Simplifying a set and proving it Let $A$ be a set defined as the intersection
$$\bigcap_{n\in\mathbb N}\{x \in\mathbb R | -1/(2^n) < x < 1+ n| n \in N\}$$
Not I believe it's $(0, 1]$ but I could only prove that $(0,1]$ is a subset of the intersection above(A), but I can't prove that A is a subset of $(0,1]$. 
I proved that for basically for any natural number:
$-1/(2^n) > 0$
and $1 + n\geq 1 $
But I am stuck from here on
 A: You are incorrect. The set $A$ is $[0,1)$ (if you consider $0$ an element of $\mathbb N$) or $[0,2)$ (if you do not consider $0$ an element of $\mathbb N$.

Assuming $0\in\mathbb n$, then here is a sketch of the proof. Let's call $A_n=\{x\in\mathbb R| -\frac{1}{2^n}<x<1+n\}$.
First, $A\subseteq [0,1)$:
Take any $x\in A$. Therefore, $x\in A_n$ for all values of $n$, so, for all $n$, $x>-\frac{1}{2^n}$ which means $x\geq 0$ (*this statement actually should be proven in your final solution)
Also, $x\in A_0$, which means $x<1+0=1$.
So, we conclude $0\leq x<1$, meaning $A\subseteq[0,1)$.
Second, $[0,1)\subseteq A$:
Let $x\in A$ and let $n\in\mathbb N$. Then, $-\frac{1}{2^n} < 0\leq x$ meaning $x>-\frac{1}{2^n}$. Also, $x<1\leq 1+n$, so $x<1+n$. Therefore, $x\in A_n$.
Since $n$ was arbitrary, we conclude $x\in A_n$ for all $n$, which means $x\in A$.
A: $$A=\bigcap_{n\in\Bbb N}(-1/2^n,n+1)=\bigcap_{n\in\Bbb N}(-1/2^n,0]\cup (0,n+1)=\bigcap_{n\in\Bbb N}(-1/2^n,0]~\bigcup~\bigcap_{n\in\Bbb N}(0,n+1)$$
The individual intersections are $\{0\}$ and $(0,1)$ respectively if $0\in\Bbb N$, otherwise they are $\{0\}$ and $(0,2)$ (notice that $(0,n+1)\subset (0,n+2)$ for all $n\in\Bbb N$, so the intersection of the second part is obtained for the smallest element of $\Bbb N$)
Hence, $$A=\begin{cases}\{0\}\cup (0,1)=[0,1)~\textrm{if }0\in\Bbb N\\ \{0\}\cup (0,2)=[0,2)~\textrm{otherwise}\end{cases}$$
Note: We're allowed to break down the intersection into two intersections because $(-1/2^n,0]$ and $(0,m+1)$ are disjoint for all $m,n\in\Bbb N$
More precisely, $\bigcap_n A_n\cup B_n=\bigcap_n A_n~\bigcup~\bigcap_n B_n$ only when the $A_i$'s and $B_j$'s are disjoint $\forall~i,j$
