Timeline for special extremally disconnected spaces with only finite isolated points
Current License: CC BY-SA 3.0
9 events
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| Nov 26, 2015 at 23:03 | comment | added | Joseph Van Name | Furthermore, the extremal disconnectedness of $[\kappa]^{<\omega}$ is equivalent to the Prikry lemma which states that for every statement $\sigma$ and every stem $s$ (the stems are the elements in the space $[\kappa]^{<\omega}$) there is some $A\in M$ where $(s,A)$ decides $\sigma$. I have not seen any papers or investigations where this relation between Prikry forcing and general topology has been seriously investigated before even though it seems very interesting. | |
| Nov 26, 2015 at 23:03 | comment | added | Joseph Van Name | This is a really nice topological space which is very much related to Prikry forcing. Let's take a normal ultrafilter $M$ on a measurable cardinal $\kappa$ and define a topology on $[\kappa]^{<\omega}$ where a subset $U\subseteq[\kappa]^{<\omega}$ is open if and only if for each $(a_{1},...,a_{n})\in U$ there is some $A\in M$ where if $a\in A,a>a_{n}$, then $(a_{1},...,a_{n},a)\in U$ as well. Then $[\kappa]^{<\omega}$ is extremally disconnected, so the Boolean algebra of clopen sets of $[\kappa]^{<\omega}$ is a complete and it is the completion of the Prikry forcing. | |
| Nov 25, 2012 at 1:39 | comment | added | Goldstern | Whenever $s\in X$, and $\bar A=(A_t:s \le t)$ is a family of sets in $U$, then $\bar A$ naturally defines a clopen neighborhood $O=O_{\bar A}$ of $s$, and these neighborhoods form a base: $s\in O$, all elements of $O$ extend $s$, and for each $t\in O$ we have $(t,i)\in O$ iff $i\in A_t$. | |
| Nov 25, 2012 at 1:34 | history | edited | Goldstern | CC BY-SA 3.0 |
deleted wrong claim
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| Nov 25, 2012 at 1:29 | comment | added | Goldstern | You are right. Thank you. For some stupid reason I thought that there are only countably many finite sequences. Thinking too much about Laver trees. | |
| Nov 25, 2012 at 0:36 | comment | added | Joseph Van Name | I don't think the sets $O_{s,F}$ really form a neighborhood basis for $s$. It seems like a way to get around this is to only take increasing sequences from $\kappa$ and assume that the ultrafilter is a normal ultrafilter. Or you can just take another basis. | |
| Nov 23, 2012 at 8:44 | history | edited | Goldstern | CC BY-SA 3.0 |
corrected definition of base
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| Nov 22, 2012 at 23:43 | history | edited | Goldstern | CC BY-SA 3.0 |
typos
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| Nov 22, 2012 at 23:31 | history | answered | Goldstern | CC BY-SA 3.0 |