# Verification of a semialgebra

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 $$\mathcal{I_0}=\underbrace{\{(a,b]\;|\;-\infty\le a\le b<+\infty\}}_{:=U}\cup\underbrace{\{(a_1,+\infty)\;|\;-\infty The family $$\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 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 $$2.$$?

Thanks!

$$\newcommand{I}{\mathcal{I}_0}$$ 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.)

• @Pepjin de Maat thanks for your answer. Yes, $E^c$ means the complement of $E$. – Jack J. Nov 12 '18 at 19:04
• @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 $V$. – Pepijn de Maat Nov 12 '18 at 19:11