For n $\in \mathbb{N},$define $A_{n}$ to be the set of all natural numbers divisible by n. Find with proofs:
(a)$\bigcup_{n=2}^{\infty}A_{n}.$
(b)$\bigcap_{n=1}^{\infty}A_{n}.$
I do not know How to organize my thoughts so that I gave a general formula for the answer and How to proof? Could anyone help me?
Thanks.
Best Answer
For a) This would be the set of all numbers divisible by 2, or by 3, or by 4, ... etc. Since every number is divisible by itself, that means that this is the set of all numbers greater or equal to 2.
For b) This would be the set of all numbers divisible by 1, and by 2, and by 3, and by 4, ... etc. No number is like that, so this is the empty set.
To prove this you need to prove the following set equalities:
$\bigcup_{n=2}^{\infty}A_{n} = \{ n | n \ge 2\}$
$\bigcap_{n=1}^{\infty}A_{n} = \{ \}$
So for the first one, prove that any $n$ that is in $ \bigcup_{n=2}^{\infty}A_{n}$ is also in $\{ n | n \ge 2\}$, and vice versa.
The same for the second one ... except of course there is no such $n \in \{ \}$ ... so prove that there is no $n$ in $\bigcap_{n=1}^{\infty}A_{n}$ either. And that you could do by a proof by contradiction: suppose $n \in \bigcap_{n=1}^{\infty}A_{n}$. Then in particular $n \in A_{n+1}$, i.e. $n$ is divisible by $n+1$. But that is impossible, so we have a contradiction. So there is indeed no $n$ in $\bigcap_{n=1}^{\infty}A_{n}$.