$\lambda$ (Dynkin) system equivalent definitions, proof?

According to wikipedia (https://en.wikipedia.org/wiki/Dynkin_system#Definitions), there are two equivalent definitions of a $\lambda$-system. How are these two definitions equivalent? To prove equivalence, do I have to show that one definition implies all of the conditions of the other definition and vice versa? For example, let's start with the second definition:

$D$ is a Dynkin system if

(S1) $\Omega \in D$,

(S2) if $A \in D$, then $A^c \in D$,

(S3) $D$ is closed under countable disjoint union: if $A_1, A_2, \cdots \in D$ and $A_i \cap A_j = \emptyset$ for all $i \neq j$, then $\cup_{n \in \mathbb{N}} A_n \in D$.

We want to show that the above definition implies the first definition:

(F1) $\Omega \in D$,

(F2) If $A, B \in D$ and $A \subset B$ then $B \setminus A \in D$.

(F3) If $A_1 \subset A_2 \subset \cdots$ with each element being in $D$, then $\cup_{n \in \mathbb{N}} A_n\in D$.

Showing (F1) is true is trivial since it is the same as (S1). Showing (F2) is true can be done as follows: Note that $B \setminus A = B \cap A^c = (B^c \cup A)^c$ is in $D$ because it is the complement of the set $B^c \cup A$ which is in $D$ because it is the union of two disjoint sets $A$ and $B^c$ both of which are in $D$.

Question 1) How can I complete the above proof by showing that (F3) is true?

Question 2) How can I prove that (F1)-(F3) implies (S1)-(S3)?

Let $A_1 \subset A_2 \subset \cdots$ be as described in (F3). Let $B_1=A_1$ and $B_i := A_i \setminus A_{i-1}$ for $i \ge 2$. Apply (S3) to $B_1,B_2,\ldots$ and note that $\bigcup_i A_i = \bigcup_i B_i$.

(S1) = (F1)

(S2): Apply (F2) with $B=\Omega$.

(S3): Let $A_1,A_2,\ldots$ be as described in (S3). Let $B_n:=\bigcup_{i=1}^n A_i$. Apply (F3) to $B_1,B_2,\ldots$ and note that $\bigcup_n A_n = \bigcup_n B_n$.

• How do you (in the last step) that $B_n$ is in D? It appears to me that F1,F2,F3 do not imply S3, but I don't have a counter-example. Jul 21 '17 at 10:28
• @KaviRamaMurthy $B_1=A_1 \in D$ and for $n \ge 2$, $B_n = A_n \cup B_{n-1} = (A_n^c \setminus B_{n-1})^c \in D$, using (F2). Jul 21 '17 at 16:54
• To see F2 from S1,S2,S3 is still an argument. First note that $\emptyset \in D$ from S1 and S2, so that we can stuff finite disjoint unions from $D$ with $\emptyset$ to see that their unions too are in $D$. Then if $A \subseteq B$, then $A \cup B^c \in D$ by their disjointness and so $(A \cup B^c)^c = B \setminus A \in D$ as well, applying S2. Jun 30 '18 at 22:40

Consider the class consisting of (0,1),(1,2), empty set and the entire real line.This class (of just 4 sets) satisfies F1,F2,F3 but not imply S3.

• This does not satisfy (F2), since it does not contain $\mathbb{R} \setminus (0,1)$. Jul 21 '17 at 16:49
• Sorry! angryavian's proof above is correct and the equivalence holds. Jul 27 '17 at 8:04

Proof of F2.

Let $\mathscr{R}$ be a $\lambda$-systen and $A, B \in \mathscr{R}$, $A \subset B$.

$B-A= B \cap A^* = (B^* \cup A)^*$ (1)

Applying S3, and since $A^*$ (by S2 and $A \in \mathscr{R}$ you have $A^* \in \mathscr{R}$) and $B$ are disjoint substets ($A^* \cap B = \emptyset$), you get $A^* \cup B \in \mathscr{R}$. So applying S2 and (1) you get F2.