Why isn't the family of semi-algebras (aka semi-rings) of sets closed under intersection? 
*

*A link says: 

Any type of algebraic structure on subsets of $S$ that is defined
  purely in terms of closure properties will be preserved under
  intersection. Examples are σ-algebras, π-systems, λ-systems, or monotone classes of subsets.
...
Note however, this does not apply to semi-algebras,
  because the semi-algebras is not defined purely in terms of closure
  properties (the condition on $A^c$ is not a closure property).
...
$S$ is said to be a semi-algebra if it is closed under intersection
  and if complements can be written as finite, disjoint unions:
  
  
*
  
*If $A,B∈S$ then $A∩B∈S$.
  
*If $A∈S$ then there exists a finite, disjoint collection $\{B_i:i∈I\}⊆S$ such that $A^c=⋃_{i∈I} B_i$.
  

In "the condition on $A^c$ is not a closure property", 


*

*what does "the condition on a set operation such as taking complement is not a closure property" mean?

*What is the meaning of "closure properties"? 
How do you see the family of semi-algebras (aka semi-rings) of sets
isn't closed under intersection?

*Michael Greinecker also commented:  The family of semi-rings on a
set are not closed under intersections. 
BTW, if I am correct,
the concept of a semi-algebra of sets is the same as semi-ring of
sets in Wikipedia.
Thanks and regards!
 A: Closure properties can be formulated in terms of concepts from universal algebra. Let $X$ be the underlying set (in our examples, $X$ is a famly of sets itself). Let $I$ be an index set, $(\kappa_i)_{i\in I}$ be a family of cardinal numbers and $(f_i)_{i\in I}$ a family of function satisfying $f_i:X^{\kappa_i}\to X$ for all $i$. We say that $C\subseteq X$ is closed under $(f_i)_{i\in I}$ if we have for all $i\in I$ that $f_i(x)\in C$ for all $x\in C^{\kappa_i}$. One can show that the family of sets closed under $(f_i)_{i\in I}$ forms a Moore collection. 
Let's look an an example: Let $U$ be a set and $X\subseteq 2^U$. We let $I=\{s,c,u\}$, $\kappa_s=0$, $\kappa_c=1$, and $\kappa_u=\omega$. We identify constants and nullary functions, so we can let $f_s=U$. We let $f_c(A)=A^C$ for all $A\in X$, and we let $f_u(A_0,A_1,\ldots)=\bigcup_n A_n$. That $X$ is closed under these three functions means simply that it contains $X$, is closed under complements and countable unions- it is a $\sigma$-algebra.  
Now, one cannot write down semi-algebras this way, since there is no unique decomposition of the complement into disjoint sets. If $\mathcal{S}$ is a semi-algebra and $A\in\mathcal{S}$, then there exists a number $n$ and sets $B_1,\ldots,B_n\in\mathcal{S}$ that are disjoint and such that $A_c=B_1\cup\ldots\cup B_n$. Now if there exists a unique such family and if this family only depended on $A$, we could write down this property as closure under some functions in the following way: We let $f_{c_1}=B_1,\ldots, f_{c_n}=B_n$, and for $m>n$ we let $f_{c_m}=f_{c_n}$. We use the last condition because we have no a priori bound on how many sets are needed. But these sets are not a function of $A$, so this property can not be viewed as a closure property. 
Here is an explicit example (taken from Alprantis & Border) that shows that the intersection of sem-algebras might fail to be a semi-algebra: Let $X=\{0,1,2\}$, $\mathcal{S}_1=\big\{\emptyset, X,\{0\},\{1\},\{2\}\big\}$, $\mathcal{S}_2=\big\{\emptyset, X,\{0\},\{1,2\}\big\}$, and $A=\{0\}$. We have $\mathcal{S}_1\cap\mathcal{S}_2=\big\{X,\emptyset,\{0\}\big\}$, and $A^C=\{0\}^C=\{1,2\}$ is not the disjoint union of elements of this intersection.
