Prove all finite disjoint unions of intervals in a collection of all $(a, b],(-\infty, b]$or $(a,\infty)$ ,$-\inftyDefine

*

*$\mathcal{C}_{\mathcal{I}} \equiv\{\text { all intervals }(a, b],(-\infty, b], \text { or }(a, \infty):-\infty<a<b<\infty\}$

*$\mathcal{C}_{F} \equiv\left\{\right.$ all finite disjoint unions of intervals in $\left.\mathcal{C}_{I}\right\}$.

Show that $\mathcal{C}_{F}$ is a field.

I didn't study any mathematical courses rigorously at the college level due to my curriculum design so so it is very challenging to study probability course founded on measure theory. I tried to prove this statement using the definition of a set but I am not sure if my proof is correct. Here's my proof:
$ \mathcal{C}_F$ is the set of all finite disjoint unions of intervals, that is
$$
\left(a_{1}, b_{1}\right] \cup \cdots \cup\left(a_{n}, b_{n}\right]
$$

*

*$\emptyset\in \mathcal{C}_F$ because it is the disjoint zero interval in $ \mathcal{C}_I$.

*$A\equiv \bigcup^\infty (a_n,b_n]$ where $b_i\leq a_{i+1},i\in N$, allowing for $a=-\infty,b=\infty$, it is easily seen $\Omega \in \mathcal{C}_F$.

*Suppose $A'=(a,b]$,then $A'^c=(-\infty,a]\bigcup(b,\infty)$, $(-\infty,a] = \bigcup^\infty (a_n,b_n], b_1=a,b_{i+1}=a_i$, $(b,\infty)$ can also be expressed as the union of disjoint finite interval in similar fashion.

*if $A_1,...A_n$ are finite disjoint unions of interval, let $A'_1=A_1,A'_2=A_2\backslash A_1, .... A_n=A_n\backslash \cup^{n-1}A'_n$,$A'_i$ can be still expressed as the union of disjoint finite intervals, and by constructions, $\cup^\infty A_n=\cup^\infty A'_n \in \mathcal{C_F}$.


I messed up the definition of a field and a definition of a $\sigma$-field, as pointed by @YuvalFilmus. Following what was suggested I changed my proof:

*

*Suppose $A=(a,b]$,then $A^c=(-\infty,a]\bigcup(b,\infty)$. $(-\infty,a] = \bigcup(a_n,b_n],n\in \mathbb{N}$, where $b_1=a,b_{i+1}=a_i$, $(b,\infty)= \bigcup(a_n,b_n],n\in \mathbb{N}$, where $a_1=b,a_{i+1}=b_i$. Therefore,$A^c\in \mathcal{C_F}$.

*Let $A,B \in \mathcal{C_F}$, 1) $A \cap B=\emptyset, A\cup B\in \mathcal{C_F}$; 2) $A\cap B=A$ or $B, A \cup B\in \mathcal{C_F}$; 3) Let $A=(a_1,b_1], B=(a_2,b_2]$, if $ a_1<a_2<b_1<b_2$, then $A\cup B=(a_1,a_2]\cup (a_2,b_1] \cup (b_1,b_2]\in \mathcal{C_F}$; similarly, if $a_2<a_1<b_2<b_1$, $A \cup B\in \mathcal{C_F}$.

*Since $ \mathcal{C_F}$ is closed under finite union, it must also be closed under finite interval by DeMorgan's Law, hence $A \cap A^c=\emptyset\in \mathcal{C_F}$. $\Omega=\emptyset^c\in \mathcal{C_F}$.

Two additional questions:

*

*If indeed my proof is correct are there ways to prove the statement more concisely?

*To my knowledge $\emptyset$ is always in a set but why is it the case? Is it by definition?

*When I prove $ \mathcal{C_F}$ is a field I think of the infinite many unions are still in $ \mathcal{C_F}$ and it is very convincing to me. But being a field means the infinite intersections are also in the field, that implies that $ \mathcal{C_F}$ also includes every single point $\{x\}, x\in\mathbb{R}$. But I don't quite get how can a single point be in $ \mathcal{C_F}$? Could someone please express a point as a finite disjoint union of interval?

 A: According to Wikipedia, a field of sets is a collection of sets which is closed under complementation and finite unions and intersections. In particular, there is no requirement to be closed under infinite unions and intersections. A related concept, $\sigma$-algebra allows countable unions and intersections.
Now regarding your proof:

*

*A non-empty field $F$ always contains the empty set: if $A \in F$ then $\emptyset = A \cap \overline{A} \in F$.

*If a collection of sets is closed under complement and finite unions, then it is also closed under finite intersections, due to de Morgan's laws:
$$ A_1 \cap \cdots \cap A_n = \overline{\overline{A_1} \cup \cdots \cup \overline{A_n}}. $$

*I'm not sure what "$A \equiv \bigcup^\infty (a_n,b_n]$" means. I have never seen the notation $\bigcup^\infty$, for example.

*The collection $\mathcal{C}_F$ is not closed under infinite unions and intersections, even countably many. Indeed, as you mention, $\{0\} = \bigcap_{n \geq 0} (-2^{-n},0]$, yet $\{0\} \notin \mathcal{C}_F$.

Showing that $\mathcal{C}_F$ is a field unfortunately requires some lengthy case analysis, with much more detail than is given in your proof.

Here is a complete proof. We start with closure under union.
Lemma 1. If $A \in \mathcal{C}_F$ and $B \in \mathcal{C}_I$ then $A \cup B \in \mathcal{C}_F$.
Proof. Let us say that $A$ has complexity $n$ if it is the union of $n$ disjoint intervals from $\mathcal{C}_I$. The proof is by induction on $n$. If $n = 0$ then $A \cup B = B \in \mathcal{C}_F$.
Assuming the lemma holds for $n$, we prove it for $n+1$. Suppose that $A$ has complexity $n+1$, and so it can be written as a disjoint union of $n+1$ intervals $I_1,\ldots,I_{n+1} \in \mathcal{C}_I$. Let $I_1$ be the interval with minimal endpoint. We consider three cases:

*

*The interval $B$ is entirely to the left of $I_1$. In that case, $B,I_1,\ldots,I_{n+1}$ are disjoint, and so clearly $A \cup B \in \mathcal{C}_F$.


*The interval $B$ is entirely to the right of $I_1$. By induction, $(I_2 \cup \cdots \cup I_{n+1}) \cup B \in \mathcal{C}_F$. Appealing to the preceding case shows that $I_1 \cup (I_2 \cup \cdots \cup I_{n+1} \cup B) \in \mathcal{C}_F$.


*The intervals $B$ and $I_1$ intersect. Write $B = (a,b]$ and $I_1 = (c,d]$, where possibly $a,c = -\infty$ or $b,d=+\infty$. Then $J := B \cup I_1 = (\min(a,c),\max(b,d)]$. If $J \in \mathcal{C}_I$ then $A \cup B = J \cup I_2 \cup \cdots \cup I_{n+1} \in \mathcal{C}_F$ by induction. Otherwise, $J = (-\infty,\infty)$, and so $A \cup B = (-\infty,\infty) = (-\infty,0] \cup (0,\infty) \in \mathcal{C}_F$. $\quad\square$
Lemma 2. If $A,B \in \mathcal{C}_F$ then $A \cup B \in \mathcal{C}_F$.
Proof. The proof is by induction on the complexity $n$ of $B$. If $B = \emptyset$ then $A \cup B = A \in \mathcal{C}_F$. Otherwise, write $B = C \cup D$, where $C$ has complexity $n-1$ and $D \in \mathcal{C}_I$. By induction, $A \cup C \in \mathcal{C}_F$, and so $A \cup B = (A \cup C) \cup D \in \mathcal{C}_F$ by Lemma 1. $\quad\square$
We proceed to closure under complementation.
Lemma 3. If $A \in \mathcal{C}_F$ then $\overline{A} \in \mathcal{C}_F$.
Proof. If $A = \emptyset$ then $\overline{A} = (-\infty,0] \cup (0,\infty) \in \mathcal{C}_F$. Otherwise, write
$$
A = (a_1,b_1] \cup \cdots \cup (a_n,b_n],
$$
where $b_i \leq a_{i+1}$ for all $i \in \{1,\ldots,n-1\}$, and possibly $a_1=-\infty$ and $b_n=\infty$. Then
$$
\overline{A} = (-\infty,a_1] \cup (b_1,a_2] \cup \cdots \cup (b_{n-1},a_n] \cup (b_n,\infty).
$$
If $a_1 = -\infty$, we can remove the first interval. If $b_n = \infty$, we can remove the last interval. If $b_i = a_{i+1}$ for some $i$, we can remove the interval $(b_i,a_{i+1})$. After removing these empty intervals, we get a representation of $\overline{A}$ as a disjoint union of intervals from $\mathcal{C}_I$, hence $\overline{A} \in \mathcal{C}_F$. $\quad\square$
