If closed under countable intersection, then also closed closed under countable union

Question

I am at looking at Klenke's Probability Theory. And he starts if the class of sets $\mathcal{A}$ is $\setminus$ - closed (i.e. closed under difference), then:

$$\bigcap_{n=1}^{\infty} A_n = \bigcap_{n=2}^{\infty} A_1 \cap A_n = \bigcap_{n=2}^{\infty} A_1 \setminus (A_1 \setminus A_n ) = A_1 \setminus \bigcup_{n=2}^{\infty}(A_1 \setminus A_n ) \in \mathcal{A}$$

I can follow up until the very last $=$.

If I write out the term before the last one and the last one, I get something that is different, so I do not follow:

$$\bigcap_{n=2}^{\infty} A_1 \setminus (A_1 \setminus A_n ) = A_1 \setminus (A_1 \setminus A_2) \cap A_3 \cap A_4 \cap \cdots$$

$$A_1 \setminus \bigcup_{n=2}^{\infty}(A_1 \setminus A_n ) = A_1 \setminus(A_1 \setminus A_2) \cup A_3 \cup A_4 \cup A_5 \cup\cdots$$

EDIT: I think I am now getting what he is trying to do but I don't get it. He using the property of $\setminus$ - closed there at the very end. But how come those two things are the same... I must be expanding the cups and the caps incorrectly

Answer 1

I believe you're missing some parentheses.

$$\bigcap_{n=1}^\infty A_n = A_1 \cap \bigcap_{n=2}^\infty A_n = \bigcap_{n=2}^\infty (A_1 \cap A_n) = \bigcap_{n=2}^\infty \Big(A_1 \setminus (A_1 \setminus A_n)\Big)$$

Now recall De Morgan's laws

$$\bigcap_{i\in I} B_i^c = \left(\bigcup_{i\in I} B_i\right)^c$$

So, if we consider the complement with respect to $A_1$, we get

$$\bigcap_{n=1}^\infty A_n = \bigcap_{n=2}^\infty (A_1 \setminus A_n)^c = \left(\bigcup_{n=2}^\infty (A_1 \setminus A_n)\right)^c = A_1 \setminus \left(\bigcup_{n=2}^\infty (A_1 \setminus A_n)\right)$$