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I am attempting to work through the problem below and am experiencing some difficulty in solving it, and moreover, in understanding it. I would appreciate any insights.

Problem

Use the fact that $\mathbb{R}$ is a countable union of open intervals to show that $\mathcal{B}_{\mathbb{R}}$ is generated by the collection of all open intervals.

Confusions

My naive belief is that this statement just follows from the definition of $\mathcal{B}_{\mathbb{R}}$ - "Let $E$ be a topological space, then the $\sigma$-algebra generated by the collection of all open subsets of $E$ is called the Borel $\sigma$-algebra on $E$; it is denoted $\mathcal{B}_{\mathbb{E}}$" - as each open interval of $\mathbb{R}$ is an open subset of $\mathbb{R}$. For this problem should I verify that for any open interval in $\mathcal{B}_{\mathbb{R}}$, the complement of that interval is also in $\mathcal{B}_{\mathbb{R}}$? As you can see, I am somewhat lost.

Attempt Following Assistance

$\text{ }$ Let $\mathcal{C}$ be the collection of all open intervals of $\mathbb{R}$. By definition, $\sigma \mathcal{C}$ is the smallest $\sigma$-algebra containing $\mathcal{C}$. Let $\mathcal{D}$ denote the collection of all open subsets of $\mathbb{R}$. We know that $\mathcal{B}_{\mathbb{R}}$ is the smallest $\sigma$-algebra containing $\mathcal{D}$.

$\text{ }$ To begin, consider any open interval in $\mathcal{C}$. Every open interval of $\mathbb{R}$ is also an open subset of $\mathbb{R}$, this open interval is contained in $\mathcal{D}$, thus $\mathcal{C} \subseteq \mathcal{D}$.

$\text{ }$ Now consider any open subset in $\mathcal{D}$. Since every open subset in $\mathcal{D}$ can we written as a countable union of open intervals of $\mathbb{R}$ and $\sigma \mathcal{C}$ is closed under countable unions of open intervals of $\mathbb{R}$, we have that $\mathcal{D} \subseteq \sigma \mathcal{C}$.

$\text{ }$ Since $\mathcal{C} \subseteq \mathcal{D} \subseteq \sigma \mathcal{C}$, we have that $\sigma \mathcal{C} = \mathcal{B}_{\mathbb{R}}$, i.e. $\mathcal{B}_{\mathbb{R}}$ is generated by all open intervals.

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  • $\begingroup$ See the difference? Generated by all open intervals; generated by all open sets. $\endgroup$
    – GEdgar
    Apr 15, 2021 at 20:13

2 Answers 2

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Let $$\Delta = \{(a,b): a,b \in \mathbb{R}, \; a < b\}$$ What the problem asks essentially is to show that $\sigma(\Delta) = \mathcal{B}(\mathbb{R})$. What follows form the definition, as you said, is that $\sigma(\Delta) \subseteq \mathcal{B}(\mathbb{R})$. That's because every interval $(a,b) \in \Delta$ is an open subset of $\mathbb{R}$ so it belongs to $\mathcal{B}(\mathbb{R})$. Since $\sigma(\Delta)$ is the minimal $\sigma$-algebra that contains $\Delta$ and $\mathcal{B}(\mathbb{R})$ is a $\sigma$-algebra that contains $\Delta$ from the minimality of $\sigma(\Delta)$ we get $\sigma(\Delta) \subseteq \mathcal{B}(\mathbb{R})$.

For the other inclusion remember that every open subset of $\mathbb{R}$ can be written as a countable union of open intervals. This should give you that the collection of open subsets of $\mathbb{R}$ is a subset of $\sigma(\Delta)$. Then use a similar minimality argument like the above.

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  • $\begingroup$ This makes sense. The idea of showing $\sigma \mathcal{C} \subseteq \mathcal{B}_{\mathbb{R}}$ and the converse to show equality just didn't really cross my mind. I haven't full read through your answer but will use this idea to try to attempt it myself and add it as a comment. I have upvoted the question and will check it a bit later to motivate others to participate. $\endgroup$ Apr 15, 2021 at 20:29
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    $\begingroup$ @rodeo_flagellum Happy to have helped! If something is still unclear after attempting a proof do let me know. $\endgroup$ Apr 15, 2021 at 20:35
  • $\begingroup$ I attempted the proof. What are your thoughts on it? Also, thank you again for this thorough response and the time you spent helping me learn, I appreciate it. Also, question: I am a bit confused on the second part --> is there a way to go directly from saying that every open subset is a countable union of open intervals so each element of $\mathcal{D}$ is also is $\sigma \mathcal{C}$ to saying that $\mathcal{B}_{\mathbb{R}} \subseteq \sigma \mathcal{C}$? $\endgroup$ Apr 15, 2021 at 21:18
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    $\begingroup$ @rodeo_flagellum I don't think you can say it directly without using such a minimality argument. It's quite short though. Let $U$ be the collection of open sets. $\mathcal{B}(\mathbb{R}) = \sigma(U)$ and since $\sigma(U)$ is the minimal $\sigma$-algebra that contains $U$ and $\sigma(\Delta)$ contains $U$ we have the required inclusion. $\endgroup$ Apr 15, 2021 at 21:24
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    $\begingroup$ That explanation is great. Once again, thank you. $\endgroup$ Apr 15, 2021 at 21:26
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Recall that the $\sigma$-algebra generated by a collection $F$ of subsets of $X$ is, by definition, the intersection of all $\sigma$-algebras containing $F$, (this intersection is a $\sigma$-algebra).

So you would have to show that if $\Sigma$ is a $\sigma$-algebra of subsets of $\mathbb{R}$ containing every open interval, then $\Sigma$ contains every open subset of $\mathbb{R}$. Indeed, $\mathcal{B}_\mathbb{R}$ is defined as the $\sigma$-algebra generated by the collection of open subsets of $\mathbb{R}$.

One way to prove this is to note that an open subset of $\mathbb{R}$ is a union of open intervals with rational endpoints. And since $\mathbb{Q}$ is countable this is necessarily a countable union. So you are done since $\sigma$-algebras are closed under countable unions.

N.B. I'm not sure what the problem as stated wants you to do, after all $\mathbb{R}$ is not just a countable union of open intervals, it is itself an open interval.

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  • $\begingroup$ Thank you for your thoughts and comments! $\endgroup$ Apr 15, 2021 at 21:19

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