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This a just a small question about the definition of a sigma-algebra and what is eaxactly meant by countable? Would be grateful for any clarification on this. Most texts define a sigma-algebra ($\mathcal{A}$) on a set $X$ as a collection of subsets of $X$ that satisfy the following properties

(i) $\emptyset\in\mathcal{A}$

(ii) $A\in\mathcal{A}\Rightarrow X\backslash A\in\mathcal{A}$

(iii) If $A_1,A_2...\in\mathcal{A}$ then $\bigcup\limits_{i=1}^{\infty}A_i\in\mathcal{A}$, i.e. $\mathcal{A}$ is closed under countable unions.

If a set (say $A$) is countable, I understand it to mean that $|A|\leq\aleph_0$, so finite sets are countable (or do some texts just mean infinitely countable by countable), so my 1st question is why do some texts prove that sigma-algebras are closed under finite intersections, by defining for a collection of n sets the $n+k$ set, ($k\in\mathbb{N}$) to be the empty set, as surley if we take countable to be possibly finite then by (iii) we must have $((\mathcal{A}'\subset\mathcal{A}\space$ & $|\mathcal{A}'|<\aleph_0)\Rightarrow\bigcup\mathcal{A}'\in\mathcal{A})$. So then wouldn't (i) be redundant as $\emptyset$ is a countable (finite) subset of any collection of sets and $\bigcup\emptyset=\emptyset$. Any feedback to clarify this is most welcome. Thank You

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  • $\begingroup$ Originally, countable meant finite or countably infinite. The meaning seems to be slowly shifting. It is not unreasonable to make a proof that works for speakers of either dialect. $\endgroup$ – André Nicolas Jun 29 '14 at 19:19
  • $\begingroup$ (i) is not redundant, your assumption is that the $A_i\in\mathcal{A}$, so if $\varnothing\not\in\mathcal{A}$ you cannot use (iii). In particular the empty set itself would satisfy the axioms (ii), (iii) but does not constitute a $\sigma$-algebra. $\endgroup$ – Adam Hughes Jun 29 '14 at 19:20
  • $\begingroup$ What if (iii) is replaced by this statement $(\forall\mathcal{A}')(\mathcal{A}'\in 2^{2^X}\space\&\space|\mathcal{A}'|\leq\aleph_0\Rightarrow\bigcup\mathcal{A}' \in\mathcal{A})$ which is almost the same as (iii) but I see the distinction as you brought up that so we are not assuming $A_i\in\mathcal{A}$, so (i) would then be redundant now? $\endgroup$ – Jannick Jun 29 '14 at 20:47
  • $\begingroup$ Sorry I made a typo, the statement should read $(\forall\mathcal{A}')(\mathcal{A}'\in 2^{\mathcal{A}}\space\&\space|\mathcal{A}'|\leq\aleph_0\Rightarrow \bigcup\mathcal{A}' \in\mathcal{A})$ $\endgroup$ – Jannick Jun 29 '14 at 21:02
  • $\begingroup$ When (iii) is only finite rather than countably finite, the structure is an algebra, not a $\sigma$-algebra. $\endgroup$ – Batman Jul 21 '17 at 21:28

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