Exercise 1.4.7 (Elementary Boolean algebra is generated by boxes)

Question

Let $\mathcal{E}(\mathbb{R}^d)$ be the elementary Boolean algebra. Show that it is generated by the boxes $\mathcal{B}(\mathbb{R}^d)$ in $\mathbb{R}^d$.

Elu Thingol · Accepted Answer

We demonstrate that
$$\mathcal{E}(\mathbb{R}^d) = \bigcap \left\{ \mathcal{F} \in 2^{\mathbb{R}^d}: \mathcal{B}(\mathbb{R}^d) \subseteq \mathcal{F} \right\}.$$
\begin{itemize}
\item[$\subseteq$)] Let $E \in \mathcal{E}(\mathbb{R}^d)$. Then $E = \bigcup_{i=1}^{n}B_n$ for some disjoint collection of boxes $B_1,\ldots,B_n\in\mathcal{B}(\mathbb{R}^d)$. By the definition of a Boolean algebra, any Boolean algebra $\mathcal{F}$ from the right-hand side containing $\mathcal{B}$, and thus $B_1,\ldots,B_n$, must also contain their union $\bigcup_{i=1}^{n}B_n = E$. Thus, $E$ is also contained in the intersection of any such $\mathcal{F} \supseteq \mathcal{B}(\mathbb{R}^d)$.
\item[$\supseteq)$] Notice that the elementary Boolean algebra $\mathcal{E}(\mathbb{R}^d)$ is itself a Boolean algebra on $\mathbb{R}^d$ by \href{./exercise_1-4-1}{Exercise 1.4.1}, and it contains $\mathcal{B}(\mathbb{R}^d)$ by definition. By the properties of intersection, $\bigcap \{ \mathcal{F} \in 2^{\mathbb{R}^d}: \mathcal{B}(\mathbb{R}^d) \subseteq \mathcal{F} \} \subseteq \mathcal{E}(\mathbb{R}^d)$.
\end{itemize}

Exercise 1.4.7 (Elementary Boolean algebra is generated by boxes)

Answers

Comments

Exercise 1.4.7 (Elementary Boolean algebra is generated by boxes)

Answers

Comments

Add answer