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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.

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up vote 5 down vote accepted

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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I am a bit lost in your proof. According to the problem: $A_{m+1}=\left \{ (m+2),(m+2)*2, (m+2)*3,... \right \}$. Also, you said that $m$ is not in the intersection. Which two sets are intersecting? – C. Lambda Aug 30 '12 at 3:29
@C.Lambda William probably means $A_m$ instead of $A_{m+1}$. As for your question "Which two sets are intersecting?" He means the intersection of all $A_n$. – Alex Becker Aug 30 '12 at 3:31
@C.Lambda Sorry, it should $A_m$. It is not the case that $m \in A_k$ for all $k$ (in particular $k = m$). By definition $m \notin \bigcap_{k \in \mathbb{N}} A_k$. – William Aug 30 '12 at 3:32
@Alex Becker: Now, I can see it. Thanks to both of you: William and Alex Becker – C. Lambda Aug 30 '12 at 3:32

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