset{\widetilde{\hphantom{v}}}{v}[\{ p, q \}]$

My answer:

$$ \underset{\widetilde{\hphantom{v}}}{v}[\{ p \}] = \{ \underset{\dot{\hphantom{p}}}{p}[\{ p \} ], \underset{\widetilde{\hphantom{z}}}{z}[\{ p \}] \} = \{ \underset{\dot{\hphantom{p}}}{p}[\{ p \} ], \{ \emptyset, \{ \emptyset \} \} \}$$

Because $$\begin{align} &\underset{\widetilde{\hphantom{x}}}{x}[\{ p \}] = \{ \emptyset \}\\ &\underset{\widetilde{\hphantom{y}}}{y}[\{ p \}] = \emptyset \\ & \underset{\widetilde{\hphantom{z}}}{z}[\{ p \}] = \{ \underset{\widetilde{\hphantom{y}}}{y}[\{ p \}], \underset{\widetilde{\hphantom{x}}}{x}[\{ p \}]\} = \{ \emptyset, \{ \emptyset \} \} \end{align}$$

The next one is

$$ \underset{\widetilde{\hphantom{v}}}{v}[\{ q \}] = \{ \underset{\widetilde{\hphantom{u}}}{u}[\{ q \}] \} = \{ \emptyset \}$$

Because

$$ \underset{\widetilde{\hphantom{u}}}{u}[\{ q \}] = \{ \underset{\dot{\hphantom{p}}}{p}[\{ q \} ] \}= \emptyset$$

And the last one:

$$ \begin{align*} &\underset{\widetilde{\hphantom{v}}}{v}[\{ p, q \}] = \\ &\{ \underset{\widetilde{\hphantom{u}}}{u}[\{ p, q \}], \underset{\dot{\hphantom{p}}}{p}[\{ p, q \}], \underset{\widetilde{\hphantom{z}}}{z}[\{ p, q \}] \} = \\ &\{  \{  \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}, \underset{\dot{\hphantom{q}}}{q}[\{ p, q \} ] , \underset{\dot{\hphantom{p}}}{p}[\{ p, q \} ] \}, \underset{\dot{\hphantom{p}}}{p}[\{ p, q \}], \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \} \} \end{align*} $$

Because

$$\begin{align} &\underset{\widetilde{\hphantom{u}}}{u}[\{ p, q \}] = \{  \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}, \underset{\dot{\hphantom{q}}}{q}[\{ p, q \} ] , \underset{\dot{\hphantom{p}}}{p}[\{ p, q \} ] \} \\ &\underset{\widetilde{\hphantom{z}}}{z}[\{ p, q \}] = \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \} \end{align}$$

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1  
This question is due to typos in the assignment. For every $x[G]$ they omitted $0 \in G$. Now that there is a corrected version of the assignment, this question becomes obsolete. –  Rudy the Reindeer Dec 1 '11 at 8:43

1 Answer 1

up vote 2 down vote accepted

(Again, I will use the Jech-ian notation of $\check x$ for $\mathbb P$-names)

Note that $\check x[\{p\}] = \{\check y[\{p\}] \mid \langle\check y,p\rangle\in\check x\}$. Now we can determine the interpretation by $\{p\}$ of each set in a recursive algorithm:

$$\begin{align} \check v[\{p\}] &= \Bigg\{p[\{p\}], \check z[\{p\}]\Bigg\}\\ &=\Bigg\{p,\bigg\{\check y[\{p\}],\check x[\{p\}]\bigg\}\Bigg\}\\ &=\Bigg\{p,\bigg\{\varnothing,\{\varnothing\}\bigg\}\Bigg\} \end{align}$$

For the $\check v[\{q\}]$ your answer is mostly correct, note that $\check u[\{q\}] = \{p\}$ though.

Your answer for the final has the same issue, but it looks okay.

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So $ \underset{\dot{\hphantom{p}}}{p}[\{ p \} ] = p$? Yesterday you said $\underset{\dot{\hphantom{p}}}{p}[\{ p \} ] $ didn't make any sense... –  Rudy the Reindeer Dec 1 '11 at 7:44
    
I can kind of believe you that $\underset{\dot{\hphantom{p}}}{p}[\{ p \} ] = p$ but $\underset{\widetilde{\hphantom{u}}}{u}[\{ q \} ] = \{ \underset{\dot{\hphantom{p}}}{p}[\{ q \} ]\} = \{ p \}$ seems weird. –  Rudy the Reindeer Dec 1 '11 at 7:58
    
@Matt: I thought about it and I understood what was the question. As for $p[\{p\}]=p$ it is exactly what I said before, you stop distincting between an element of the ground model and its canonical name. –  Asaf Karagila Dec 1 '11 at 8:02
    
Yes! And what about my second comment? –  Rudy the Reindeer Dec 1 '11 at 8:03
    
Sorry, this is all due to a typo where the omitted $0$ in all the sets $G$. They just uploaded a corrected version of the assignment. –  Rudy the Reindeer Dec 1 '11 at 8:42

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