2
$\begingroup$

I know that a sigma algebra generated by a subset a, $\sigma(a) = \{\emptyset,a,a^c,E\}$.

But what about $\sigma({a,b})$? Would it be $\{\emptyset,a,a^c, b, b^c,E\}$?

  • 3
    A $\sigma$-algebra is, in particular, closed under finite intersections and unions. Your claimed example is not.2011-12-11
  • 0
    You are missing, potentially, $a\cap b$, $a\triangle b$, $a-b$, $b-a$, $(a-b)\cup b^c$, and lots more...2011-12-11
  • 0
    thanks, so the only way to find \sigma({a,b}) is to check $a\cap b$...?2011-12-11
  • 0
    @Thomas: I don't understand what you mean by "check $a\cap b$". Note that each of the "basic 4" sets will be in your $\sigma$-algebra: $a\cap b$, $a-b$, $b-a$, $a^c\cap b^c$ (these correspond to the four collections in the most general situation: things in both $a$ and $b$; in $a$ but not in $b$; in $b$ but not in $a$; and neither $a$ nor $b$. Then you have all possible unions of these four regions (which gives 16 possibilities). If $a\subseteq b$, $a\subseteq b^c$, $b\subseteq a$, or $b\subseteq a^c$, then some of these possibilities are repeats, but in general, you may need all of them.2011-12-11
  • 2
    $\sigma(\{a\})$ generally has 4 elements; $\sigma(\{a,b\})$ generally has 16 elements.2011-12-11
  • 0
    Thanks arturo, sorry for not making it clear. I meant checking as in seeing if they are repeats and which are not repeats.2011-12-11
  • 1
    And more generally, if you start with $n$ sets, the $\sigma$-algebra generated by those $n$ sets will generally have $2^{2^n}$ elements, always has at most that many, and you can always find an example where it has exactly that many elements.2011-12-11
  • 0
    thanks again. So i take it that its generally not practical to list out all the elements/construct the elements for large sets? Are there any applications/examples where we would need to?2011-12-11
  • 0
    @Thomas: Depends on what you mean. It is relatively easy to list all the $2^{2^n}$ possibilities. I don't know about any particular applications or examples, though.2011-12-11
  • 0
    hmmm, okay. Thanks for all your help again!2011-12-11

1 Answers 1