Formal proof of limit of chain of nested sets towards empty set

Question

Given a countable collection of mutually disjoint sets $A_1, A_2,…$, I would like to utilize the Axiom of Continuity for the sake of completing some other proof. If we let $B_k=\bigcup_{i=k}^\infty A_i$, then the fact that $B_{i+1}\subset B_{i}$ is trivial. But now to utilize the axiom we also have to establish that $B_k \downarrow \emptyset$ (the limit of the infinite sequence $B_1\supset B_2\supset B_3\supset \cdots$ is $\emptyset$), which is easy to see intuitively. However, I'm having trouble coming up with a formal argument to why this is true. Any tips?

Answer 1

Suppose $x\in B_k$ for all $k$. In particular, then, $x\in B_1=\bigcup_{i=1}^\infty A_i$. So, there exists some $i$ such that $x\in A_i$. Now can you find a value of $k$ such that $x\not\in B_k$, to reach a contradiction?

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Since the sets $A_k$ are pairwise disjoint, $x\in A_i$ implies $x\not\in A_j$ for all $j\neq i$. In particular, $x\not\in A_j$ for all $j\geq i+1$, and thus $x\not\in B_{i+1}$. This contradicts the assumption that $x\in B_k$ for all $k$, and thus no such $x$ exists. That is, $\bigcap B_k=\emptyset$.