Timeline for some arguments concerning forcing over V
Current License: CC BY-SA 3.0
12 events
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| May 2, 2011 at 5:38 | comment | added | Andrés E. Caicedo | As for the example, yes, that;s what I mean. The existential lemma ensures us we can restrict ourselves to names that $1$ forces to be an open cover. | |
| May 2, 2011 at 5:37 | comment | added | Andrés E. Caicedo | 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 $x=x_q$. | |
| May 2, 2011 at 4:39 | comment | added | beginner | Sorry, I made a mistake, it should be "p forces $\dot{B}$ has the property" there. | |
| May 2, 2011 at 4:37 | comment | added | beginner | 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" ? | |
| May 1, 2011 at 16:54 | comment | added | Andrés E. Caicedo | (Sorry about the multiple edits, your question was not displaying correctly). | |
| May 1, 2011 at 16:54 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
added 10 characters in body
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| May 1, 2011 at 16:53 | comment | added | Andrés E. Caicedo | 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. | |
| May 1, 2011 at 16:48 | history | edited | Andrés E. Caicedo |
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| May 1, 2011 at 16:47 | history | rollback | Andrés E. Caicedo |
Rollback to Revision 1
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| May 1, 2011 at 8:43 | history | edited | Jason |
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| May 1, 2011 at 8:39 | answer | added | Jason | timeline score: 4 | |
| May 1, 2011 at 7:01 | history | asked | beginner | CC BY-SA 3.0 |