Proof that every open set in $\mathbb{R}$ is a countable union of disjoint open intervals using equivalence classes

Question

We were told to prove that every open set in $\mathbb{R}$ is a countable union of disjoint open intervals using equivalence classes in the following way:

1. Consider the following relation on a given open set $\mathcal{O}$: $x \sim y$ iff $[x,y] \subseteq \mathcal{O}$ or $[y,x] \subseteq \mathcal{O}$ depending on whether $x \leq y$ or $y \leq x$. Show that it is an equivalence relation. DONE
2. Denote the equivalence class of $x \in \mathcal{O}$ by $I_{x}$. Consider the extended reals $a_{x} = \inf I_{x}$, $b_{x} = \sup I_{x}$. Show that $a_{x}, b_{x} \notin I_{x}$. (Hint: Use $\mathcal{O}$ being open.) Conclude $I_{x} = (a_{x}, b_{x})$; i.e., that $I_{x}$ is an open interval. THIS IS WHAT I AM CURRENTLY WORKING ON AND NEED HELP WITH - I believe the remaining two parts of the proof will fall into place easily for me once I have established this part.

I am having the following difficulties with Part 2:

• $I_{x} = \left \{ y | y \in X, x \sim y \right \}$. By reflexivity, $x \in I_{x}$, so $I_{x}$ is nonempty. I was looking at the proof of this theorem in another text, and it said that since $\mathcal{O}$ is open, the $I_{x}$ are nonempty. I have a feeling this is a better justification than the reason why I gave that $I_{x}$ must be nonempty, so if someone could please explain to me why this is true, I would be most appreciative.
• How do I know there must exist another element in $I_{x}$ distinct from $x$? (Again, this might go along with the question I just asked in the above bullet point.)
• Is it immediately obvious that $I_{x} \subseteq \mathcal{O}$ or do I have to show that it is? If so, how?
• I have successfully been able to show that $b_{x} \notin I_{x}$. However, I am having difficulty showing that $\mathbf{a_{x} \notin I_{x}}$ I believe that if I assume that $a_{x} \in I_{x}$, in order to show a contradiction, if $x \in I_{x}$ I need to show that $(a_{x}+s, x] \subseteq \mathcal{O}$, where $s > 0$, so that way, $a_{x}$ will no longer be the greatest lower bound. I have attempted the following thus far:

Let $w \in I_{x}$ so that $a_{x} < w < x$. Then, by definition of $a_{x}$ as the infimum, $\exists y_{1}$ such that $(y_{1}, x) \subseteq [y_{1}, x] \subseteq \mathcal{O}$, so $w \in \mathcal{O}$.

Suppose that $a_{x} \in \mathcal{O}$, then since $\mathcal{O}$ is open, $\exists s > 0$ for which $(a_{x}-s,a_{x}+s) \subseteq \mathcal{O}$.

Now is where I run into my difficulty: we have that $a_{x} < w < x$, so $a_{x} - s < w - s < x - s$ and $a_{x} + s < w + s < x + s$. I need $a_{x} + s < x$ (actually, I suppose I need $a_{x} + x < w$), but (in either case), I am unable to establish that bound. Could somebody please help me with this? I've been trying to figure it out since yesterday with no luck.

Thank you!!

Answer 1

Daniel answered most of your questions in his comment, so I'll just address thet last one here. We proceed by contradiction as you suspected.

If $a_x \in I_x$, then $a_x \sim x$ and $a_x \in \mathcal{O}$. Since $\mathcal{O}$ is open, there is some $s > 0$ such that $(a_x - s, a_x+s) \subset \mathcal{O}$. We then have that $$ \left[a_x - \frac{s}{2}, a_x \right] \subset (a_x - s, a_x+s) \subset \mathcal{O}. $$ Then $a_x - \frac{s}{2} \sim a_x$, so $a_x - \frac{s}{2} \sim x$ (by transitivity). This shows that $a_x - \frac{s}{2} \in I_x$, but this contradicts the definition of $a_x$, as $a_x - \frac{s}{2} < a_x$.

Thus $a_x \notin I_x$, and an analogous argument shows that $b_x \notin I_x$ also.