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In some papers, the author wants to show 1 forces (A implies B), i.e., for every generic G, (A implies B) holds in V[G], as follows: Suppose 1 forces A, then let G be a generic filter over V and show B holds in V[G], done. (For example, to show Lindeloffness is preserved, suppose $\dot{U}$ is a name for an open cover of $X$ ($X$ is a Lindelof space in the ground model, here, the author assumes 1 forces ($\dot{U}$ is an open cover of $X$). Then prove $\dot{U}_{G}$ has a countable subcover for a given G, done.)

I don't know why this proof works:

  1. Fix a single G, if A holds in V[G], then certainly A is forced by some p in G. But can we say A is forced by 1?

  2. As I know, to prove 1 forces (A implies B), one must show {p $\in$ P: p forces (($\neg A$) \vee B)} is dense in P. So let q in P. Suppose no $p\lt q$ forces $\neg A$,i.e.,q forces A. Then find a $p\lt q$ such that p forces B by letting G is a generic filter contains q. Done.

Could anyone kind enough to explain it to me?

share|cite|improve this question
beginner: A key basic result in forcing (in the presence of choice), sometimes called the "existential lemma", explains how, if it is forced (by 1) that there is an $A$ with some property, then we can actually find a name $\dot A$ forced by 1 to have that property; moreover, if $p$ forces $\dot B$ to have this property, we can ensure that $p$ forces that $\dot A=\dot B$. This essentially takes care of your first issue. – Andrés E. Caicedo May 1 '11 at 16:53
(Sorry about the multiple edits, your question was not displaying correctly). – Andrés E. Caicedo May 1 '11 at 16:54
Thank you! I know the first part of the "existential lemma", but how to prove the second part? fixed a name $\dot{A}$ such that 1 forces $\dot{A}$ has the property, and a name $\dot{B}$ such that 1 forces $\dot{B}$ has the property, how to ensure p forces $\dot{A}=\dot{B}$? Back to the example of "preserving Lindeloffness", do you mean it is sufficient to "Suppose 1 forces $\dot{U}$ is an open cover of ..." by the "existential lemma" ? – beginner May 2 '11 at 4:37
Sorry, I made a mistake, it should be "p forces $\dot{B}$ has the property" there. – beginner May 2 '11 at 4:39
beginner: This "existential" or "fullness" property is covered in most introductory books. Jech's "Set theory" explains it in terms of complete Boolean algebras. Kunen's book explains it using partial orders. I think it is better if you read the details in one of these sources or some other introductory text. The usual statement is that, given a maximal antichain $M$ of conditions, and associated with each condition $q\in M$ a name $x_q$, we can form a name $x$ such that any $q\in M$ forces $x=x_q$. – Andrés E. Caicedo May 2 '11 at 5:37
  1. No, if $A$ holds in a forcing extension $V[G]$, it need not be forced by $1$ in general. But this is not what is done. Instead, the argument can proceed as follows: In order to show that $1$ forces a statement to be true, we may show it is true in all forcing extensions. Consider an arbitrary forcing extension $V[G]$. If $V[G] \nvDash A$, then $V[G] \models A \rightarrow B$. Consequently, it suffices to restrict our attention to forcing extensions $V[G]$ for which $A$ is true and then show that $B$ will also be true in $V[G]$.

  2. I'm still not sure which part of the proof you find problematic, but let me try to break it down step by step.

  1. First note that $\lnot A \vee B$ is tautologically equivalent to $A \rightarrow B$ so: $1 \Vdash A \rightarrow B$ $\Leftrightarrow$ $1 \Vdash \lnot A \vee B$. Specifically, $A \rightarrow B$ will be true in every forcing extension if and only if $\lnot A \vee B$ is true in every forcing extension.

  2. But to show $1 \Vdash \lnot A \vee B$, it is sufficient to show that the set $D$ of conditions from $P$ forcing $\lnot A \vee B$ is dense in $P$. This is because if $G$ is a $V$-generic filter for $P$, then it must meet every dense set. Consequently, if $D$ is dense, then there will be a condition $p$ in $G \cap D$. In this case, $p \in G$ and $p \Vdash \lnot A \lor B$ so $V[G] \models \lnot A \lor B$, or equivalently, $V[G] \models A \rightarrow B$.

  3. Fix a $q$ in $P$. If any $p \lt q$ forces $\lnot A$, then it will also force $\lnot A \lor B$ because $\lnot A \vee B$ is a tautological consequence of $\lnot A$. Specifically, if every forcing extension $V[G]$ such that $p \in G$ models $\lnot A$, then every such extension will also model $\lnot A \vee B$.

  4. We may therefore assume that all $p \lt q$ do not force $\lnot A$ so that $q \Vdash A$. Notice that if $q$ did not force $A$, then we'd have $V[G] \models \lnot A$ for some $G$ where $q \in G$. In this case, we would have some condition $r \in G$ forcing $\lnot A$. Now notice that if $r \Vdash \lnot A$ and $s < r$, then also $s \Vdash \lnot A$ since every filter containing $s$ will contain $r$ by the upward closure of filters. But also if $q \in G$ and $r \in G$, then we may find such a condition $s$ in $G$ below both $q$ and $r$ by virtue of $G$ being a filter. This $s$ will then be below $q$ and will force $\lnot A$ by virtue of being less than $r$, contrary to assumption.

  5. It therefore suffices to show that we have a condition $p$ below $q$ forcing $B$. If we let $q \in G$, and $V[G] \models B$, then we will have a condition $r \Vdash B$ such that $r \in G$. As in 4., we may then find an $s \in G$ below $q$ that forces $B$ (and hence also $\lnot A \lor B$), and so we are done.

As a technical point, it is implicitly assumed throughout this argument that there is no minimal condition in $P$. Otherwise, replace $\lt$ with $\leq$.

share|cite|improve this answer
Thank you! question 2 is now completed. – beginner May 2 '11 at 4:30
Glad I could help. It looks like I missed your comment before I typed up my answer to 2, but maybe it'll still be useful to you or someone else. – Jason May 2 '11 at 5:14

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