Properties of $\mathscr{C}$, a $Boolean$ algebra:
- $Ω∈\mathscr{C}$
- If $A∈\mathscr{C}⇒A^c∈\mathscr{C}$
- If $A,B∈\mathscr{C}⇒A∪B∈\mathscr{C}$
Demostrated properties of $\mathscr{C}$:
- $∅∈\mathscr{C}$
- If $A,B∈\mathscr{C}⇒A\cap B∈\mathscr{C}$
- If $A_{1},...,A_{n}∈\mathscr{C}⇒\left(\bigcup_{i=1}^{i=n}A_{i}\right)∈\mathscr{C}$ and $\left(\bigcap_{i=1}^{i=n}A_{i}\right)∈\mathscr{C}$ with $n∈N^*$
Every $σ-Algebra$ has $(1)$, $(2)$, $(3)$ and the following:
- If $A_{1},...,A_{n}∈\mathscr{C}⇒\left(\bigcup_{i=1}^{i=\infty}A_{i}\right)∈\mathscr{C}$
So, I want to prove that the following laws of $De$ $Morgan$ are fulfilled:
$$\left(\bigcap_{i=1}^{\infty}A_{i}\right)^c=\bigcup_{i=1}^{\infty}A^c_{i}$$
and
$$\left(\bigcup_{i=1}^{\infty}A_{i}\right)^c=\bigcap_{i=1}^{\infty}A^c_{i}$$
Can I say that...
- $X=\bigcap_{i=1}^{i=n}A_{i}$
- $Y=\bigcap_{i=n}^{\infty}A_{i}$
- $\left(\bigcap_{i=1}^{\infty}A_{i}\right)^c=\left( X \cap Y \right)^c$
To do this?
Let $A_{j} \in (X \cap Y)^c$. Then, $A_{j} \not\in X \cap Y$.
Because $X \cap Y = \{A_{k} | A_{k} \in X $ and $ A_{k} \in Y\}$, it must be the case that $A_{j} \not\in X$ or $A_{j} \not\in Y$.
If $A_{j} \not\in X$, then $A_{j} \in X^c$, so $A_{j} \in X^c \cup Y^c$.
Similarly, if $A_{j} \not\in Y$, then $A_{j} \in Y^c$, so $A_{j} \in X^c\cup Y^c$.
Thus, $\forall A_{j}( A_{j} \in (X\cap Y)^c \rightarrow A_{j} \in X^c \cup Y^c)$;
that is, $(X\cap Y)^c \subseteq X^c \cup Y^c$.
HINT: the proofs related to equality of sets (by example $A=B$) in general follow this strategy structured in two parts:
First we prove that $A\subseteq B$
After we prove the other direction, i.e. that $A\supseteq B$
Both statements together imply that $A=B$. Explicitly for the first part we choose some arbitrary point of $A$ and we show that it belong to $B$, hence $A\subseteq B$.
The other part is the same but reversed: we choose an arbitrary point of $B$ and we shows that it belong to $A$, hence $A\supseteq B$.
I think that with this information you can prove now the De Morgan laws for $\sigma$-algebras.