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Problem:

For each natural number $n$, we define: $A_{n}=\left \{ (n+1)k:k\in \mathbb{N} \right \}$. It seems obvious to me that $\bigcap_{n=1}^{\infty }A_{n}=\varnothing $ where $\varnothing$ is the empty set. However, I couldn't come up with a proof for that. Any help please? Thanks.

1 Answers 1

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Using $\mathbb{N} = \{1, 2, ...\}$.

For any $m \in \mathbb{N}$, $m \notin A_{m}$ since $A_{m} = \{m + 1, (m + 1)(2), ... \}$. Hence $m$ is not in the intersection $\bigcap_{n = 1}^\infty A_n$. Thus the intersection does not contain any $m \in \mathbb{N}$.

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    @Alex Becker: Now, I can see it. Thanks to both of you: William and Alex Becker2012-08-30