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I want to show that the set $X = \{ \{\omega\} : \omega \in \Omega \}$ for $\Omega \neq \emptyset$ generates the $\sigma$-algebra $$\sigma(X) = \{ A \subset \Omega : A \text{ countable or } A^C \text{ countable}\}.$$

I don't know if the idea is correct: In $X$ we have all one-element subsets of $\Omega$. Now we do some set operations:

  • $\Omega = \bigcup_{w \in \Omega} \{w\}$, so $\Omega$ would be part of $\sigma(X)$.
  • Let $w_1, \dots, w_n \in \Omega$, then $\bigcup_{i=1}^n \{w_i\} = \{w_1, \dots, w_n\}$ is countable, so countable sets are part of $\sigma(X)$.
  • For each countable set $\{w_1, \dots, w_n\}$ the complement $\{w_1, \dots, w_n\}^C$ is part of $\sigma(X)$. If $\Omega$ is not countable, the complement of a countable set is uncountable. If a set is uncountable, it's complement must be countable, otherwise we cannot generate such a set with using $X$ (because we are only allowed to join countable many subsets).

My questions:

  • Is the idea correct that in order to generate $\sigma(X)$, we take the elements of $X$, do set $\sigma$-algebraic set operations on them and conclude that the resulting sets are in $\sigma(X)$?
  • How do I know the resulting $\sigma(X)$ is really the 'smallest' $\sigma$-algebra that contains $X$ (We have $X = \bigcap_{X \subset \mathcal{A}_i} \mathcal{A}_i$)?
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I guess you mixed some stuff up so let's start a bit slower… first of all: you don't need to show that $\sigma(X)$ is a sigma algebra because it's defined that way. It's the smalles sigma algebra that contains $X$, so especially it's a sigma algebra.

Now you have a set given $$\mathcal{C} = \{ A \subset \Omega : A \text{ countable or } A^C \text{ countable}\}$$

and now you want to show that $\mathcal{C}$ equals the sigma algebra generated by $X$, namely $\sigma(X)$.

Think about that $\mathcal{C}$ really defines a sigma algebra!

So you want to show that $\mathcal{C} = \sigma(X)$ holds. Equality of sets are always shown by showing that each set contains the other one, so here you have to show $\sigma(X) \subseteq \mathcal{C}$ and $\mathcal{C} \subseteq \sigma(X)$. Then the equation follows.

The first one is easy to verify: $X\subseteq \mathcal{C}$ holds because each element of $X$ is one-elemented so especially a countable one, so in $\mathcal{C}$. But $\sigma$ is a so called closure operator so it's increasing, so from $X\subseteq \mathcal{C}$ it follows that $\sigma(X)\subseteq \sigma(\mathcal{C})$. But because $\sigma(\mathcal{C})$ is the smallest sigma algebra that contains $\mathcal{C}$ and $\mathcal{C}$ itself is a sigma algebra it follows $\sigma(\mathcal{C}) = \mathcal{C}$. So indeed $\sigma(X)\subseteq \mathcal{C}$

The other way around is almost that easy: To show $\mathcal{C} \subseteq \sigma(X)$ you have to take an element $C\in \mathcal{C}$ and show that it's also element of $\sigma(X)$. But due to the fact that $C\in \mathcal{C}$ we know that either $C$ or it's complement is countable. w.l.o.g let it be $C$ itself. But because it's countable we can write it as $$C = \{\omega_1,\omega_2,\ldots\} = \bigcup_{n\in\Bbb N} \{\omega_n\}$$

So we can describe $C$ by only using elements of $X$ and "sigma operations" so it's definetly in $\sigma(X)$. So $\mathcal{C}\subseteq \sigma(X)$ and we are done.

Last but not least I want to have a look at some of your mistakes and questions above:

$\Omega = \bigcup_{w \in \Omega} \{w\}$ so $\Omega$ would be part of $\sigma(X)$

There is no assumption made that $\Omega$ is countable so your "operation" may be an uncountable one and so not allowed within a sigma algebra. $\Omega$ is part of $\sigma(X)$ because $\sigma(X)$ is a sigma algebra by definition and has to contain $\Omega$! Hope you got the difference…

Is the idea correct that in order to generate $\sigma(X)$, we take the elements of $X$, do set $\sigma$-algebraic set operations on them and conclude that the resulting sets are in $\sigma(X)$?

The idea is correct but it's false in general to assume that each element of $\sigma(X)$ can be written in Terms of sigma operations of elements of $X$. There can be sigma algebras where this is not true.

How do I know the resulting $\sigma(X)$ is really the 'smallest' $\sigma$-algebra that contains $X$

By definition! You have $$X = \bigcap_{X \subset \mathcal{A}_i} \mathcal{A}_i$$ where you intersect over all sigma algebras which contains X. So if you assume there is a sigma algebra "smaller" then $\sigma(X)$ it's part of the intersection and then $\sigma(X)$ has to be a subset of it. So assuming a "smaller" one, let's call it $\mathcal{D}$, hence $\mathcal{D} \subseteq \sigma(X)$, implies directly by definition also $\sigma(X) \subseteq \mathcal{D}$ and so $\mathcal{D} = \sigma(X)$.

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