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$$ S_n = \mathscr P (\{ -n, -n+1, \ldots, n-1, n\}) $$ $$ R_n = \{r : \Omega - r \in S_n\} $$ $$ T_n = S_n \cup R_n$$

I need to check whether

  1. $T_n$ is an algebra, semi-algebra or sigma algebra.

  2. $T_n \subset T_{n+1}$.

  3. If $T = \bigcup_n T_n$, whether $T$ is algebra, semi-algebra, sigma algebra.

I considered an example for this:

Let $ S_1 = \mathscr P \{-1, 0, 1\} $ $$ S_1 = \{\{-1\}, \{0\}, \{1\} ,\{-1,0\}, \{0,1\}, \{-1,1\}, \{-1,0,1\},\{ \varnothing\}\}$$

So, for 1) I feel that $T_n = \Omega$, so it can either be any of the algebras.

Please advise.

  • 0
    Which part of the definitions are you having trouble checking?2012-09-16
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    Still, I want to check my idea for 1) is correct.2012-09-16
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    What is $\Omega$? Is it $\mathbb{Z}$? And do you mean that $T_n = \mathscr{P}(\Omega)$ for all $n$? If so, why?2012-09-16
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    Does what $\Omega$ is matter here? Since $T_n = S_n \cup R_n$ doesn't $T_n$ automatically becomes $\Omega$.2012-09-16
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    I think you may be confused about what the definitions in the problem mean. The symbol $\mathscr{P}$ means "power set," so for example $S_1$ consists of all eight subsets of the set $\{-1,0,1\}$.2012-09-16
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    ...and $R_n$ is the set of complements of elements of $S_n$, which is not the same as the complement of $S_n$ itself.2012-09-16
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    Yeah, I have edited the problem to include that example. So R becomes the complement sets of S. So when we take their union again...doesn't that make up to $\Omega$ again. I am confused!2012-09-16
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    $S_n$ and $R_n$ both consist of subsets of $\Omega$, not elements of $\Omega$. Their union $T_n$ also consists of subsets of $\Omega$, not elements of $\Omega$. Maybe if you write down some of the elements of $R_1$ it will become clearer.2012-09-16

1 Answers 1

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  1. $T_n$ is stable by complementation (almost by definition) and by intersections: if $A, B\in T_n$ and $A,B\in S_n$ then $A\cap B\in S_n$, if $A,B\in R_n$ then $A\cap B\in R_n$ and if $A\in R_n$ and $B\in S_n$ then $A\cap B=\emptyset$. So $T_n$ is a semi-algebra and since $\Omega\in T_n$, an algebra. $T_n$ is a $\sigma$-algebra as a finite algebra.

  2. Since $S_n\subset S_{n+1}$, if $A\in T_n$, either $A\in S_n$, then $A\in S_{n+1}\subset T_{n+1}$ or $A\in R_n$, then $\Omega\setminus A\in S_n\subset S_{n+1}$ hence $A\in T_{n+1}$, which proves inclusion.

  3. An increasing sequence of (semi-)algebras is a (semi-)algebra so $T$ is a (semi-)algebra. It's not a $\sigma$-algebra as $\{2k,k\in\Bbb Z\}$ is a countable union of elements of $T$ but not an element of $T$.