# Sigma algebra and algebra difference

1. An algebra is a collection of subsets closed under finite unions and intersections.

2. A sigma algebra is a collection closed under countable unions and intersections.

Whats the difference between finite and countable unions and intersections? Does "countable" mean it implies there can be infinitely many unions and intersections?

Secondly, I was reading a definition

For an algebra on a set: By De Morgan's law, $A \cap B = (A^c \cup B^c)^c$, thus an algebra is a collection of subsets closed under finite unions and intersections.

What law are they using here to get $A \cap B = (A^c \cup B^c)^c$? I thought de morgan's law was $(A\cap B)^c = A^c \cup B^c$?

Finally, what exactly do they mean by "closed under finite unions and intersections?

• Take your version of the De Morgan law, then take the complement of both sides. – André Nicolas May 27 '12 at 22:33
• Ahh okay, thanks! – Steven May 27 '12 at 22:51

The word 'countable' is the same as 'in bijection with the natural numbers' or 'in bijection with the integers.' There are infinitely many integers, so it's "bigger" than finite. But it's also somehow the smallest infinity.

A common case where this might come up is with respect to open and closed sets. A finite union of closed sets is closed. But an infinite union of closed sets might not be closed. For example, if we consider the sets $I_n = [\frac{1}{n}, 1 - \frac{1}{n}]$, then each $I_n$ is closed. But $\cup_{n \in \mathbb{N}} I_n = (0,1)$, an open set.

With respect to your De Morgan's law question: It is a fundamental fact that $A = B \iff A^c = B^c$, and that $(A^c)^c = A$. So they complemented your De Morgan's law to get that statement.

Finally - algebras and sigma algebras are collections of sets. To be closed under finite intersections means that taking any number of finite intersections of elements of the algebra yields an element (another set) that is in the algebra. But maybe this isn't true for an infinite intersection, etc.

• Thanks mixedmath, appreciated! – Steven May 27 '12 at 22:52
• I have a question (not a mathematician): if an algebra of sets $\Sigma$ is closed under finite unions, i.e $A_i \cup A_{i-1} = B_i \in \Sigma$, then couldn't I keep "iterating" $A_i \cup B_{i-1}$ forever, always producing a set in the algebra $\Sigma$? Why is that not the same as an infinite countable union? Like in your example of the closed interval being summed into an open interval, if I knew that I could nevertheless iterate finite unions over the closed sets forever, why would that not imply that $(0,1) \in \Sigma$? – Marses Oct 8 at 17:45
• @Marses If you have a question, you should ask a separate question. But if your claim is that $(0, 1)$ is a finite union of sets in $\Sigma$, then I would ask you to give me the finite decomposition that form the union. – davidlowryduda Oct 9 at 2:29
• @davidlowryduda sorry, the reason I didn't ask it as a question is because I'm not sure i know enough formal maths to word it properly (or at least to be sure it's a good question). (0,1) is not a finite union of closed sets in $\Sigma$; but, if every finite union of sets in $\Sigma$ is also in $\Sigma$, why can't it be shown by induction that an infinite union of sets in $\Sigma$ is also in $\Sigma$? I'm not claiming this is the case, but I want to find out where exactly I'm going wrong with my thinking I guess. – Marses Oct 9 at 10:01

Ok, am not a mathematician or a student - BUT Doob - measure theory had a pretty terrific answer for it. I was actually looking for the answer and got in here, and obviously could not understand the answer. So, according to Doob, here is the answer:- The algebra S is $\sigma$ algebra if, S contains the limits of every monotone sequence in S. Notice that this is the same idea - we use for complete metric space.

So, now I believe I get it. There is a one to one similarity between this measure stuff and metric stuff, and what is a complete space there, becomes sigma algebra here.

Hope I do understand it.