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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<q forces $\neg A$,i.e.,q forces A. Then find a p<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?

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?