Exercise 1.4.1 (Elementary Boolean algebra)

Question

Let $\mathcal{E}(\mathbb{R}^d)$ be the elementary Boolean algebra of $\mathbb{R}^d$, i.e., the collection of subsets of $\mathbb{R}^d$ which are either elementary  or co-elementary (complements of the elementary sets). Show that $\mathcal{E}(\mathbb{R}^d)$ is a Boolean algebra.

Elu Thingol · Accepted Answer

We verify the criteria from the Definition 1.4.1 (Boolean algebra).
\begin{enumerate}[label=(oman*)]
	\item (Empty set)
ewline
	The empty set is contained in $\mathcal{E}(\mathbb{R}^d)$ by definition as a trivial box $(a,a)^d=\emptyset$.

\item (Complement)
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	If $B\in\mathcal{E}(\mathbb{R}^d)$ then $B$ is either an elementary set, or a complement of an elementary set. In both cases $\mathbb{R}^d/B$ is contained in $\mathcal{E}(\mathbb{R}^d)$ by the definition given in the exercise.

\item (Finite unions)
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	Let $B,C \in \mathcal{E}(\mathbb{R}^d)$. We have three cases
	\begin{itemize}
		\item $B,C$ are both elementary
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		Then $B\cup C$ is elementary by \href{./exercise_1-1-1}{Exercise 1.1.1}.
		
		\item $B,C$ are both co-elementary
ewline
		Then we have $(B\cup C)^c = B^c \cap C^c$ is elementary, since $B^c,C^c$ are elementary, and the intersection of the elementary sets is again elementary by \href{./exercise_1-1-1}{Exercise 1.1.1}.
		
		\item $B$ is elementary, and $C$ is co-elementary
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		Then $B \cup C$ is co-elementary, since $B \cup C = (C^c/B)^c$. To see why
		\begin{align*}
		\mathbb{R}^d/(C^c/B) &= \mathbb{R}^d \cap (C^c \cap B^c)^c = \mathbb{R}^d \cap (C \cup B) = (C \cup B)
		\end{align*}
		By \href{./exercise_1-1-1}{Exercise 1.1.1}, $C^c/B$ is elementary.
	\end{itemize}
	
\end{enumerate}

Exercise 1.4.1 (Elementary Boolean algebra)

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Exercise 1.4.1 (Elementary Boolean algebra)

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