• $\mathcal{C}_{\mathcal{I}} \equiv\{\text { all intervals }(a, b],(-\infty, b], \text { or }(a, \infty):-\infty<a<b<\infty\}$
  • $\mathcal{C}_{F} \equiv\left\{\right.$ 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:

  1. If indeed my proof is correct are there ways to prove the statement more concisely?
  2. To my knowledge $\emptyset$ is always in a set but why is it the case? Is it by definition?
  3. 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?
  • $\begingroup$ @Ramiro I think I am fairly clear with the idea of the field and $\sigma$field... but what the relationship between a field and an algebra and similar I am not quite sure about the difference between the $\sigma$-field and $\sigma$-algebra... $\endgroup$
    – JoZ
    Commented Sep 17, 2020 at 3:59
  • $\begingroup$ In fact, field (of sets) is the same as algebra (of sets), and $\sigma$-field is the same as $\sigma$-algebra. They are synonyms. $\endgroup$
    – Ramiro
    Commented Sep 17, 2020 at 14:38
  • $\begingroup$ The FOUR related concepts I mentioned in a previous comment are actually: ring (of sets), algebra (of sets), $\sigma$-ring and $\sigma$-algebra. Rings and $\sigma$-rings are not required to be closed under complement, just closed under difference. $\endgroup$
    – Ramiro
    Commented Sep 17, 2020 at 14:45

1 Answer 1


According to Wikipedia, a field of sets is a collection of sets which is closed under complementation and finite unions and intersections (also known as algebra). 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:

  1. 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 $A \cup B \in \mathcal{C}_F$.

  2. 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 $I_1 \cup (I_2 \cup \cdots \cup I_{n+1} \cup B) \in \mathcal{C}_F$.

  3. 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 $\overline{A} \in \mathcal{C}_F$. $\quad\square$

  • $\begingroup$ Oh yes... I am confused $\endgroup$
    – JoZ
    Commented Sep 16, 2020 at 9:36
  • $\begingroup$ I changed my proof. Would it be a valid one? $\endgroup$
    – JoZ
    Commented Sep 16, 2020 at 10:29
  • $\begingroup$ I’m afraid this is not enough. You’re only considering single intervals, whereas you should be considering unions of disjoint intervals, per the definition. $\endgroup$ Commented Sep 16, 2020 at 11:16

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