What is an example of a field, that is not a $\sigma$-field?
I read a $\sigma$-field requires closure of it's elements (which are sets) under countable unions, countable intersection, and complement; and a field only requires closure under finite unions, finite intersections, and complement.
3 Answers
$\begingroup$A field is closed under finite operations. For example take the field generated by open sets $(a,b)\in\mathbb{R}$. By definition, this will include all finite intersections, unions, complements, etc of such sets. Then such a field will never contain a single point, say $\{0\}$, since you'd need an infinite intersection of open sets to generate it.
$\endgroup$ 3 $\begingroup$Fix an infinite set $X$, and define $\scr{F}$ to be the set of $E\subset X$ for which either $E$ or $X\setminus E$ is finite.
$\endgroup$ $\begingroup$Another example of a field, that is more complicated but also important: the $\bf \Sigma^0_\lambda$ sets for any limit ordinal $\lambda$.
Recall: the $\bf \Sigma^0_1$ sets are the open sets, the $\bf \Sigma^0_{\alpha+1}$ sets are the sets which can be written as a countable union of complements of $\bf \Sigma^0_\alpha$ sets, and a set is $\bf \Sigma^0_\lambda$ ($\lambda$ limit) if it is $\bf \Sigma^0_\alpha$ for some $\alpha<\lambda$. In the usual topology on $\mathbb{R}$, $\bf \Sigma^0_{\omega_1}$ is the smallest $\sigma$-field containing every open sets - that is, the $\bf \Sigma^0_{\omega_1}$ sets are the Borel sets.
However, even though (in the usual topology on $\mathbb{R}$) the $\bf \Sigma^0_\alpha$ sets do not form a $\sigma$-field for any $\alpha<\omega_1,$ there are lots of ordinals $\alpha<\omega_1$ for which the $\bf \Sigma^0_\alpha$ sets do form a field - namely, the limit ordinals.
Note that even the smallest of these - the set of $\bf \Sigma^0_\omega$ sets - is still significantly larger than the field described in Alex's answer above. So don't think of these as "small" examples; rather, think of them as natural examples of fields which arise in the process of trying to build a $\sigma$-field "from below".
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