Show that the "algebra generated by a semi-algebra" is actually an algebra We know that algebra F generated by semi-algebra A of a set X is the collection of subsets of X which can be written as finite disjoint union of sets from A. The question was to prove this is an algebra.
I was able to show all things except the property that it is closed under complements. Idea is clear to me as it seems I have to take union of many sets from A but I am not getting how to put it down on paper.
 A: Let me know if your definition of semi-algebra and algebra is different; but my definition is (from here)

$\mathcal{A}$ is a semi-algebra if:
  
  
*
  
*$\varnothing$ and $X$ are in $\mathcal{A}$
  
*$\mathcal{A}$ is closed under finite instersection
  
*If $S \in \mathcal{A}$ then $S^c$ (the complement) is a finite disjoint union of sets in $\mathcal{A}$.

I understand that you defined $\mathcal{F}$ as the set of finite disjoint unions of sets from $\mathcal{A}$, and you showed that $\mathcal{F}$ satisfies all of the properties of an algebra, except that it remains to show closure under complement.
For this, consider an element $S \in \mathcal{F}$; we know
$$
S = S_1 \cup S_2 \cup \cdots \cup S_k,
$$
where $S_i \in \mathcal{A}$ for all $i$, and these sets are disjoint. So, that gives us
$$
S^c = (S_1^c \cap S_2^c \cap \cdots \cap S_k^c) = \bigcap_i S_i^c.
$$
Well that isn't quite helpful, because the $S_i^c$ are not necessarily elements of $\mathcal{A}$ (since $\mathcal{A}$ isn't necessarily closed under complement). But we haven't used property (3) above of a semi-algebra -- this tells us each $S_i^c$ can be written as a finite disjoint union of sets in $\mathcal{A}$. So we get
that each $S_i^c$ can be written as $S_{i,1} \cup S_{i,2} \cup \cdots \cup S_{i,k_i}$, where this union is disjoint, and where $k_i$ depends on $i$, and where each $S_{i,j}$ is in the semi-algebra $\mathcal{A}$.
$$
S^c = \bigcap_{i=1}^k \bigcup_{j=1}^{k_i}S_{i,j}.
$$
To show $S^c$ is in fact in $\mathcal{F}$, we want to show that it's a disjoint union of sets from $\mathcal{A}$, and right now we have an intersection. So we should distribute the intersection inside the union to get a union (the notation gets a bit messy)
$$
S^c = \bigcup_{j_1, j_2, \ldots, j_k} \bigcap_{i=1}^k S_{i,j_i}, \tag{1}
$$
where that union is taken over all possible choices of $j_1$ to $j_k$ (try a small example if that is confusing.)
I claim that (1) is a disjoint union of elements in $\mathcal{A}$. It's clearly a union, and each element is an intersection of $S_{i,j_i}$ terms, which is in $\mathcal{A}$ since $\mathcal{A}$ is closed under intersection. So what remains for you to show is that any two terms $\cap_{i=1}^k S_{i,j_i}$ are disjoint.
