If $A_n \downarrow A.$ then $A_1 - An \uparrow A_1 - A$? Set theory.

Question

Let $A_1, A_2 , \dots$ be subsets of a set $\Omega$. If $ A_1 \subset A_2 \subset \dots$ and $\bigcup_{n = 1}^{\infty} A_n = A $ then we write $A_n \uparrow A.$

$ A_1 \supset A_2 \supset \dots$ and $\bigcap_{n = 1}^{\infty} A_n = A $ then we write $A_n \downarrow A.$

I want to prove that if $A_n \downarrow A.$ then $A_1 - An \uparrow A_1 - A$ where the minus is the set minus but I am having some troubles with the second requirement.

Answer 1

Since $A_1 \supset A_2 \supset \ldots$, it follows that $A_1-A_1 \subset A_1-A_2 \subset \ldots$. Further \begin{align*} \bigcup_\limits{n=1}^{\infty}(A_1-A_n) &= \bigcup_\limits{n=1}^{\infty}(A_1\cap A_n^c)\\ &= A_1 \cap \left(\bigcup_\limits{n=1}^{\infty}A_n^c\right)\\ & = A_1 \cap \left(\bigcap_\limits{n=1}^{\infty}A_n\right)^c & & \text{(De Morgan's Law)}\\ &=A_1-A. \end{align*}

Answer 2

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