Specific part (not entire question) of the proof that O can be written as a union of disjoint open intervals I have completed a proof of the following statement:
Prove that any bounded open subset of R is the union of disjoint open intervals
Let O $\subset$ R
We want to write O as a disjoint union of open intervals
Since O is bounded, we can define $\forall x \in O$
\begin{align}
a_x &= \inf\{ y \in R: (y,x) \subset O\}\\
b_x &= \sup\{z \in R: (x,z) \subset O\}
\end{align}
Let
$$\ U = \bigcup_{x \in O} (a_x,b_x),$$
where $(a_x,b_x) \subset O$.
There is more to the proof, but this is the part I need help on. I have been told by my instructor the proof is correct. However, he says it is not evident that $(a_x,b_x) \subset O$. He wants me to prove this specific part, and I cannot figure out how to do it. I know I need to take an element in $(a_x,b_x)$, and show that it is in O, but I am not sure how to do this and how to connect it. I've been trying this for a week, and I've decided it is about time I ask for help. Can anyone help me here?
 A: So $a_x = \inf\{y \in \Bbb R\mid (y,x) \subseteq O\}$
and $b_x = \sup\{y \in \Bbb R\mid (x,y) \subseteq O\}$
which are well-defined (there are $a' < x < b'$ such that $(a',b') \subseteq O$ as $x$ is in the interior of $O$ so the sets are non-empty); and $O$ is bounded so any lower bound for $O$ is one for the set defining $a_x$ etc.
Why $(a_x, b_x) \subseteq O$? Let $z \in (a_x, b_x)$. If $z=x$ we are done, so assume (first case) that $z < x$. Because $a_x$ is the greatest lower bound for the set $A_x:=\{y \in \Bbb R\mid (y,x) \subseteq O\}$, we are sure that $z$ is not a lower bound for $A_x$, so there is some $z' \in A_x$ such that $z' < z$. But then $z' \in A_x$ implies $(z',x) \subseteq O$ and as $z \in (z',x)$, we know $z \in O$.
A similar argument can be held for $b_x$ and $B_x = \{y \in \Bbb R\mid (x,y) \subseteq O\}$ when $z > x$ instead, using that $b_x$ is the smallest upper bound for $B_x$ etc.
A: Hint: prove that $(a_x+\frac1n,\, b_x-\frac1n)\subseteq O$ for all $x\in O$ and big enough $n\in\Bbb N$.
A: Let $x\in O.$
Consider any $v\in (x,b_x).$ We have $v<b_x.$ Now  $$ v\not \in O\implies \forall z\in \Bbb R\,(\,(x,z)\subset O\implies z\le v)\implies b_x\le v$$ contrary to $v<b_x.$ So $v\in O.$
So $(x,b_x)\subset O.$
Similarly we obtain $(a_x,x)\subset O.$
So $(a_x,b_x)=(a_x,x)\cup \{x\}\cup (x,b_x)\subseteq O.$
