Let $\mathcal{B}([0, 1])$ be the Boolean algebra of measurable subsets of $[0, 1]$ modulo almost everywhere equivalence, i.e., two measurable sets which differ only by a Lebesgue null set are identified. For each $t \in [0, 1]$, let $\mathcal{D}_t$ be the filter on $\mathcal{B}([0, 1])$ generated by open neighborhoods of $t$. Does these exists a filter $\mathcal{F}$ on $\mathcal{B}([0, 1])$ s.t. for each $t \in [0, 1]$, the filter generated by $\mathcal{D}_t$ and $\mathcal{F}$ is an ultrafilter?
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3 No. Suppose $\mathcal{F}$ is such a filter. Clearly each $X \in \mathcal{F}$ has positive measure intersection with every positive-length interval. Then neither $[0,1/2]$ nor $[1/2, 1]$ are in the filter generated by $\mathcal{D}_{1/2}$ and $\mathcal{F},$ contradiction.
- $\begingroup$ That seems right. Do you know if such a $\mathcal{F}$ could exist if I modify my question as follows: So suppose instead of defining $\mathcal{D}_t$ I define a left version $\mathcal{L}_t$ generated by all intervals $(t - \epsilon, t)$ for $t > 0$ and a right version $\mathcal{R}_t$ generated by all intervals $(t, t + \epsilon)$ for $t < 1$. Is it possible for the filter generated by $F$ and $\mathcal{L}_t$ to be an ultrafilter for all $t > 0$ and the filter generated by $F$ and $\mathcal{R}_t$ to be an ultrafilter for all $t < 1$? $\endgroup$David Gao– David Gao2023-06-04 08:43:40 +00:00Commented Jun 4, 2023 at 8:43
- $\begingroup$ I doubt any simple modification of this question will lead to a positive answer. E.g., neither $\bigcup [2^{-1} - 2^{-2n}, 2^{-1} - 2^{-2n-1}]$ nor its complement will be in the filter generated by $\mathcal{F}$ and $\mathcal{L}_{1/2}.$ $\endgroup$Elliot Glazer– Elliot Glazer2023-06-04 08:52:48 +00:00Commented Jun 4, 2023 at 8:52
- $\begingroup$ I see. I guess I was overly optimistic about this and thinking some kind of Zorn’s lemma argument might produce a filter of this sort. In any case thank you! $\endgroup$David Gao– David Gao2023-06-04 09:01:06 +00:00Commented Jun 4, 2023 at 9:01