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

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.

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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).

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$S$ 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?

2. 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!

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The family of all bounded intervals is a semi-ring on $\mathbb{R}$ but not a semi-algebra (since the complement is unbounded and hence not a finite union of bounded sets). – Michael Greinecker Jan 6 '13 at 0:47

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.

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Thanks! (1) How is $C\subseteq X$ closed under $(f_i: X^{\kappa_i} \to X)_{i \in I}$ related to the example for semi-algebras? (2) For semi-algebras, why do "there do no unique decomposition of the complement into disjoint sets (even then, one would need a hack)" and "a function can have only one value" matter? What is "a function"? – Tim Jan 6 '13 at 14:33
@Tim I've edited it and hope have clarified these issues. – Michael Greinecker Jan 6 '13 at 22:19
Thanks! I edited some places I thought were typos. Feel free to edit again if I am wrong. I was wondering if the definition of a set operation having closure property comes from some references? – Tim Jan 7 '13 at 14:02
@Tim All things closure operations I learned from the Handbook of Analysis and its Foundations by E. Schechter. – Michael Greinecker Jan 7 '13 at 14:13
Thanks! Besides the two books, what other books in analysis do you recommend? – Tim Jan 11 '13 at 21:36