Union of infinite sets of integers Problem:
For every $n$ in $\mathbb{N}$, we consider the sets $A_{n}:=\left \{  (2n+1)\lambda :\ \lambda \in \mathbb{N}\right \}$. The question is to find $\bigcap_{n=1}^{\infty }A_{n}$, and $\bigcup_{n=1}^{\infty }A_{n}$.
For the intersection, I think it is the empty set, because for every $n$ in $\mathbb{N}$; we have $n\notin A_{n}\Rightarrow n\notin \bigcap_{n=1}^{\infty }A_{n}$. For the union, I tried the first few sets, but I can't see exactly what the union of all the sets should be. Any help is appreciated.
 A: These are all the positive integers that have a factor of the shape $2n+1$. That's the collection of all positive integers that have an  odd factor $\ge 3$.  Every positive integer qualifies, except for the powers $1,2,4,8,\dots$ of $2$.
So our set is the complement of the set of powers of $2$.  
A: I’m assuming that you’re using $\Bbb N$ for $\Bbb Z^+$, the set of positive integers, rather than for the set of non-negative integers. In that case $A_1$ is the set of positive multiples of $3$, $A_2$ is the set of positive multiples of $5$, $A_3$ is the set of positive multiples of $7$, and so on. Clearly no positive integer is a multiple of all of the odd numbers $3,5,7,9,11,\dots$ greater than $1$, so $\bigcap_{n\in\Bbb N}A_n$ is indeed empty. Your argument is also perfectly sound: for each $n\in\Bbb N$ the smallest member of $A_n$ is $2n+1>n$, so $n\notin A_n$, and therefore $n\notin\bigcap_{n\in\Bbb N}A_n$.
Every positive integer $n$ can be written uniquely in the form $n=2^km$, where $k$ is a non-negative integer, and $m$ is a positive odd integer. (This is worth trying to prove; induction on $n$ will work.) You shouldn’t have much trouble showing that if $m=1$, so that $n$ is a power of $2$, then $n\notin\bigcup_{k\in\Bbb N}A_k$, while if $m>1$, then $n\in A_m$.
