Results 1 to 5 of 5

Math Help - sigma-algebra

  1. #1
    Junior Member
    Joined
    Jun 2010
    Posts
    32

    sigma-algebra

    Hi!

    Trying to get a grasp on \sigma-algebra.
    Letīs say I have a the set  \Omega = \left\{1,2,3}\right\} .

    Now, if I want to find the smallest \sigma-algebra \, F that includes both 1 and 2. Would this be:

     F = \left\{\emptyset, \Omega, 1, 2, 3, \left\{1,2\right\}\right\} ?

    The empty set is there, and also its complement  \Omega . Both 1 and 2 are there and their complement 3. Also the union of 1 and 2 are is there.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2009
    Posts
    95
    Thats not quite right.

    First, here is a result for \sigma algebras that is quite useful to know:

    If X is a set and \Sigma \subset \mathcal{P}\{X\} is a \sigma-algebra on X then
    \Sigma is closed under countable intersections.

    Proof:

    Let \bigcap_{n=1}^\infty A_n for A_i \in \Sigma. Then note that for each A_i \in \Sigma we have that \bar{A}_i \in \Sigma where \bar{A}_i is the complement of A_i. It follows that \bigcap_{n=1}^\infty A_n = X \setminus{ \bigcup_{n=1}^\infty \bar{A}_n } \in \Sigma. _\Box


    But other than that I can point out that 1,2,3 are not themselves sets and can not possibly be in the \sigma-algebra. However, I think you meant \{1\},\{2\},\{3\}, which is correct.

    You are however, missing some sets in this. Consider this corollary to the previous result: if A,B\in \Sigma then A \setminus B \in \Sigma. This means that for example \{2,3\} = \Omega \setminus\{1\}\in \Sigma along with a few other sets.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2010
    Posts
    32
    I think I understood your claim and proof. You say that if we have a sequence of sets in a \sigma-algebra F their intersection must also lie in F. Because all the complements of the sets must lie in F, we get that the union of all complements must lie in F, and because of this the complement of the union of complements must lie in F. And this is equal to the intersection of our original sets.

    I donīt think I understand your corollary though.

    I updated:
    F= \left\{\emptyset,\Omega, \left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{1,2\right\}, \left\{2,3\right\}, \left\{1,3\right\} \right\}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2010
    From
    Texas
    Posts
    48
    Quote Originally Posted by ecnanif View Post
    I donīt think I understand your corollary though.

    I updated:
    F= \left\{\emptyset,\Omega, \left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{1,2\right\}, \left\{2,3\right\}, \left\{1,3\right\} \right\}
    The corollary stating: If A,B \in \Sigma, then A \setminus B \in \Sigma means
    That since \Omega, \{1\} \in \Sigma, by the corollary this implies \Omega \setminus \{1\} = \{2,3\} \in \Sigma .
    He was just showing you how you could get your missing sets in your \sigma-algebra. Your updated set is now correct.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2010
    Posts
    32
    Ah okay =)
    Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sigma-algebra
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 26th 2011, 12:03 PM
  2. Replies: 1
    Last Post: February 4th 2011, 08:39 AM
  3. Show intersection of sigma-algebras is again a sigma-algebra
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 20th 2010, 07:21 AM
  4. sigma algebra over R
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: August 8th 2009, 03:35 PM
  5. Sigma-Algebra
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 7th 2009, 11:18 PM

Search Tags


/mathhelpforum @mathhelpforum