# Every open subset $U\subseteq\mathbb{R}$ is countable union of disjoint open intervals.

Theorem. Every open subset $$U\subseteq\mathbb{R}$$ is countable union of disjoint open intervals.

I was looking for proofs for this result and I came to this interesting post: Any open subset of $$\Bbb R$$ is a at most countable union of disjoint open intervals. [Collecting Proofs].

Among all the proofs I started from the simplest one: the one written by G.T. https://math.stackexchange.com/a/1949873/554978.

However, I am not convinced of the proof that $$I_x$$ is an interval. I do not know if I did not understand correctly or the proof in question is not valid, so I propose one myself and I would like you to tell me which one is valid.

Definition. An interval is a subset $$I\subseteq\mathbb{R}$$ such that, for all $$a in $$\mathbb{R}$$, if $$a,b\in I$$ then $$c\in I$$.

Let $$x\in U$$ and we suppose that $$x\in\mathbb{Q}$$, then define \begin{align} I_x = \bigcup\limits_{\substack{I\text{ an open interval} \\ x~\in~I~\subseteq~U}} I,\end{align} we must prove that $$I_x$$ is an interval. About that let $$a,b\in I_x$$ such that $$a, then we must prove that $$c\in I_x$$.

Since $$a,b\in I_x$$, then $$a,b\in I$$ for same open interval $$I$$ which contains $$x$$. If $$a$$ and $$b$$ belong to the same $$I$$, since $$a and $$I$$ is an interval, $$c\in I$$, therefore $$c\in I_x$$.

Now, we denote with $$I_a$$ the open interval of $$I_x$$ which contains $$a$$, but not $$b$$ and we denote with $$I_b$$ the open interval of $$I_x$$ which contains $$b$$, but not $$a$$.

First case: $$[c=x]$$.

If $$c=x$$, then $$c\in I_x$$ by definition of $$I_x$$;

Second case: $$[c.

If $$c, then either $$a or $$a.

$$(i)$$ If $$a, since $$a\in I_a$$ and $$x\in I_a$$ and $$I_a$$ is an interval, then $$c\in I_a$$, therefore $$c\in I_x$$.

$$(ii)$$ If $$a, since $$x\in I_a$$ and $$a\in I_a$$, and $$I_a$$ is an interval we have that $$b\in I_a$$, absurd.

Third case $$[c>x]$$.

If $$c>x$$, then either $$a or $$x.

$$(i)$$ If $$a, since $$a\in I_a$$, $$x\in I_a$$ and $$I_a$$ is an interval we have that $$c\in I_a$$ therefore $$c\in I_x$$.

$$(ii)$$ If $$x, since $$x\in I_b$$ and $$b\in I_b$$, we have that $$a\in I_b$$ absurd.

Then in general $$c\in I_x$$, this prove that $$I_x$$ is an interval.

Thanks!

The third case is solved incorrectly.

Third case $$[c>x]$$

If $$c>x$$, then either $$a or $$x.

It should be $$a < x < c < b$$, not $$a < c < x < b$$.

So, the proof of the first part should now be:

(i) If $$a < x < c < b$$, since $$b \in I_b$$, $$x \in I_b$$ and $$I_b$$ is an interval we have that $$c \in I_b$$ and therefore $$c \in I_x$$.

With this correction, your proof is alright. Well done! :)