For the sequence of sets $\{A_n\}_{n=1}^{\infty}$I have that $A_n\uparrow A$ i.e. $A_1\subseteq A_2 \subseteq \cdots $ and $\bigcup\limits_{i=1}^{\infty} A_{i}$ = A.
I want to prove that for $B_n=A-A_n$ we have that $B_n\downarrow\emptyset$.
What I have is:
$$\bigcap\limits_{i=1}^{\infty} (A-A_{i})=(A-\bigcup\limits_{i=1}^{1} A_{i})\bigcap (A-\bigcup\limits_{i=1}^{2} A_{i}) \bigcap \cdots \bigcap (A-\bigcup\limits_{i=1}^{\infty} A_{i})=\\(A-\bigcup\limits_{i=1}^{1} A_{i})\bigcap (A-\bigcup\limits_{i=1}^{2} A_{i}) \bigcap \cdots \bigcap \emptyset=\emptyset$$
Is this formally and rigorously correct? Am I allowed to use the dot notation like that and go to infinity in that manner?
Thanks in advance!
I think it is totally fine. If there's a lack of rigor, it's trivial to fill in.
If we broke down the logic we could say $$x \in \bigcap_{i=1}^{\infty}(A - A_i) \iff x\in A \ \wedge \ \forall i, x \notin A_i \iff x \in A \ \wedge \ x \notin \bigcup_i^{\infty} A_i \iff x \in A \ \wedge \ x \notin A$$