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Can someone explain how this $\sigma$-algebra is attained? It's mainly the $X\cup Y$ bit which I don't understand.

Question: If $\Omega = \{1, 2, 3, 4\}$ and we have a collection of sets $\mathcal C = \{\{3\},\{2, 3, 4\}\}$ what is the smallest $\sigma$-algebra generated by $\mathcal C$?

Answer: The ”atoms” that $\mathcal C$ generates are a bit more complex, there are in fact three:

  • Clearly $X = \{3\}$ is one atom.
  • If we set $A =\{2, 3, 4\}$, then $A^c =\{1\} = Y$ is an atom.
  • $X\cup Y = \Omega\setminus\{1, 3\} = \{2, 4\} = Z$.

Given these three atoms, the $\sigma$-algebra is $\{\emptyset, X, Y, Z, X\cup Y, X\cup Z, Y\cup Z,\Omega\}$.

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It is likely a typo: what is intended is $(X \cup Y)^c = \Omega \setminus \{1,3\} = \{2,4\} =: Z$.

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  • $\begingroup$ That makes more sense now but I still don't understand why we even find Z. My initial answer was {∅, Ω, A, B, Ac, Bc, A∪B, A∪Bc, Ac∪B, Ac∪Bc}. Where A ={3} and B={2,3,4}. $\endgroup$ – Murdo Apr 19 '13 at 13:29
  • $\begingroup$ What do you mean by that? An attempt at clarification: Since $X$ and $Y$ do not suffice to create $\{2,3,4\} \in C$, we need a further atom. $Z$ (the complement of the union of all preceding atoms) is our next attempt at adding an atom. Because $X,Y,Z$ suffice to create all sets in $C$, we needn't split up $Z$ into smaller atoms. $\endgroup$ – Lord_Farin Apr 19 '13 at 13:38
  • $\begingroup$ Why can we not just use A = {2,3,4} in our sigma-algebra? I don't really understand the atom stuff. $\endgroup$ – Murdo Apr 19 '13 at 13:51
  • $\begingroup$ Please join mathematics chat; I'll try to explain it to you there. $\endgroup$ – Lord_Farin Apr 19 '13 at 13:53

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