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I am trying to prove (or disprove) the following statement

For arbitrary countable collection of sets $\{A_i\}_{i \in \mathbb{N}}$ that is non-increasing, i.e. $A_i \supseteq A_{i+1}$ for all $i \in \mathbb{N}$, and have a non-empty intersection, i.e. $\bigcap\limits_{i \in \mathbb{N}} A_i \neq \emptyset$, it holds that: If $B \subsetneq A_j$ for some $j \in \mathbb{N}$ and $\bigcap\limits_{i \in \mathbb{N}} A_i \subsetneq B$, then there exists some $k \in \mathbb{N}$ such that $A_k \subsetneq B$

However, all my efforts so far have failed. Intuitively it seems that the statement must be true, given that the intersection is (sort of) a limit for the sequence $A_i \supseteq A_{i+1} \supseteq \ldots$ and therefore one should be able to find $A_i$'s arbitrarily 'close' to the intersection, but I can't seem to find the right argument that it must be true (making we wonder if the statement is actually false), but I can't find any counterexamples either.

Any help would me much appreciated.

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The result is false. Take $A_i=\{0\}\cup\{n \in \mathbb{N}~:~n \geq i\}$. These sets are non-increasing with nonempty intersection $\{0\}$. Let $B=\{0,1\}$. Then $B\subsetneq A_1$, $\bigcap_{i\in\mathbb{N}}A_i\subsetneq B$, yet there does not exist $k$ such that $A_k \subsetneq B$.

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