Timeline for Does a monotone subadditive $f: \mathcal{P}(\bf N)\to [0,1]$ admit a finite partition with values in $(0,1)$?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Oct 8, 2015 at 7:26 | vote | accept | Paolo Leonetti | ||
| Oct 7, 2015 at 17:27 | answer | added | Ilya Bogdanov | timeline score: 7 | |
| Oct 7, 2015 at 13:40 | comment | added | Jochen Wengenroth | Okay, I see the problem. For $k=2$ and $A_2=A_1^c$ it may happen that $f(A_2)=1$. | |
| Oct 7, 2015 at 13:39 | comment | added | Salvo Tringali | @JochenWengenroth. Of course. But only if you can prove that $0 < f((A_1 \cup \cdots \cup A_k)^c) < 1$ for some $k$, in your construction. In principle, you don't have any information on the behaviour of $f(A^c)$ for $A \subseteq \mathbf N$ and $f(A) > 0$ (if $f(A) = 0$, it is seen that $f(A^c) = 1$). | |
| Oct 7, 2015 at 13:30 | comment | added | Jochen Wengenroth | If you want to stop at a given $k$, just take $A_k=\mathbb N \setminus (A_1\cup\cdots\cup A_{k-1})$. | |
| Oct 7, 2015 at 12:24 | comment | added | Paolo Leonetti | Why do you expect to stop in a finite number of steps? | |
| Oct 7, 2015 at 12:20 | comment | added | Jochen Wengenroth | What's wrong with this rather trivial construction: Choose $A_1\subseteq \mathbb N$ with $f(A_1)=1/2$. Because of subadditivity, $1=f(A_1 \cup A_1^c)\le f(A_1)+f(A_1^c)$ so that $f(A_1^c)\ge 1/2$. Next choose $A_2\subseteq A_1^c$ with $f(A_2)=1/4$. As above $f(A_1^c \cup A_2^c)\ge 1/4$. Continue in that way. | |
| Oct 7, 2015 at 9:06 | history | edited | Paolo Leonetti | CC BY-SA 3.0 |
added 185 characters in body
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| Oct 7, 2015 at 9:01 | history | asked | Paolo Leonetti | CC BY-SA 3.0 |