2
$\begingroup$

page containing Lemma 15.19 from Jech

As the proof in the picture, the author says that we can assume that every condition forces that $\dot{f}$ is a function form $\lambda$ to $A$.

I guess that here he means that we can assume $A=M\cap V_\alpha$ for some ordinal $\alpha$, and that $\left \| \dot{f}\text{ is a function } \lambda \rightarrow A \right \|=1$.

And that if $\dot{f}$ don't satisfy this, we can take another name of f satisfy this. But I can't see why. This is my question. Why we can do this assume?

$\endgroup$
2
  • 3
    $\begingroup$ Generally, it would be better to add texts rather than pictures as it increases the visibility of your post in search engines and redirects more relevant visitors to it in the future. :-) $\endgroup$ Jul 16, 2018 at 9:33
  • 2
    $\begingroup$ Cross posted. $\endgroup$
    – user57432
    Jul 16, 2018 at 14:20

1 Answer 1

8
$\begingroup$

When Jech says that $\dot f$ is a name for $f$, what he means is that the interpretation of $\dot f$ by the filter $G\times H$ is $f$. Therefore some condition $p$ in $G\times H$ forces that $\dot f$ is a function from $\check\lambda$ to the ground model $\check M$.

Other incompatible conditions might force totally different things about this particular name, but by extending $p$ to a maximal antichain and then applying the mixing lemma, we may replace $\dot f$ with another name $\dot h$, such that $p$ forces $\dot f=\dot h$, and all the other conditions in the maximal antichain force that $\dot h$ is, say, the constant zero function on $\check\lambda$. Now, every condition will force that $\dot h$ is a function from $\check\lambda$ to $\check M$, and the interpretation of $\dot h$ by $G\times H$ will agree with that of $\dot f$, since $p$ forces they are equal.

This is a very commonly used method in forcing arguments. If you have an object in the extension, then you can take a name for it, and by the mixing lemma, you can assume without loss that every condition forces that it has the property you assumed about that object, provided that you can easily manufacture names for such objects to use with the other conditions in an antichain.

Next, from $\dot h$, we can look at the possible values of $\dot h(\check\alpha)$ for $\alpha<\lambda$, meaning the objects $a$ for which there is some condition $q$ forces $\dot h(\check\alpha)=\check a$. For each $\alpha$ there are only a set of such possible values $a$ in $M$, because these are incompatible features of the function $\dot h$ and therefore the conditions must be incompatible. So if we let $A$ be the set of all such $a$ that arise, we have a set $A\in M$ such that every condition forces that $\dot h$ is a function from $\check\lambda$ to $\check A$. So we've achieved the desired situation of a name for $f$ and a set $A$ in $M$, such that every condition forces that that name is a function from $\lambda$ to $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.