Csdrhrt

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

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

$1.$ For each $E, Finmathcal{S}$ we have $Ecap Finmathcal{S}$;

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

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

Case 1.[$E, Fin U$]

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

Case 2.[$Ein U$ and $Fin V$]

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

Case 3.[$E,Fin V$]

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

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

Thanks!

$newcommand{I}{mathcal{I}_0}$
Given $Ein I$, since $Ucap V=emptyset$, we have two cases.

Case 1.[$Ein U$]
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 $F_1cap F_2=emptyset$ and
begin{align}
F_1&in UsubsetI\
F_2&in VsubsetI.
end{align}

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

Of course, $F_1in UsubsetI$.

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.)