Timeline for countably complete filters
Current License: CC BY-SA 3.0
5 events
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| Apr 7, 2012 at 20:39 | comment | added | KP Hart |
I mentally inserted ultra', hence my singleton'.
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| Apr 6, 2012 at 10:34 | comment | added | Douglas Somerset | In fact, if we take $X=[0,1]$ then for each subset $Y$ there is a largest countably complete filter of dense $G_{\delta}$s with intersection $Y$ (namely all dense $G_{\delta}$s containing $Y$) and a smallest countably complete filter with intersection $Y$ (namely all co-countable dense $G_{\delta}$s containing $Y$). In the case when $Y=\emptyset$, the question is to describe the countably complete filters of dense $G_{\delta}$s which contain the filter of co-countable sets. | |
| Apr 5, 2012 at 20:50 | comment | added | Douglas Somerset | Also if there are no isolated point in the original space $X$ then for any non-empty subset $Y\subseteq X$ the set of dense $G_{\delta}$s containing $Y$ is a countably complete filter whose intersection is $Y$. Thus there are at least $2^X$ distinct countably complete filters. | |
| Apr 5, 2012 at 20:13 | comment | added | Andreas Blass | There can also be some smaller filters. For example, if there are no isolated points, then the co-countable sets form a countably complete filter in the lattice of dense $G_\delta$ sets. And there are lots of minor variants; for example, fix an uncountable subset $A$ of the space and consider the filter of those dense $G_\delta$ sets that contain all but countably many of the points in $A$. Or, if there's a set $A$ of size $\aleph_1$ whose closure is nowhere dense, fix a bijection $f$ from $\omega_1$ to it and form the filter of sets that include $f$ of a club. | |
| Apr 5, 2012 at 19:39 | history | answered | KP Hart | CC BY-SA 3.0 |