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Let $\Omega$ be arbitrary set. Let $Q$ be a partition of $\Omega$. I already proved that the collection of all unions of the cells in $Q$ is a complete field $\mathcal{F}$ (complete field is collection of subsets of $\Omega$ that is closed under arbitrary interesections and complements).

I have to prove that $\mathcal{F}$ defines a partition $P$ of $\Omega$ such that for every $\omega\in\Omega$, $P(\omega)$ is the interesection of all the sets in $\mathcal{F}$ that include $\omega$. ($P(\omega)$ denotes cell of $P$ that contains $\omega$).

I would be very grateful for help.

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Start by defining $P(\omega)=\bigcap\{F\in\mathscr{F}:\omega\in F\}$ for each $\omega\in\Omega$, and let $\mathscr{P}=\{P(\omega):\omega\in\Omega\}$; you need to show that $\mathscr{P}$ is a partition of $\Omega$. In order to do this, you must show two things:

  • $\bigcup\mathscr{P}=\Omega$, meaning that each element of $\Omega$ belongs to at least one member of $\mathscr{P}$, and
  • $\mathscr{P}$ is pairwise disjoint, meaning that each element of $\Omega$ belongs to at most one member of $\mathscr{P}$.

The first is trivial: we defined $\mathscr{P}$ to make it true. Thus, what you really have to show is that if $\omega_0,\omega_1\in\Omega$, and $P(\omega_0)\ne P(\omega_1)$, then $P(\omega_0)\cap P(\omega_1)=\varnothing$.

HINT: Suppose that $P(\omega_0)\ne P(\omega_1)$; without loss of generality $P(\omega_0)\setminus P(\omega_1)\ne\varnothing$. (Why?)

  • Show that there is an $F\in\mathscr{F}$ such that $F\cap\{\omega_0,\omega_1\}=\{\omega_1\}$. Note that $P(\omega_1)\subseteq F$.
  • What can you say about $\Omega\setminus F$?
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  • $\begingroup$ Could you elaborate please? I don't know how to prove it... $\endgroup$
    – luka5z
    Commented Jul 10, 2015 at 13:29
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    $\begingroup$ @luka5z: You know that $\Omega\setminus F\in\mathscr{F}$, and clearly $(\Omega\setminus F)\cap\{\omega_0,\omega_1\}=\{\omega_0\}$. What’s the relationship between $P(\omega_0)$ and $\Omega\setminus F$? $\endgroup$ Commented Jul 10, 2015 at 18:24
  • $\begingroup$ they are equal? $\endgroup$
    – luka5z
    Commented Jul 10, 2015 at 18:51
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    $\begingroup$ @luka5z: Not necessarily. Do you understand why $P(\omega_1)\subseteq F$? Apply the same reasoning to $P(\omega_0)$ and $\Omega\setminus F$. $\endgroup$ Commented Jul 10, 2015 at 19:00
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    $\begingroup$ @luka5z: You’re welcome! $\endgroup$ Commented Jul 10, 2015 at 20:18

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