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 $a<c<b$ 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 $a<c<b$, 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 $a<c<b$ 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: $[c<x]$.
If $c<x$, then either $a<c<x<b$ or $a<c<b<x$.
$(i)$ If $a<c<x<b$, 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 $a<c<b<x$, 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 $a<c<x<b$ or $x<a<c<b$.
$(i)$ If $a<c<x<b$, 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 $x<a<c<b$, 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!
Best Answer
The third case is solved incorrectly.
It should be $a < x < c < b$, not $a < c < x < b$.
So, the proof of the first part should now be:
With this correction, your proof is alright. Well done! :)