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Thread: sigma-algebra

  1. #1
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    sigma-algebra

    Hi!

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

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

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

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

    Thanks!
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  2. #2
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    Thats not quite right.

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

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

    Proof:

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


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

    You are however, missing some sets in this. Consider this corollary to the previous result: if $\displaystyle A,B\in \Sigma$ then $\displaystyle A \setminus B \in \Sigma.$ This means that for example $\displaystyle \{2,3\} = \Omega \setminus\{1\}\in \Sigma$ along with a few other sets.
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  3. #3
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    I think I understood your claim and proof. You say that if we have a sequence of sets in a $\displaystyle \sigma$-algebra $\displaystyle F$ their intersection must also lie in $\displaystyle F$. Because all the complements of the sets must lie in $\displaystyle F$, we get that the union of all complements must lie in $\displaystyle F$, and because of this the complement of the union of complements must lie in $\displaystyle F$. And this is equal to the intersection of our original sets.

    I donīt think I understand your corollary though.

    I updated:
    $\displaystyle 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\} $
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  4. #4
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    Quote Originally Posted by ecnanif View Post
    I donīt think I understand your corollary though.

    I updated:
    $\displaystyle 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 $\displaystyle A,B \in \Sigma, then A \setminus B \in \Sigma$ means
    That since $\displaystyle \Omega, \{1\} \in \Sigma$, by the corollary this implies $\displaystyle \Omega \setminus \{1\} = \{2,3\} \in \Sigma $.
    He was just showing you how you could get your missing sets in your $\displaystyle \sigma$-algebra. Your updated set is now correct.
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  5. #5
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    Ah okay =)
    Thank you!
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