Prove or disprove: $\bigcap_{n\in\mathbb{N}}\left[-\frac{1}{n!},1+\frac{1}{2^{n}}\right]=\left \{ 1,2 \right \}$ 
Prove or disprove:
  $\bigcap_{n\in\mathbb{N}}\left[-\frac{1}{n!},1+\frac{1}{2^{n}}\right]=\left
\{ 1,2 \right \}$

I have problems understanding the symbols.
$\bigcap_{n\in\mathbb{N}}$ stands for intersection, right? If so, what actually is intersected here?
What's meant by $\left[-\frac{1}{n!},1+\frac{1}{2^{n}}\right]$, especially by those brackets?
For me it looks like this is an interval, starting from $-\frac{1}{n!}$ going till $1+\frac{1}{2^{n}}$.
What is meant by $\left \{1,2 \right \}$? This seems to be a set, right?
So the complete thing in words is saying: the intersection of the interval $\left[-\frac{1}{n!},1+\frac{1}{2^{n}}\right]$ equals the set $\left \{1,2 \right \}$, is that correct?

I would solve it like that:
We know that $n\in\mathbb{N}$ which is very important info.
Starting from $1$, I would insert several values for $n$ in the interval:
For $-\frac{1}{n!}$ we will always get negative rational numbers which is bad (bad in that case that this will most likely lead to a contradiction.
For $1+\frac{1}{2^{n}}$ we will get positive rational numbers which is also bad.
Since there is no way to even get a single positive natural number, we can never get to the set $\left \{1,2 \right \}$. So the statement is wrong.
I hope I understood most things correctly?
 A: Your (bold-faced) statement is not quite correct. 
Instead, what it is saying is that the intersection of the collection of intervals $\bigl[ - \frac{1}{n!}, 1 + \frac{1}{2^n} \bigl]$ (for $n \in \mathbb{N}$) equals the set $\{1,2\}$. Perhaps you might understand the intersection better on an intuitive level if you wrote out the first few intervals that are being intersected, one by one:
$$\bigl[ - \frac{1}{1!}, 1 + \frac{1}{2^1} \bigl] \,\, \cap  \,\, \bigl[ - \frac{1}{2!}, 1 + \frac{1}{2^2} \bigl]  \,\, \cap  \,\, \bigl[ - \frac{1}{3!}, 1 + \frac{1}{2^3} \bigl]  \,\, \cap  \,\, \bigl[ - \frac{1}{4!}, 1 + \frac{1}{2^4} \bigl]  \,\, \cap  
\,\, \cdots
$$
You can then simplify the expression, if you wish, by doing some arithmetic:
$$\bigl[ - 1, 1 + \frac{1}{2} \bigl] \,\, \cap  \,\, \bigl[ - \frac{1}{2}, 1 + \frac{1}{4} \bigl]  \,\, \cap  \,\, \bigl[ - \frac{1}{6}, 1 + \frac{1}{8} \bigl]  \,\, \cap  \,\, \bigl[ - \frac{1}{24} , 1 + \frac{1}{16} \bigr] \,\, \cap \cdots
$$
So, now, ask yourself, is the set of all numbers $x$ which is in this intersection equal to the set $\{1,2\}$?
A: Counterexample for $n=1$:
$$2\not\in\left[-1,1+\frac12\right]\implies2\not\in\left[-\frac{1}{1!},1+\frac{1}{2^{1}}\right]\implies2\not\in\bigcap_{n\in\mathbb{N}}\left[-\frac{1}{n!},1+\frac{1}{2^{n}}\right]$$
A: \begin{align}
  & {{A}_{1}}=\left[ -1\,,\frac{3}{2} \right] \\ 
 & {{A}_{2}}=\left[ -\frac{1}{2}\,,\frac{5}{4} \right] \\ 
 & {{A}_{3}}=\left[ -\frac{1}{6}\,,\frac{9}{8} \right] \\ 
 & {{A}_{4}}=\left[ -\frac{1}{24}\,,\frac{17}{16} \right] \\ 
 & \vdots  \\ 
 & \underset{n\to \infty }{\mathop{\lim }}\,{{A}_{n}}=[0,1] \\ 
\end{align}
because
\begin{align}
  & \underset{n\to \infty }{\mathop{\lim }}\,-\frac{1}{n!}=0 \\ 
 & \underset{n\to \infty }{\mathop{\lim }}\,\left( 1+\frac{1}{{{2}^{n}}} \right)=1 \\ 
\end{align}
$$\bigcap\limits_{n=1}^{+\infty }{{{A}_{n}}}=[0,1]$$
