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Suppose $X$ is a set and $\mathcal{A}$ is the set of subsets of $X$ that consist of exactly one element: $$\mathcal{A} = \{\{x\}:x \in X\}.$$ Prove that the smallest $\sigma$-algebra on $X$ containing $\mathcal{A}$ is the set of all subsets $E$ of $X$ such that $E$ is countable or $E^c$ is countable.

To prove this, I first want to show that $\mathcal{S}$ is a $\sigma$-algebra where $$\mathcal{S} = \{E\subset X: E \text{ is countable or } X\setminus E \text{ is countable}\}.$$

My question: I've already shown that $\mathcal{S}$ is a $\sigma$-algebra. Now, I think I need to show that $\mathcal{A}$ is contained in $\mathcal{S}$. How can I show this (since both $\mathcal{A}$ and $\mathcal{S}$ are sets of sets)? So, how can $\mathcal{A}$ be contained in $\mathcal{S}$? Also, lastly, I think I need to show that $\mathcal{S}$ is the intersection of all $\sigma$-algebras on $X$ that contain $\mathcal{A}$. Can someone show how this can be proven?

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    $\begingroup$ $\mathcal A$ is contained in $\mathcal S$ means $\mathcal A \subseteq \mathcal S$. You can proceed with the smallest $\sigma$-algebra part by showing that any $\sigma$-algebra containing $\mathcal A$ must contain $\mathcal S$. $\endgroup$ Commented Mar 10, 2021 at 17:17
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    $\begingroup$ @IzaakvanDongen To show that $\mathcal{A}\subset\mathcal{S}$, I have: Let $P\in\mathcal{A}$. Then $P=\{x\}$ for some $x\in X$. Since $P$ is finite, and hence countable, $P\in\mathcal{S}$. Is this all that was needed to show that $\mathcal{A}\subset\mathcal{S}$? $\endgroup$ Commented Mar 10, 2021 at 17:24
  • $\begingroup$ Yes, that's exactly right! $\endgroup$ Commented Mar 10, 2021 at 17:26
  • $\begingroup$ @Ricky_Nelson That works, but that's not the interesting part of the proof. What you need to show is that if $\mathcal A\subset\mathcal M$, and $\mathcal M$ is a $\sigma$-algebra, then $S\subset\mathcal M$. $\endgroup$ Commented Mar 10, 2021 at 17:26
  • $\begingroup$ @DonThousand Can you please elaborate on how this can be shown? Also, according to your argument, would we not need to show that $\mathcal{A}\subset\mathcal{S}$? $\endgroup$ Commented Mar 10, 2021 at 17:31

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As you said you have already proved that $$\mathcal{S} = \{E\subseteq X: E \text{ is countable or } X\setminus E \text{ is countable}\}$$ is in fact a $\sigma$-algebra.

Now, we have $\mathcal{A} = \{\{x\}:x \in X\}$ and let us prove that the smallest $\sigma$-algebra containing $\mathcal{A}$ is equal to $\mathcal{S}$.

Since for each $x \in X$ , $\{x\}$ is countable, we have that $ \mathcal{A} \subseteq \mathcal{S}$.

Since $\mathcal{S}$ is a $\sigma$-algebra, it follows that $$ \sigma(\mathcal{A}) \subseteq \mathcal{S} \tag{1}$$ where $\sigma(\mathcal{A})$ is the smallest $\sigma$-algebra containing $\mathcal{A}$ (the $\sigma$-algebra generated by $\mathcal{A}$).

Now let us prove that $ \mathcal{S} \subseteq \sigma(\mathcal{A}) $.

We will begin by proving the claim:

Every countable set $F\subseteq X$ is in $\sigma(\mathcal{A})$.

Proof of the claim: Since for every $x \in X$ , $\{x\} \in \mathcal{A} \subseteq \sigma(\mathcal{A})$ and every countable set $F\subseteq X$ is a countable union of single-point sets, we have that $F \in \sigma(\mathcal{A})$ (because $\sigma(\mathcal{A})$ is a $\sigma$-algebra). So, every countable subset of $X$ is in $\sigma(\mathcal{A})$ . $\square$

Now, for any $E \in \mathcal{S}$, we have two possibilities:

  1. $E$ is countable. In this case, it follows directly from the claim that $E \in \sigma(\mathcal{A})$.
  2. $X\setminus E$ is countable. In this case, it follows directly from the claim that $X\setminus E \in \sigma(\mathcal{A})$. Since $\sigma(\mathcal{A})$ is a $\sigma$-algebra, we have $X\setminus(X\setminus E) \in \sigma(\mathcal{A})$. But $X\setminus(X\setminus E) =E$, so we have $E \in \sigma(\mathcal{A})$.

So, in both cases, we have that $E \in \sigma(\mathcal{A})$. So, we have:

$$ \mathcal{S}= \{E\subseteq X: E \text{ is countable or } X\setminus E \text{ is countable}\} \subseteq \sigma(\mathcal{A}) \tag{2}$$

From $(1)$ and $(2)$, $$ \sigma(\mathcal{A}) = \mathcal{S} $$ It means, the smallest $\sigma$-algebra containing $\mathcal{A}$ is equal to $\mathcal{S}$.

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  • $\begingroup$ Thanks for your answer! One question: how does $X\setminus E$ being countable imply that $X\setminus E \in \sigma(\mathcal{A})$? (This is right above (2) in your answer.) $\endgroup$ Commented Mar 10, 2021 at 23:38
  • $\begingroup$ @Ricky_Nelson Just before that sentence, we have proved that : "Now, since for every $x \in X$ , $\{x\} \in \mathcal{A} \subseteq \sigma(\mathcal{A})$ and every countable set $E\subseteq X$ is a countable union of single-point sets, we have that $E \in \sigma(\mathcal{A})$ (because $\sigma(\mathcal{A})$ is a $\sigma$-algebra). " So, every countable set is in $\sigma(\mathcal{A})$. $\endgroup$
    – Ramiro
    Commented Mar 11, 2021 at 0:52
  • $\begingroup$ @Ricky_Nelson I have updated the answer to include more details. Please, take a look and let me know if you have any further question. $\endgroup$
    – Ramiro
    Commented Mar 11, 2021 at 1:17
  • $\begingroup$ Makes perfect sense now, thanks so much! $\endgroup$ Commented Mar 11, 2021 at 1:17

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