Timeline for Does the Lebesgue measure induce a finitely additive measure on the Boolean algebra of regular open subsets of (0,1)?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 7, 2016 at 5:37 | vote | accept | Marcus Pivato | ||
| Jul 7, 2016 at 5:35 | comment | added | Marcus Pivato | Then the closure of $L$ is the set $K \sqcup L \sqcup R'$. (Proof: Any sequence in $L$ either converges to a point in $L$, or it converges to a point in $K$ (by pushing the "initial 1" further and further down the quaternary expansion until it vanishes altogether), or it converges from the left to a point in $R'$.) Since $K$ and $R'$ are both nowhere dense, the interior of $K \sqcup L \sqcup R'$ is simply $L$, which shows that $L$ is regular. Finally, there is a self-homeomorphism of [0,1] transforming my sets $L$ and $R$ into your $L$ and $R$, so they are regular also, as you claimed. | |
| Jul 7, 2016 at 5:33 | comment | added | Marcus Pivato | In your notation, $L$ is the set of all points in $U$ such that the first "1" appears before the first "2", while $R$ is the set of all points in $U$ such that that the first "2" appears before the first "1". Clearly, $U=R\sqcup L$, and $L$ and $R$ are open (because a small enough perturbation will not change the first "1" into a "2" or vice versa...) Let $R'$ be the set of left endpoints of $R$-intervals; equivalently, $R'$ is the set of all tetradic rationals whose quaternary expansion contains no "1" and ends with "20000..." [continued...] | |
| Jul 7, 2016 at 5:31 | comment | added | Marcus Pivato | Consider the Cantor set $K$ obtained by deleting all numbers which contain a "1" or a "2" anywhere in their quaternary expansion, except for tetradic rationals which have only a single "1" and end in "10000...". (Yes, I know this is not a "fat" Cantor set, but I'm only interested in the topological properties here, not the measure.) In this case, $U$ is the set of all points in $[0,1]$ containing a "1" or a "2" anywhere in their quaternary expansions (except the aforementoned tetradics). [continued] | |
| Jul 7, 2016 at 5:29 | comment | added | Marcus Pivato | Thank you very much, Joel. This is a beautiful argument. But I think it would help to add a bit more detail on why $L$ and $R$ are regular. I suggest something like the following. Every number in $[0,1]$ has a unique quaternary (i.e. base-4) expansion. (That is, "unique" once we exclude expansions ending in "3333333...".) Let us say that a number is tetradic rational if it is a rational number whose denominator is a power of 4 ---equivalently, its quaternary expansion ends in an infinite sequence of zeros. [Continued in next comment...] | |
| Jul 6, 2016 at 16:35 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Fixed a minor issue
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| Jul 6, 2016 at 14:08 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 3 characters in body
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| Jul 6, 2016 at 14:01 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |