I got this homework in my forcing class:

  1. Let $G\subseteq P$ be generic over M. Show that there is a cardinal of M, $\lambda$ such for every set of ordinals $X\in M\left[G\right]$ there is a set $Y\in M$ s.t. $X\subseteq Y$ and $\left|Y\right|^{M}\le\max\left(\lambda,\left|X\right|\right)$. Infer from this fact that large enough cardinals of $M$ remain cardinals in $M\left[G\right]$.
  2. Let $G\subseteq P$ be a generic filter over $M$. Show that there exist a cardinal of $M$, $\lambda$ such for every set of ordinals $X\in M\left[G\right]$ there is family $\mathcal{F}\in M$ of sets of ordinals s.t. $\left|\mathcal{F}\right|\le\lambda $ and $X$ is the union of a subfamily $\mathcal{F}^{\prime}\subseteq\mathcal{F}$ such that $\mathcal{F}^{\prime}\in M\left[G\right]$.

I finished (I think) the first question:

In the whole exercise, if $x\in M\left[G\right]$ we will denote by $x^{\circ}$ the P-name s.t. $\left[x^{\circ}\right]_{G}=x$. If $x\in M$ we will sometimes use the notation $\check{x}$ instead of $x^{\circ}$. For any P-name $n$, $C_{n}$ would be the coresponding constant symbol in the language of forcing.

Let $\lambda$ be a cardinal in $M$ s.t. every anti-chain of $P$ in $M$ is of length at most $\lambda$. Let $X$ be a set of ordinals in $M\left[G\right]$, $\mu=\left|X\right|^{M\left[G\right]}$ and $f\in M\left[G\right]$ the function that shows this cardinality, $f:\mu\rightarrow X$ and onto $X$. Denote by $\psi\left(x,y,z\right)$ the first order formula that states that $x$ is a function which domain is $y$ range is $z$ and it is onto $z$. Then $M\left[G\right]\models\psi\left(C_{f^{\circ}},C_{\check{\mu}},C_{X^{\circ}}\right)$ and thus, from the truth lemma, exist $p\in G$ s.t. $p\Vdash\psi\left(C_{f^{\circ}},C_{\check{\mu}},C_{X^{\circ}}\right)$. Now, for every $\alpha\in\mu$ define $E_{\alpha}=\left\{ \beta:\beta\in On\land\left(\exists q\le p\right)\left(q\Vdash C_{\check{\left\langle \alpha,\beta\right\rangle }}\in C_{f^{\circ}}\right)\right\} $. From the definability lemma, $\left(\exists q\le p\right)\left(q\Vdash\beta\in rng\left(C_{f^{\circ}}\right)\right)$ is definable in $M$ and thus $E_{\alpha}\in M$. For every $\alpha\in\mu$ and $\beta\in E_{\alpha}$ define $q_{\alpha,\beta}$ to be a whitness of $\beta\in E_{\alpha}$ meaning that $q_{\alpha,\beta}\le p$ and $q_{\alpha,\beta}\Vdash C_{\check{\left\langle \alpha,\beta\right\rangle }}\in C_{f^{\circ}}$. Then $D_{\alpha}=\left\{ q_{\alpha,\beta}:\beta\in E_{\alpha}\right\} $ is an anti-cahin. Indeed, assume in contradiction that exist $\gamma,\delta\in E_{\alpha}$ s.t. $\gamma\neq\delta$ and exist $r\le q_{\alpha,\gamma}$ with $r\le q_{\alpha,\delta}$. Then $r\Vdash C_{\check{\left\langle \alpha,\gamma\right\rangle }}\in C_{f^{\circ}}$ because $r\le q_{\alpha,\gamma}$ and $r\Vdash C_{\check{\left\langle \alpha,\delta\right\rangle }}\in C_{f^{\circ}}$ because $r\le q_{\alpha,\delta}$ in contradiction to the fact $r\Vdash"C_{f^{\circ}}\text{ is a function"}$ (this is because $r\le p$). Thus $\left|D_{\alpha}\right|\le\lambda$ and from that we conclude that $\left|E_{\alpha}\right|\le\lambda$ as well ($q_{\alpha,\beta}\neq q_{\alpha,\gamma}$ for $\beta\neq\gamma$ as well because $q_{\alpha,\beta}\perp q_{\alpha,\gamma}$ and thus $\left|E_{\alpha}\right|=\left|D_{\alpha}\right|$). Now, define $Y=\bigcap_{\alpha<\mu}E_{\alpha}$. Then for every $\beta\in X$ exist $\alpha\in\mu$ s.t. $M\left[G\right]\models C_{\check{\left\langle \alpha,\beta\right\rangle }}\in C_{f^{\circ}}$ meaning that exist $r\in G$ s.t. $r\Vdash C_{\check{\left\langle \alpha,\beta\right\rangle }}\in C_{f^{\circ}}$ and thus (because $G$ is a filter) exist $q\le r$, $q\le p$ and then $q\Vdash C_{\check{\left\langle \alpha,\beta\right\rangle }}\in C_{f^{\circ}}$ meaning that $\beta\in E_{\alpha}\subseteq Y$ and from the generality of $\beta$ follows $X\subseteq Y$. Finally, $$\left|Y\right|\le\lambda\cdot\mu=\max\left(\lambda,\mu\right)=\max\left(\lambda,\left|X\right|\right)$$ Now, let $\kappa>\lambda$ be a cardinal and assume in contradiction it is not a cardinal in $M\left[G\right]$. Then $\left|\kappa\right|^{M\left[G\right]}<\kappa$ and from what we proved above, exist $Y\in M$ s.t. $\left|Y\right|^{M}\le\max\left\{ \lambda,\left|\kappa\right|^{M\left[G\right]}\right\} $ and $\kappa\subseteq Y$. Then $$\left|\kappa\right|^{M}\le\left|Y\right|^{M\left[G\right]}\le\max\left\{ \lambda,\left|\kappa\right|^{M\left[G\right]}\right\} <\kappa$$ in contradiction to the fact $\kappa$ is a cardinal. $\square$

But I am completely lost with the second question. It is supposed to be similar in some sense, but I just can't see why. I tried to do the same trick with function from $\mu$ to X (I tried with a bijection from $\mu$ to X and with a bijection from X to $\mu$) But I am not able to create any infinite set in M that I can ensure is entirely included in X.

Could someone help me with that please?

  • $\begingroup$ Something is a bit off in the second question. (1) $\lambda$ is not mentioned after the first line; and (2) taking singletons would work. Perhaps you had meant that $\mathcal F$ has size $\lambda$? $\endgroup$ – Asaf Karagila Jun 6 '15 at 12:23
  • $\begingroup$ @AsafKaragila - of course, I editted $\endgroup$ – sss89 Jun 6 '15 at 12:32
  • 1
    $\begingroup$ By the way, \mathring{x} will produce $\mathring x$ for $P$-names. $\endgroup$ – Asaf Karagila Jun 6 '15 at 12:50

HINT: Fix a name $\mathring X$ in $M$. For each $p\in P$, consider $X_p=\{\alpha\mid p\Vdash\check\alpha\in\mathring X\}$. What can you say about $\mathcal F=\{X_p\mid p\in P\}$?


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