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

• Maybe $(0,1)$ itself? Aug 1 '18 at 19:23
• i clearly excluded the trivial case.
– Shew
Aug 1 '18 at 19:30
• It not just the trivial case, it is the only case! Aug 2 '18 at 2:51

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.

• i clearly mentioned that excluding the trivial case of (0,1).
– Shew
Aug 1 '18 at 19:30
• I assumed that you mentioned the trivial case because you were confused. In answer to that question, the answer is "no", no other partition is possible. This is because $(0,1)$ is connected. Aug 1 '18 at 19:31
• If you wish to exclude the trivial case and all things equivalent to it (appending a bunch of empty sets) then the answer is "no" for connectivity reasons. Aug 1 '18 at 19:31

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).

{(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?