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 $(0,\frac{1}{2}+\epsilon) \cup (\frac{1}{2}-\epsilon,1)$?
Best Answer
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:
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.