Describe smallest algebra, monotone class, $\sigma$-algebra I'm trying to understand better the concepts of monotone classes, algebras and  $\sigma$-algebras so I came into the following problem. For the family
$E := \{∅, \mathbb{N}, \{2\}, \{2, 4\}, \{2, 4, 6\}, . . .\} $
 of subsets of $\mathbb{N}$, describe: 


*

*$M(E)$ - the smallest monotone class that contains E

*$A(E)$ - the smallest algebra that contains E

*$\sigma(E)$ - the smallest $\sigma$-algebra that contains E

 A: I’ll use $\mathscr{E}$ in place of your $E$, and for $n\in\Bbb Z^+$ I’ll let $E_n=\{2k:k\in\Bbb Z^+\text{ and }k\le n\}$. I suspect that your $\Bbb N$ doesn’t include $0$; mine does, so I’ll use $\Bbb Z^+$ instead. Then $$\mathscr{E}=\{\varnothing,\Bbb Z^+\}\cup\{E_n:n\in\Bbb Z^+\}\;.$$
A monotone class $\mathscr{M}$ of subsets of $\Bbb N$ is a collection of subsets of $\Bbb N$ that is closed under countable monotone unions and intersections. That is, suppose that $M_k\in\mathscr{M}$ for $k\in\Bbb N$; if $M_k\subseteq M_{k+1}$ for each $k\in\Bbb N$, then $\bigcup_{k\in\Bbb N}M_k\in\mathscr{M}$, and if $M_k\supseteq M_{k+1}$ for each $k\in\Bbb N$, then $\bigcap_{k\in\Bbb N}M_k\in\mathscr{M}$. 
Note that $\mathscr{E}$ is a nested family: for any $A,B\in\mathscr{E}$, either $A\subseteq B$, or $B\subseteq A$. If $M_k\in\mathscr{E}$ for $k\in\Bbb N$, and $M_k\subseteq M_{k+1}$ for all $k\in\Bbb N$, then either there is an $n\in\Bbb N$ such that $M_k=M_n$ for all $k\ge n$, in which case $\bigcup_{k\in\Bbb N}M_k=M_n\in\mathscr{E}$, or $\bigcup_{k\in\Bbb N}M_k=\{2n:n\in\Bbb Z^+\}$, the set of positive even integers. If $M_k\supseteq M_{k+1}$ for each $k\in\Bbb N$, then there is an $n\in\Bbb N$ such that $M_k=M_n$ for all $k\ge n$, in which case $\bigcap_{k\in\Bbb N}M_k=M_n\in\mathscr{E}$. Thus, all of the sets to make $\mathscr{E}$ a monotone class are already in $E$ except $E=\{2n:n\in\Bbb Z^+\}$, and you can easily check that $\mathscr{E}\cup\{E\}$ is a monotone class.
An algebra of sets is closed under pairwise unions and intersections and under taking complements. $E$ contains only one pair of complementary sets, $\varnothing$ and $\Bbb N$, so we clearly have to throw in the complements of the sets $E_n$ for $n\in\Bbb Z^+$; let $C_n=\Bbb Z^+\setminus E_n$. For $m,n\in\Bbb Z^+$ with $m<n$ we have $E_m\cup E_n=E_n$, $E_m\cap E_n=E_m$, 
$$C_m\cup C_n=(\Bbb Z^+\setminus E_m)\cup(\Bbb Z^+\setminus E_n)=\Bbb Z^+\setminus(E_m\cap E_n)=\Bbb Z^+\setminus E_m=C_m\;,$$
and
$$C_m\cap C_n=(\Bbb Z^+\setminus E_m)\cap(\Bbb Z^+\setminus E_n)=\Bbb Z^+\setminus(E_m\cup E_n)=\Bbb Z^+\setminus E_n=C_n\;,$$
all of which are in $\mathscr{E}\cup\{C_n:n\in\Bbb Z^+\}$. Moreover, $C_m\cap E_n=\varnothing\in\mathscr{E}$, and $C_n\cup E_m=\Bbb Z^+\in\mathscr{E}$. However,
$$C_n\cap E_m=\{2k:m<k\le n\}\;,$$
so
$$C_m\cup E_n=(\Bbb Z^+\setminus E_m)\cup(\Bbb Z^+\setminus C_n)=\Bbb Z^+\setminus(E_m\cap C_n)=\Bbb Z^+\setminus\{2k:m<k\le n\}\;,$$
and we need to add these sets as well. For $m,n\in\Bbb Z^+$ with $m\le n$ let $D_{m,n}=\{2k:m\le k\le n\}$; then $C_n\cap E_m=D_{m+1,n}$. Note that $E_n=D_{1,n}$ for each $n\in\Bbb Z^+$.
Let 
$$\mathscr{A}_0=\{\varnothing,\Bbb Z^+\}\cup\{D_{m,n}:m,n\in\Bbb Z^+\text{ and }m\le n\}\cup\{\Bbb Z^+\setminus D_{m,n}:m,n\in\Bbb Z^+\text{ and }m\le n\}\;;$$
we’re making progress, but $\mathscr{A}_0$ still isn’t an algebra, because (for instance) $$D_{2,2}\cup D_{4,4}=\{4\}\cup\{8\}=\{4,8\}\notin\mathscr{A}_0\;,$$ and neither is the complement of this set. Let 
$$\mathscr{D}=\{D_{m,n}:m,n\in\Bbb Z^+\text{ and }m\le n\}\;,$$
and let $\mathscr{A}$ be the set of all unions of finite subfamilies of $\mathscr{D}$ together with all complements of such unions. It’s clear that every member of $\mathscr{A}$ must belong to the smallest algebra containing $\mathscr{E}$. And since $D_{1,1}\cap D_{2,2}=\varnothing$, its complement $\Bbb Z^+$, and each $E_n=D_{1,n}$ are in $\mathscr{A}$, we have $\mathscr{E}\subseteq\mathscr{A}$. Notice that the members of $\mathscr{D}$ are just the finite sets of even positive integers, so $\mathscr{A}$ contains precisely these sets and their complements. It’s now very easy to check that $\mathscr{A}$ is closed under pairwise unions and intersections and taking complements, so $\mathscr{A}$ is the smallest algebra of sets containing $\mathscr{E}$. 
$\mathscr{A}$ is not a $\sigma$-algebra because (for instance) $E=\bigcup_{n\in\Bbb Z^+}D_{n,n}\notin\mathscr{A}$. In fact if $S$ is an arbitrary set of positive even integers, we can write $S=\bigcup_{n\in S}D_{n,n}$ as a countable union of elements of $\mathscr{A}$, so any $\sigma$-algebra containing $\mathscr{E}$ must include every set of positive even integers and of course the complement of every such set. Let
$$\mathscr{A}^\sigma=\wp(E)\cup\{\Bbb Z^+\setminus S:S\subseteq E\}\;,$$
and let $O=\Bbb Z^+\setminus E$, the set of odd positive integers. A little thought reveals that 
$$\mathscr{A}^\sigma=\{S\subseteq\Bbb Z^+:O\subseteq S\text{ or }O\cap S=\varnothing\}\;.$$
It should be clear that $\mathscr{A}^\sigma$ is a $\sigma$-algebra containing $\mathscr{E}$. Any $\sigma$-algebra containing $\mathscr{E}$ must contain $\mathscr{A}$, and it’s pretty clear that any $\sigma$-algebra containing $\mathscr{A}$ must contain $\mathscr{A}^\sigma$, so $\mathscr{A}^\sigma$ is the smallest $\sigma$-algebra containing $\mathscr{E}$.
