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Suppose that $ A_1 \cap A_2 \cap \dots \cap A_n = \emptyset $ for all $n \in \mathbb{N}$

Is it also true that $ A_1 \cap A_2 \cap \dots = \emptyset$ ?

Could I have a hint as how to think about disproving this? (I'm assuming it's false with some fancy counterexample that I'm probably not going to guess...)

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    Yes, of course; if $A_1 \cap A_2=\emptyset$, then you can go on forever "adding" sets, but the "common part" to all them will remain empty, because the "common part" must be common also to $A_1$ and $A-2$. – 2017-01-06
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    if $A_1\cap\cdots\cap A_n=\varnothing$ for _even one_ $n$, then $\bigcap_{n\in\mathbb N} A_n=\varnothing$ too... – 2017-01-06
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    "for all $n \in\mathbb{N}$" includes $n=1$, so we even have $A_1 = \emptyset$. – 2017-01-06

2 Answers 2

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Adding more terms to an intersection only ever makes the result smaller (remember: $A\cap B$ is a subset of both $A$ and $B$, so in particular $A\cap B\cap C$ is a subset of $A\cap B$). So as soon as $A_1\cap A_2\cap ...\cap A_n=\emptyset$ for some $n$, evry further intersection will also be empty.

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    Thanks; this is actually rather obvious.I forgot the not equals sign in my question, so I'll ask a new one and leave this as it is... – 2017-01-06
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    @ChristopherTurnbull I can just answer that here. In the "$\not=$" case, the answer is "no": let $A_n=\{m\in\mathbb{N}: m>n\}$ . . . – 2017-01-06
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    Brilliant. I fear I wouldn't have came to that conclusion myself- how did you do it? – 2017-01-06
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Hint:

$$\bigcap_{i=1}^{\infty} A_i \subseteq \bigcap_{i=1}^{2} A_i$$

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    @fabian, thanks for the suggestion, it does make things clearer. It is arguable whether $\subset$ notation means $\subseteq$ or $\subsetneq$ though my intention was $\subseteq$. Thanks again. – 2017-01-07