Intersection of sets: $\bigcap_{n \in \mathbb{Z}^{+}} \left[ -\frac{1}{n}, 0 \right[$ Why is the following true?$$\bigcap_{n \in \mathbb{Z}^{+}} \left[ -\frac{1}{n}, 0 \right[ = \emptyset$$
The sets which are intersected get "smaller", and as $n \rightarrow \infty$, I reasoned that the interval becomes $[0,0[$. I don't exactly know how to make sense of that, but why not let this intersection be the set $\{0\}$ instead of an empty set? I understand that $[-1/n,0[$ does not contain zero, but $[0,0[$ seems to contain and not contain (?) zero.
Thanks in advance. 
 A: Suppose that set was not empty. Then, there would be $x \in \mathbb{R}$ such that $x\geq\frac{-1}{n}$ for every $n \in \mathbb{N}$, but $x<0$. Can you show why there is no such x?
A: You've got the hard part but not the easy here.  It does become $\left[0,0\right[$; but this is obviously the empty interval.  $\left[x,x\right]$ is merely $\left\{x\right\}$, but $\left[x,x\right[$ is open on the right, and so the set can't contain $x$ either.  Since that's the only thing in the set already, there's nothing left.
A: The question looks interesting, the primary purpose is to dissect the following method!
Say $x\in[-1/n,0)$
Now for any $x$ (negative) which belong to any of the above  the set, we can always find a $n_1\in Z^+$, such that $x<-1/n_1$. Hence there can't be any number which satisfies the above condition!
A: $
\newcommand{\calc}{\begin{align}\quad &}
\newcommand{\calcop}[2]{\\ #1\quad &\quad\text{"#2"}\\\quad & }
\newcommand{\endcalc}{\end{align}}
$Here is how I would prove this, directly using the definitions.  Implicitly letting $\;n \in \mathbb Z ^+\;$, we calculate the elements $\;x\;$ of this set:
$$\calc
x \in \langle \cap n :: \left[ -1/n, 0 \right[ \rangle
\calcop{\equiv}{definition of $\;\cap\;$}
\langle \forall n :: x \in \left[ -1/n, 0 \right[ \rangle
\calcop{\equiv}{definition of $\;[\cdot,\cdot[\;$}
\langle \forall n :: -1/n \leq x \lt 0 \rangle
\calcop{\equiv}{logic: move part not using $\;n\;$ out of $\;\forall n\;$; arithmetic: multiply by $\;n\;$}
\langle \forall n :: -1 \leq x \times n \rangle \;\land\; x \lt 0
\calcop{\equiv}{arithmetic: divide by $\;x\;$, allowed with changing sign by $\;x \lt 0\;$}
\langle \forall n :: -1/x \geq n \rangle \;\land\; x \lt 0
\calcop{\equiv}{arithmetic: $\;\langle \forall n :: y \geq n \rangle\;$ is false}
\text{false}
\endcalc$$
