Inclusion or equality of sigma-algebras generated by intervals I have the following problem:
Let $C_{1} = \{ ( n,n+1 ) : n \in \mathbb{Z} \}$ and $C_{2} = \{ [ n,n+1 ] : n \in \mathbb{Z} \}$
I want to check if $\sigma(C_1)=\sigma(C_2)$.
Let me mention that we know, the for any sets $A,B \subset X$ we have: $A \subset \sigma(B) \implies \sigma(A) \subset \sigma(B)$
Now my proofs of inclusions:
We ask if $C_1 \subset \sigma(C_2)$, so: let $z\in \mathbb{Z}$, $(z,z+1)\in C_1$
$ [z,z+1]^C \setminus [z+1,z+2]^C = ((-\infty, z) \cup (z+1,\infty))\setminus ((-\infty, z+1) \cup (z+2,\infty)) = (z+1,z+2] $
$(z+1,z+2] \cap ((-\infty, z+2) \cup (z+3,\infty))=(z+1,z+2] \cap [z+2,z+3]^C = (z+1,z+2) \in \sigma (C_2)$
So we have inclusion: $C_1 \subset \sigma(C_2)$and so $\sigma(C_1)\subset \sigma(C_2)$ and I think that proof is ok, but I have problem with proof of another inclusion. I can get sets like: $(\bigcup_{z\in \mathbb{Z}}(z,z+1))^C=\mathbb{Z} \in \sigma(C_1)$ and add open intervals to it, but i can't get rid of the rest of integers to get one particulat interval: $[z,z+1]$.
Any idea? Maybe this inclusion is wrong? If it is, how can I show it?
 A: Having shown $\sigma(C_1) \subset \sigma(C_2)$ — which you correctly did, but I find using $$(n,n+1) = [n,n+1] \setminus ([n-1,n] \cup [n+1,n+2])$$ or $$(n,n+1) = \mathbb{R}\setminus \biggl(\bigcup_{k \neq n} [k,k+1]\biggr)$$ more obvious — one can see $\sigma(C_1) \subsetneqq \sigma(C_2)$ by noting that for every $A \in \sigma(C_1)$ one has either $\mathbb{Z} \subset A$ or $\mathbb{Z}\cap A = \varnothing$, while of course $\{n\} \in \sigma(C_2)$ for all $n \in \mathbb{Z}$. Intuitively, that is clear once it has been stated. But of course we need a proof, there are enough things that are "intuitively clear" but wrong.
So how do we set about proving it? One way is to inductively construct $\sigma(C_1)$. We set
$$U_0 = \Bigl\{ \bigcup \mathscr{C} : \mathscr{C}\subset C_1\},\quad V_0 = \{\mathbb{R}\setminus A : A \in U_0\}, \quad \text{and} \quad S_0 = U_0 \cup V_0\,.$$
Then one defines $U_{\alpha+1}$ as the family of all unions of countable subfamilies of $S_{\alpha}$, $V_{\alpha+1}$ as the family of complements of members of $U_{\alpha+1}$ and $S_{\alpha+1} = U_{\alpha+1} + V_{\alpha+1}$, and for a limit ordinal $\lambda$ one defines $S_{\lambda} = \bigcup_{\alpha < \lambda} S_{\alpha}$. This way, $S_{\omega_1} = \sigma(C_1)$, where $\omega_1$ is the first uncountable ordinal. And along the way one can also see that for all $\alpha$ one has either $\mathbb{Z}\subset A$ or $\mathbb{Z}\cap A = \varnothing$ for all $A \in S_{\alpha}$.
Okay, that works but isn't pretty. And requires a bit more set theory than it should. What else can we do? Well, $C_1$ consists of subsets of $X := \mathbb{R}\setminus \mathbb{Z}$. Let's use that and define $\mathscr{A}$ as the $\sigma$-algebra on $X$ generated by $C_1$. Denote the inclusion of $X$ in $\mathbb{R}$ by $\iota$ and set
$$\mathscr{S} = \{ \iota(A) : A \in \mathscr{A}\} \cup \{ \iota(A) \cup \mathbb{Z} : A \in \mathscr{A}\}\,.$$
Then $\mathscr{S}$ is a family of subsets of $\mathbb{R}$ containing $C_1$, and it has the property that either $\mathbb{Z}\subset B$ or $\mathbb{Z}\cap B = \varnothing$ for every $B \in \mathscr{S}$.
It remains to see that $\mathscr{S}$ is a $\sigma$-algebra (that actually $\mathscr{S} = \sigma(C_1)$ isn't needed, $\sigma(C_1) \subset \mathscr{S}$ suffices).
So let's check:


*

*$\varnothing = \iota(\varnothing) \in \{ \iota(A) : A \in \mathscr{A}\} \subset \mathscr{S}$.

*$\mathbb{R} \setminus \iota(A) = \iota(X\setminus A) \cup \mathbb{Z} \in \mathscr{S}$ and $\mathbb{R} \setminus (\iota(A)\cup \mathbb{Z}) = \iota(X\setminus A) \in \mathscr{S}$ for all $A \in \mathscr{A}$.

*Let $A_n \in \mathscr{S}$ for $n \in \mathbb{N}$. For each $n \in \mathbb{N}$ let $B_n = A_n \cap X$ and $C_n = A_n \cap \mathbb{Z}$. Then $B_n \in \mathscr{A}$ for all $n$, and $C_n \in \{\varnothing, \mathbb{Z}\}$ for all $n$. Hence
$$\bigcup_{n \in \mathbb{N}} A_n = \iota\biggl(\bigcup_{n \in \mathbb{N}} B_n\biggr) \cup \bigcup_{n \in \mathbb{N}} C_n$$
belongs to $\mathscr{S}$ too.


The family $\mathscr{S}$ contains the empty set and is closed under complements and countable unions, i.e. it is a $\sigma$-algebra.
I think this one's nicer.
