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I’m reading A Probability Path by S. Resnick, and I got stuck on one problem:

Problem 1.16 (pp. 22-23) Suppose $C$ is a class of subset of $\Omega$ such that $\Omega \in C$, $A \in C $ implies that $A^c \in C$, and $C$ is closed under disjoint unions. Using an example, show that $C$ does not have to be a field.

A hint is provided in book:

Try $\Omega = \{1,2,3,4\}$ and let $C$ be the field generated by two point subsets of $\Omega$.

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    $\begingroup$ I am adding the [homework] tag not because it is your homework, but because the question meets our general guidelines for when this tag should be added. $\endgroup$ – cardinal Nov 4 '12 at 15:20
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The hint actually takes you most way home.

Consider the collection $$ \mathscr{C} =\{\{1,2,3,4\}, \{1,2\}, \{2,3\}, \{3,4\}, \{4,1\}, \{1,3\}, \{2,4\},\emptyset\} $$

It is easy to see that this collection satisfies all the properties that are stated in the problem, but is not a field, for example, $\{1,2\}\cup \{2,3\}\notin \mathscr{C}$.

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    $\begingroup$ While the essence of this answer is correct, the concrete example given is not. If you were to remove one particular element from the class, then it would be correct. $\endgroup$ – cardinal Nov 4 '12 at 15:06
  • $\begingroup$ Missed $\{2,4\}$ didn't I? Well-spotted. $\endgroup$ – tchakravarty Nov 4 '12 at 15:12
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    $\begingroup$ Or, just remove $\{1,3\}$. It doesn't match the hint as closely, but has the virtue of being an even simpler example. (+1) $\endgroup$ – cardinal Nov 4 '12 at 15:14
  • $\begingroup$ The "hint" as given in the question is kind of bogus anyway. I hope it's not written like that in the book. $\endgroup$ – cardinal Nov 4 '12 at 15:17
  • $\begingroup$ @cardinal. This exactly how it is written in the book. $\endgroup$ – user9292 Nov 4 '12 at 20:40

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