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Given the set $S =\{1, 2, 3\}$ and the fact that $\sigma$-algebra is generated by its single subset $A = \{1\} \subset S$ and is denoted by $\sigma(\{1\})$ or by $\sigma(A)$. I don't really understand how $\sigma(A)$ is constructed as well as whether the power set of $S$, $\mathcal{F}(S) = 2^S$, is a $\sigma$-algebra?

I have tried to check the $\sigma$-algebra $\mathcal{F}$ properties such as: $$\emptyset \in \mathcal{F}$$ $$A \in \mathcal{F} \Rightarrow A^c \in \mathcal{F}$$

$$ A_1,A_2, \dots, A_n, \dots \in \mathcal{F} \Rightarrow \bigcup\limits_{k=1}^{\infty} A_k \in \mathcal{F} $$

Therefore, I assumed that $\sigma(\{1\})$ satisfies the properties and therefore may be called $\sigma(A)$. In regard to the power set, I assumed that a legit power set is in the form of $\mathcal{F}(S) = 2^\Omega$ so the question is whether $S=\Omega$? Here I'm a bit confused.

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2 Answers 2

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The $\sigma$-algebra $\sigma(\{1\})$ on $S = \{1,2,3\}$is the least subset of $\mathscr{P}(\{1,2,3\})$ which

  1. has $\emptyset$ as a member,
  2. is closed under $S$-complements and countable unions, and
  3. has $\{1\}$ as a member.

It's easy to check that the following collection meets those three conditions: $$ \mathscr{S} = \{\emptyset, \{1\}, \{2,3\}, \{1,2,3\} \} \text{.} $$ So $\sigma(\{1\}) \subseteq \mathscr{S}$. But clearly every $X \in \mathscr{S}$ is also a member of $\sigma(\{1\})$, so $\sigma(\{1\}) = \mathscr{S}$.

Because for example $\{2\} \notin \sigma(\{1\})$, $\sigma(\{1\}) \neq \mathscr{P}(\{1,2,3\})$.

It's trivial (yes?) to confirm that for any set $X$, $\mathscr{P}(X)$ is a $\sigma$-algebra: $\emptyset$ is a member; it's closed under complements relative to $X$, and closed under countable unions (the union of any family of subsets of $X$ is a subset of $X$).

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  • $\begingroup$ I'm sorry, but for me its not as trivial. I dont understand why every $X$ is a member of $\sigma$ $(\{1\})$ because you show that $\{2\}$ its not a member of it... $\endgroup$
    – Mamba
    Commented Oct 29, 2015 at 17:18
  • $\begingroup$ @Mamba I said nothing of the sort. That's two misunderstandings in one sentence, actually. I did not say that every $X$ is a member of $\sigma(\{1\})$. Every X from where? Please reread my last sentence. There, think of $X$ the underlying set $\{1,2,3\}$. "The collection of all subsets of any set $X$ is a $\sigma$-algebra": just check that it contains $X$, is closed under complement relative to $X$, and closed under countable unions. This really should be obvious: the result of each of those operations, applied to subsets of $X$, is a subset of $X$. $\endgroup$
    – BrianO
    Commented Oct 29, 2015 at 17:26
  • $\begingroup$ ah ok I'm sorry. In that context, it implies that $2^s$ is a $sigma$-algebra as well because the power set of $S= \{\},\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}$ $\endgroup$
    – Mamba
    Commented Oct 29, 2015 at 17:33
  • $\begingroup$ @Mamba That's right :) $\endgroup$
    – BrianO
    Commented Oct 29, 2015 at 17:35
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The power set of $S$ is indeed a $\sigma$-algebra containing $S$. It is the largest such $\sigma$-algebra.

But the $\sigma$-algebra generated by $S$ is the smallest $\sigma$-algebra containing $S$ (in particular, it is the intersection of all such $\sigma$-algebras).

In this case, they are not equal, since, for example, $\{2\}$ lies in the power set of $\{1,2,3\}$ but not in $\sigma(\{1\})$.

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  • $\begingroup$ Sorry, I'm bit confused. Why are you interested in the power set of $A$ rather in the power set of $S$? $\endgroup$
    – Mamba
    Commented Oct 29, 2015 at 17:07
  • $\begingroup$ Hehe...Im also lost :) $\endgroup$
    – Googme
    Commented Oct 29, 2015 at 17:21
  • $\begingroup$ I'm not. I meant $S$. My bad. $\endgroup$
    – MPW
    Commented Oct 29, 2015 at 18:09

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