Open set $(0,1)$ as union of disjoint open sets One theorem says "Any open subset of  $\mathbb{R}$ can be written as the countable union of disjoint open intervals ".
I am trying to see how one can partition the open set $(0,1)$ such that it is the union of disjoint open sets? (other than the trivial partition of $(0,1)$ itself).  Is it possible?
For example, if we take the sets $(0,\frac{1}{2})$ and $(\frac{1}{2},1)$, clearly the point $\frac{1}{2}$ does not lie in both open sets, and it can not be in the union either. After the union we get a set $(0,1)\setminus\{\frac{1}{2}\}$. What is a meaningful partition for the open set $(0,1)$ such that it partitions are open?  Can I do something like $(0,\frac{1}{2}+\epsilon) \cup (\frac{1}{2}-\epsilon,1)$?
 A: One possible approach is to write down the most general interpretation of the definition, and see what happens.  A countable union of disjoint open sets is a set of the form
$$ \bigcup_{n=1}^{\infty} U_n $$
where $U_m\cap U_n = \emptyset$ whenever $m\ne n$ and each $U_n$ is open.  
Note that the emptyset itself is open and that the definition does not require that the sets in the union be nonempty.  So, for example, we can write
$$ (0,1) = \bigcup_{n=1}^{\infty} U_n, $$
where $U_1 = (0,1)$ and $U_n = \emptyset$ for all $n > 1$.  You should check that this really is a disjoint union of open sets, and that it really gives you $(0,1)$, but this check should be pretty straight-forward.

Alternatively, if we understand the word "countable" to mean "any natural number or countably infinite" then the union
$$ \bigcup_{n=1}^{1} (0,1) $$
gets the job done.

As per the comments, a part of the question that I did not address is the following:

Do there exist two (or more) disjoint open sets $U_1$ and $U_2$ such that $(0,1) = U_1 \cup U_2$?

The answer is "No."  The interval $(0,1)$ is a connected set (see this question for an argument that justifies this statement; we lose no generality replacing the closed unit interval with the open unit interval).  If there were two such open sets, then they would form a disconnection of the interval $(0,1)$, which is a contradiction.  Note that the non-existence of a nontrivial disjoint cover of $(0,1)$ by open sets does not violate the original theorem in any way, via the reasoning given in the first part of this answer.
A: A union of one term counts as a "disjoint" union.  If you could write $(0,1)$ as a union the way you want to, it would be disconnected (it's not).
A: {(0,1)} is a countable collection of pairwise disjoint,
open intervals whose union is (0,1);  the trivial partition.  
Since R, (a,oo), (-oo,a) are open and connected, the term
open interval has to include the open rays and R itself.
Riddle of the day.  How would you partition the open
empty set into pairwise disjoint open intervals?
