Represent the interval [0,1) by union I can represent the set $(0,1]$ like: 
$(0,1] = \bigcap_{n=1}^{\infty} (0, \frac{n+1}{n}) = \bigcup_{n=1}^{\infty} [\frac{1}{n},1]$. 
But how can I represent $[0,1)$? I have a task, which requires this of me. 
$[0,1) = \bigcap_{n=1}^{\infty} (\frac{-1}{n},1) = \bigcup_{n=1}^{\infty} [0, \frac{1}{n}]$. 
Is this $\bigcup_{n=1}^{\infty} [0, \frac{1}{n}]$ correct form? 
 A: Your representation of $[0,1)$ as the intersection of open sets is correct, but not that as the union of closed sets. This can be obtained by "reversing" the corresponding representation of $(0,1]$:
$$\bigcap_{n=1}^\infty\left[0,\frac{n-1}n\right]$$
A: Presumably $\bigcup_{n=1}^{\infty} [0, \frac{1}{n}]$ means something like this:
$$ [0, \tfrac{1}{1}] \cup [0, \tfrac{1}{2}] \cup [0, \tfrac{1}{3}] \cup \cdots$$
where the "$\cdots$" represents the limit as we take unions of more sets of the form 
$[0, \frac{1}{n}]$ on the right side of the expression.
Since $[0, \frac{1}{n}] \subseteq [0, \frac{1}{1}]$ whenever $n \geq 1,$
the union comes out to simply $[0, \tfrac{1}{1}].$
This has $1$ as an element, which is not what you want.
Starting at a larger value of $n$ doesn't help; it eliminates $1$ but also eliminates many elements you want to include, such as $\frac34.$
Here's an alternative idea: instead of having the right-hand end of the interval start at some positive value and approach $0$, have it approach the open end of the desired set instead; that is, have it approach $1.$
