The question concerns the verification that a set family is a semialgebra.

Definition. A non empty family $\mathcal{S}\subseteq\mathcal{P}(X)$ is said semialgebra on $X$ if:

$1.$ For each $E, F\in\mathcal{S}$ we have $E\cap F\in\mathcal{S}$;

$2.$ For each $E\in\mathcal{S}$ exist $F_1,\dots F_n\in\mathcal{S}$ disjoint such that $E^c=\cup_{k=1}^n F_k$.

We consider the family $$\mathcal{I_0}=\underbrace{\{(a,b]\;|\;-\infty\le a\le b<+\infty\}}_{:=U}\cup\underbrace{\{(a_1,+\infty)\;|\;-\infty<a_1<+\infty\}}_{:=V}.$$ The family $\mathcal{I}_0$ is a semialgebra on $\mathbb{R}$. We observe that if $a=b$, then $(a,b]=\emptyset$, therefore $\emptyset\in \mathcal{I_0}$ by definition. We suppose that $a<b$ and we prove $1$.

Case 1.[$E, F\in U$]

$(a_1,b_1]\cap(a_2,b_2]=(\sup\{a_1,a_2\}, \min\{b_1,b_2\}]\in\mathcal{I_0}$.

Case 2.[$E\in U$ and $F\in V$]

$(a_1, b_1]\cap(a_2,+\infty)=(\sup\{a_1,a_2\}, b_1]\in \mathcal{I_0}$

Case 3.[$E,F\in V$]

$(a_1,+\infty)\cap (a_2,+\infty)=(\sup\{a_1,a_2\},+\infty)\in \mathcal{I_0}$.

Question. Could someone help me to prove the property $2.$?



$\newcommand{I}{\mathcal{I}_0}$ Given $E\in \I$, since $U\cap V=\emptyset$, we have two cases.

Case 1.[$E\in U$] \begin{align} E&=(a,b]\\ E^C&=\underbrace{(-\infty,a]}_{F_1}\cup \underbrace{(b,+\infty)}_{F_2} \end{align}

Notice that if $a=-\infty$, then $F_1=\emptyset$. It's trivial that $F_1\cap F_2=\emptyset$ and \begin{align} F_1&\in U\subset\I\\ F_2&\in V\subset\I. \end{align}

Case 2.[$E\in V$] \begin{align} E&=(a,+\infty)\\ E^C&=\underbrace{(-\infty,a]}_{F_1} \end{align}

Of course, $F_1\in U\subset\I$.


From the context, I guess that $E^c$ means the complement of $E$ (not the closure). In that case, since the lower boundary may be $-\infty$ for the sets in $U$, I think that writing out the complement as $(-\infty, a] \cup (b, \infty)$ or $(-\infty, a_1]$ (resp.) should give property 2 in a quite straight-forward manner.

(I wanted to give this as a comment, but I do not have sufficient reputation.)

  • $\begingroup$ @Pepjin de Maat thanks for your answer. Yes, $E^c$ means the complement of $E$. $\endgroup$ – Jack J. Nov 12 '18 at 19:04
  • $\begingroup$ @JackJ. In that case, note that $(-\infty, a], (-\infty, a_1) \in U$ and $(b, \infty) \in V$ and that for $a \leq b$, we have $(-\infty, a]$ is disjoint with $(b, \infty)$; hence property 2 has been satisfied. The main exceptions are boundary cases like $b = -\infty$, which can be solved by $(-\infty, 0] \cup (0, \infty)$ which is again a combination of a subset in $U$ and one in $V$. $\endgroup$ – Pepijn de Maat Nov 12 '18 at 19:11

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