Union and intersection of the family of sets $[-1,1-\frac1n]$; describe and prove! 
Problem: Describe $\bigcap_{n=1}^{\infty}[-1,1-\frac{1}{n}]$ and
  $\bigcup_{n=1}^{\infty}[-1,1-\frac{1}{n}]$. And prove that their results are true.

Let $A_{n}=\left [-1,1-\frac{1}{n}  \right ]$. We see that 
\begin{equation*}
A_{1}=\left [-1,0\right ]\qquad A_{2}=\left [-1,\frac{3}{4}\right ]\qquad \dots\qquad A_{\infty}=\left [-1,1\right ]
\end{equation*}
so we have
\begin{equation*}
\bigcap_{n=1}^{\infty}A_{n}=A_{1}\cap A_{2}\cap A_{3}\cap \dots\cap A_{\infty}=\left [ -1,0 \right ]
\end{equation*}
and
\begin{equation*}
\bigcup_{n=1}^{\infty}A_{n}=A_{1}\cup A_{2}\cup A_{3}\cup \dots\cup A_{\infty}=\left [ -1,1\right]
\end{equation*}
If I try to google something about it, the second one should have been $\bigcup_{n=1}^{\infty}A_{n}=[-1,1[$. It doesn't really make sense why. If it's true, then $A_{\infty}$ must be wrong. Could  you please elaborate? 
And how do I prove them more precisely by showing two different ways: $\subseteq $ and $\supseteq $? I've tried to look for some inspirations in See page 73, Example 3b but it doesn't really give me some ideas how to prove on my own.
 A: You interpret the notation $\bigcup_{n=1}^{\infty}[-1,1-\frac{1}{n}]$ a bit too literally.  Unlike finite unions, this infinite union does not contain a term corresponding to the "upper limit" $\infty$. To spell out this notation correctly, one does not write $A_{1}\cup A_{2}\cup A_{3}\cup \dots\cup A_{\infty}$ but rather $A_{1}\cup A_{2}\cup A_{3}\cup \dots$ (without a last term). Over the hyperreals, one can have such infinite terminating unions with a last term having an infinite index $H$ (better notation than $\infty$), but even then this last term will not contain the number 1.  This is because $1/H$ is not zero; it is a nonzero infinitesimal.
A: There is no $A_\infty$: the notation is traditional and unfortunate. If you write 
$$
\bigcap_{n\ge1}A_{n}
$$
you'll probably will figure out better what's the set you want to compute. The variable $n$ is considered to take on natural number values and $\infty$ is not a natural number.
A: I see this problem as a classic homework in Basic Set Theory. Remember,

Definition of $\displaystyle\bigcup_{i=1}^{\infty}$ : $
x\in \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]
\Longleftrightarrow
\exists \;n \in\mathbb{N}-\{0\}, \mbox{ such that } x\in \left[-1,+1-\frac{1}{n} \right]
$

Note that, for all $x\in[-1,1)$ there is $n_x\in\mathbb{N}-\{0\}$ such that 
$
-1<x<1-\frac{1}{n_x}<1.
$
Then $x\in \left[-1,1-\frac{1}{n_x}\right]$. By cause  $\left[-1,1-\frac{1}{n_x}\right]\subset \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]$ we have $x\in \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]$. Therefore, $\left[-1,1\right)\subset \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n}\right]$. For outher hand, 
$
 \left[-1,1-\frac{1}{n}\right]\subset  \left[-1,1,\right)
$
for all $n\in\mathbb{N}-\{0\}$ implies 
$\bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]\subset \left[-1,1\right)$. For the characterization of equality of sets via subset and superset can conclude,
$$
\bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]= \left[-1,1\right).
$$
The other part of the exercise is done in a similar way using the definition below and conveniently exchanging in demonstrating the existential quantifier by universal quantifier. But the difference now is that $
\bigcap_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]= \left[-1,0\right).
$

Definition of  $\displaystyle\bigcap_{i=1}^{\infty}$:
  $\quad
x\in \bigcap_{i=1}^{\infty} \left[-1,1-\frac{1}{n}\right]
\Longleftrightarrow
\forall \;n \in\mathbb{N}, \mbox{ hold } x\in \left[-1,+1-\frac{1}{n} \right]
$

