Prove or disprove: $\bigcup_{n\ge1}\left[0,\frac n{n+1}\right]=[0,1)$ and $\bigcap_{n\ge1}\left[0,\frac n{n+1}\right]=\{0\}$ 
For integers $ n \ge 1 $, let $ A _ n = \left \{ x \in \mathbb R \Biggm | 0 \le x \le \frac n { n + 1 } \right \} = \left [ 0 , \frac n { n + 1 } \right ] $. Prove or disprove the statements below:

*

*$ \bigcup _ { n \ge 1 } A _ n = [ 0 , 1 ) $

*$ \bigcap _ { n \ge 1 } A _ n = \{ 0 \} $

My textbook for my class does not contain any notes on how to solve things like this. Any help solving these or links to sites that can show me how to solve things like this would be much appreciated.
 A: In general, when you have to prove that two sets are equal (each of them has the exact same elements), the most reliable method is called "double inclusion". Basically, to show that $A$ and $B$ are equal, you have to show that:
(a) every element of $A$ is an element of $B$ (mathematically we write this as $A\subseteq B$, and say $A$ is a subset of $B$), and
(b) every element of $B$ is an element of $A$ (again that would be $B\subseteq A$, $B$ is a subset of $A$).
This of course would imply that the only elements of $A$ are elements of $B$, and vice versa, thus meaning $A=B$. So we will do that here:
1. Let's first show that $\bigcup_{n\geq 1}A_n\subseteq [0,1)$. Let $x\in\bigcup_{n\geq 1} A_n$, we have to prove that $x$ is in $[0,1)$. But this is clear, since by definition of union we have that an element of the union belongs to at least one of the sets we are taking the union of. In other words, there exists a certain $A_n$, for a certain $n\geq 1$ such that $x\in A_n$. And, by definition of $A_n$, since $A_n=[0,\frac{n}{n+1}]$, we have $0\leq x\leq \frac{n}{n+1}$. Now, because $n<n+1$, their quotient $\frac{n}{n+1}$ is less than $1$: so, in the above reasoning for $x$, we can conclude $0\leq x\leq\frac{n}{n+1}<1$. Therefore $x\in[0,1)$, like we wanted to show.
For the counterpart, let's see that $[0,1)\subseteq\bigcup_{n\geq 1} A_n$. Let $x\in[0,1)$, meaning that $0\leq x<1$. We have to show that $x$ is in the union $\bigcup_{n\geq 1}A_n$, and that means exactly that we need to prove $x$ is in $A_n$, for some $n$. However, since $$
\frac{n}{n+1}\longrightarrow 1 $$
as $n$ goes to $\infty$, and $x$ is strictly less than $1$, there exists a certain $N\geq 1$, possibly very large, such that $x\in\left[0,\frac{N}{N+1}\right]$ (remember $x$ is fixed). But that means that $x$ is in $A_N$, and by our previous reasoning this implies that $x\in\bigcup_{n\geq 1} A_n$ (because $x\in A_n$ for some $n$).
By double inclusion, because every element of $\bigcup_{n\geq 1} A_n$ is an element of $[0,1)$, and every element of $[0,1)$ is part of $\bigcup_{n\geq 1} A_n$, we conclude that both sets are equal: that is, $\bigcup_{n\geq 1}A_n=[0,1)$
2 For the second one you can do the same thing, noting that an element $x$ is in the intersection $\bigcap_{n\geq 0} A_n$ if and only if, by definition of intersection, $x\in A_n$ for every $n$. Also there must be some copying error because $\bigcap_{n\geq 1} \neq \{0\}$, I assume $n$ starts at $0$. To do this, we reason the exact same way. Let's first show $\bigcap_{n\geq 0} A_n \subseteq \{0\}$. So, let $x\in\bigcap_{n\geq 0} A_n$; we have to show that $x\in\{0\}$. This means, that for every $n\geq 0$, we have
$$
x\in A_n=\left[0,\frac{n}{n+1}\right] \Rightarrow 0\leq x \leq \frac{n}{n+1};
$$
in particular $x\in A_0$, so $x\in\left[0,0\right]=\{0\}$.
For the reverse, to show $\{0\}\subseteq\bigcap_{n\geq 0} A_n$, let $x\in\{0\}$ (this means $x=0$, by the way). Then, to see $x$ is in the intersection we have to show $x\in A_n$ for every $n\geq 0$. But this is true, since for all $n$ we have
$$
x=0\in\left[0,\frac{n}{n+1}\right],
$$
meaning that $x\in\bigcap_{n\geq 0} A_n\subseteq\{0\}$.
By double inclusion, we see $\bigcap_{n\geq 0} A_n=\{0\}$.
A: One idea to prove this problem is to see that
$$A_n \subset A_{n+1},$$
for every $n \in \mathbb{N}$. Also, for $n \rightarrow \infty$, we have $$\dfrac{n}{n+1} \rightarrow 1.$$
Hence, $\displaystyle\bigcup_{n \geq 1} A_n = [0,1)$.
Furthermore, by the same inclusion $A_n \subset A_{n+1}$, it is easy to see that $\displaystyle\bigcap_{n \geq 1} A_n = A_1$.
