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Given $M$ a transitive class model of ZFC denote $M_\beta$ the sets in $M$ of rank less thank $\beta$.

I'm trying to verify that elements of $(V[G])_\beta$, where $G$ is a generic subset of some forcing poset $P$, "have names in" $V_\beta$ whenever $|P|<\beta$. (I quote "have names in" as I'm not sure if the author of what I'm reading means that every name is in $V_\beta$ or they have at least one name in $V_\beta$.)

I tried to state and solve a simpler problem by attempting to show that every $x\in V[G]$ has a name of cardinality $|x|$. But I'm having trouble there too as I don't really see how to define names for $x$ in the ground model, given that I can't refer to $x$ there. So for example fixing a name $\tau_i$ for every $x_i\in x$ and then creating the name $\{(\tau_i,\mathbb{1})\}$ fails as I don't see how this name could be definible in $V$.

The only other strategy for forming names I've seen is using definability of the forcing relation, but I haven't been able to apply it here.

Thank you for reading.

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    $\begingroup$ That's a nice question. I'm going to have to try and remember it for when I teach forcing someday. I hope. $\endgroup$ – Asaf Karagila Feb 27 '18 at 5:51
  • $\begingroup$ Re: First paragraph. Let $(\beta+1)^*$ be the canonical $P$-name over $V$ for $\beta+1.$ Let $p\in P\setminus G.$ Let $x=y_G\in (V[G])_{\beta},$ where $y$ is a $P$-name over $V$ Let $y'=y\cup \{( \beta+1)^*,p)\}.$....... Then we have (rank$(y'))^V\geq \beta$ but $y'_G=x.$ $\endgroup$ – DanielWainfleet Feb 27 '18 at 6:09
  • $\begingroup$ Off the top of my head I suggest transfinite induction on $ \alpha <\beta$ for $x\in (V[G])_{\beta}$ and rank $(x)=\alpha,$ employing the Truth Lemma (If $x' $is a $P$-name over $ V$ and $x'_G=x$ then some $p\in G$ forces that rank$(x')=\alpha)$ and the Definability Lemma (That is,$ \{p\in P: p$ forces rank$ (x') =\alpha\}$ is a member of $V).$ $\endgroup$ – DanielWainfleet Feb 27 '18 at 6:30
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Your question is a touch confusing; do you mean to say that every element in the ground model has a name in $V[G]$? What have you assumed so far about $V[G]$? What text or notes are you reading from?

If you’re following Kunen’s “Set Theory,” one approach is to start by defining $V[G]$ as the set $\{ \tau_G : \tau \in V^P \}$ where we assume $P \in V$ and $V^P$ denotes the set of all $P$-names which are also in $V$. Here also $\tau_G$ is the image of $\tau$ under a certain evaluation function for names. We then interpret $V[G]$ as an $FL$-structure, where $FL$ is the forcing language, by interpreting names $\tau$ to the corresponding element $\tau_G$ in $V[G]$, and interpreting all other sentences in the obvious way from there. Starting from this side of things, it’s clear that every element of $V[G]$ doesn’t “have” a name in $V$, but rather is the interpretation of a name which happens to be in $V$.

After that, it takes some time to show that $V[G]$ is the smallest transitive model of of ZFC containing both $V$ and $G$, but along the way one can show:

$$rank(\tau_G ) \leq rank(\tau), \forall \tau \in V^P$$

Which is, I think, the detail you are asking about - or at least related. Here is a proof of that fact. Let $y \downarrow$ be the set of all predecessors of $y$ under $\in $.

We proceed by transfinite induction on names, along the well-founded set-like relation $x \in trcl(y)$, which we will denote as $x < y$. Suppose we have shown the theorem for $\sigma < \tau$. We now have: $$ rank(\tau_G) = sup \{ rank(x) + 1 : x \in \tau_G \downarrow\} \\ = sup \{ rank(x) + 1 : x \in \tau_G \}$$ since $V[G]$ is transitive $$ = sup \{ rank(\sigma_G) + 1 : \sigma \in dom(\tau) \}$$ since $x \in \tau_G \rightarrow x = \sigma_G$ for some $\sigma$ in $dom(\tau)$ $$ \leq sup \{ rank(\sigma) + 1 : \sigma \in dom(\tau) \} $$ by inductive hypothesis $$ \leq sup \{ rank(\sigma , p) + 1 : (\sigma, p) \in \tau \} \\ = rank(\tau)$$

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  • $\begingroup$ I think the OP is actually trying to show the opposite: that many sets have names of the least possible rank (your argument shows that small rank names can't correspond to large rank objects). $\endgroup$ – Noah Schweber Feb 28 '18 at 22:34
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    $\begingroup$ Sets in V[G] are, as you say, evaluations of names in V. My question is: Fix $x\in V[G]$ of rank $<\beta$, can you find a name in $V$ of rank $<\beta$ such that $x$ is the evaluation of that name? $\endgroup$ – JKEG Feb 28 '18 at 22:42
  • $\begingroup$ That makes much more sense. Can’t one, given $\tau_G$, find a $\sigma$ of the same rank as $\tau_G$ such that $\sigma_G = \tau_G$ just by reworking the construction of $\tau$ step by step from the ground up, eliminating elements as you go along of the form $( \tau_i, p )$ for any $p$ not in $G$, and keeping elements of the form $(\tau_i, p)$ as you go along for which $p \in G$? Then the rank should be the same and the new name $\sigma$ should still be in $V^P$. $\endgroup$ – Rachael Alvir Feb 28 '18 at 22:52
  • $\begingroup$ @RachaelAlvir I don't see how to do that without referencing $G$, which is necessary for the name to be in $V$. $\endgroup$ – JKEG Mar 18 '18 at 19:57
  • $\begingroup$ The trick is that by working on $\tau$ instead of $\tau_G$, you can reference G. $\tau$ lives outside of your forcing extension, in the “real world” where G is accessible. I’ll try to get around to writing up a more explicit explanation of this. $\endgroup$ – Rachael Alvir Mar 21 '18 at 14:16
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This is only a partial result, but it was good enough for my purposes.

Let $G$ be a generic subset of a forcing poset $\mathbb{P}$. Then, for any ordinal $\alpha$ there exists $\beta\geq\alpha$ such that for any $x\in V[G]_\beta$ there is a name $\tau\in V_\beta$ such that $\tau_G=x.$

Define $\phi:V[G]\rightarrow\text{On}$ by $$\phi(x) = \min\{\xi\in\text{On}\mid\text{There is a name }\tau\in V_\xi\text{ such that }\tau_G=x\} $$ and define $\psi:\text{On}\rightarrow\text{On}$ by $$\psi(\xi+1)=\sup\{\phi(x)\mid x\in V[G]_{\xi+1}\}+\psi(\xi)+1 $$at successor ordinals and $\psi(\lambda)=\bigcup_{\xi<\lambda}\psi(\xi)$ for limit $\lambda.$ Being strictly increasing and continuous, $\psi$ has arbitrarily large fixed points.

Now, fix an ordinal $\alpha$ and let $\beta$ be a limit ordinal $\beta≥\alpha$ such that $\beta=\psi(\beta).$ Let $x\in V[G]_\beta.$ There is a $\gamma<\beta$ such that $x\in V[G]_{\gamma+1}$. Then, $$\phi(x)≤\psi(\gamma+1)≤\psi(\beta)=\beta, $$and it follows from the definition of $\phi(x)$ that there is a $\xi≤\beta$ and a name $\tau\in V_\xi\subseteq V_\beta$ such that $\tau_G=x,$ which is what we wanted.

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