Give an example of sigma algebra in $\mathcal P(\Bbb N)$ whose order is finite, and also whose order is infinite, and whose order is $10$?

(the order means the number of elements)

  • $\begingroup$ What is the order of a $\sigma$-algebra? If it is the number of elements, there is no $\sigma$-algebra with exactly $10$ elements. $\endgroup$ Oct 20 '12 at 7:19
  • $\begingroup$ why is that?and yes the order is the number of element $\endgroup$ Oct 20 '12 at 8:02
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    $\begingroup$ Well, well. What a lazy question. Nonetheless plus one because I learned something from reading Davide's answer. $\endgroup$ Oct 20 '12 at 9:36
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    $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. $\endgroup$ Oct 21 '12 at 19:38

A example of $\sigma$-algebra contained in $\mathcal P(\Bbb N)$ which is finite is $\{\emptyset,\Bbb N\}$; an example of infinite one is $\mathcal P(\Bbb N)$ itself.

Now let $X$ a set and assume that $\mathcal B$ is a finite $\sigma$-algebra on $X$. For each $x\in X$, define $S_x:=\bigcap_{B\in\mathcal B, x\in B}B$, and define $x\sim y$ if $x\in S_y$. Then $\sim$ is an equivalence relation, which gives a finite partition of $X$ as $S_{x_1},\dots,S_{x_n}$ (each $S_{x_i}$ is measurable and the $\sigma$-algebra is assumed finite). So $\mathcal B$ has $2^n$ elements.

If we take an infinite set $S$, we can, for each $n\geq 1$, construct a $\sigma$-algebra having exactly $2^n$ elements (taking a partition of $S$ in $n$ elements).


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