Timeline for About product of Baire spaces and forcing
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 12, 2019 at 17:25 | comment | added | Gabriel Medina | The reference on this question is Lemma 1.0 of the article Products of Baire spaces by Paul Cohen. | |
| Dec 12, 2019 at 17:23 | comment | added | Gabriel Medina | Well, in this case I understand that $(P, \tau_{\leq})$ as topological space is Baire, that is, every countable family of open and dense sets in $P$ has dense intersection. | |
| Dec 12, 2019 at 10:24 | comment | added | Joel David Hamkins | For general partial orders, this isn't true, but it is exactly equivalent to the partial order being $(\omega,\infty)$-distributive. I assumed that this is the property you mean by saying the space was `Baire'. So it follows from that assumption. The usual way of stating your lemma is that a forcing notion adds no new $\omega$-sequences over the ground model if and only if it is $\omega$-distributive, which is equivalent to the assertion that the countable intersection of open dense sets is dense. | |
| Dec 11, 2019 at 20:30 | comment | added | Gabriel Medina | I have one last question, I can prove that $D_ {n}$ is open and dense at $\{q: q\leq p^{\prime} \}$, then $\bigcap _{n\in \omega}D_{n}$ is dense in $\{q: q\leq p^{\prime} \}$, but I need that $G$ intersects $\bigcap _{n\in \omega}D_{n}$, and this will happen if $\bigcap _{n\in \omega}D_{n}$ is dense. How can I get that $\bigcap _{n\in \omega}D_{n}$ is dense in $P$? | |
| Dec 11, 2019 at 17:41 | comment | added | Gabriel Medina | I understood, in addition to the proof it is enough that they are dense below $p^{\prime}$, since an open subspace of a Baire remains Baire. | |
| Dec 11, 2019 at 17:38 | comment | added | Joel David Hamkins | If you have a name for an element of a set in the ground model, then it will always be dense (below any condition forcing that) to decide which particular element it is. The reason is that otherwise you can find a G containing some condition below the given condition, in which the interpretation of the name isn't any particular one of those elements, contrary to the assumption. | |
| Dec 11, 2019 at 17:35 | comment | added | Gabriel Medina | Thanks a lot @Joel David Hamkings. I'm new studying forcing I didn't know that last facts. | |
| Dec 11, 2019 at 17:35 | vote | accept | Gabriel Medina | ||
| Dec 11, 2019 at 17:27 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |