Every non empty open set on the real line is the union of a coutable disjoint class of open interval

Proof.

Let $\mathcal{O}$$\mathcal O$ be an open subset of the real line, let $x\in \mathcal{O}$$x\in\mathcal O$, and consider the family $G_x:=\{o\subseteq \mathcal{O}|x\in o,o\text{ is open interval}\}$$G_x:=\{o\subseteq\mathcal O| x\in o,o\text{ is open interval}\}$

Denote the union of the family $G_x$$G_x$ to be $I_x$$I_x$

Then

$\diamond$$\diamond$ $I_x$$I_x$ is open, since it is the union of open sets

$\diamond$$\diamond$ $\bigcup _{x\in \mathcal{O}}{I}_x=\mathcal{O}$$\bigcup_{x\in \mathcal O} I_x=\mathcal O$

$\diamond$$\diamond$ If $y$$y$ is another point in $I_x$$I_x$, then $I_x=I_y$$I_x=I_y$

Proof.

Since $y\in I_x$$y\in I_x$, $\mathrm{\exists }(a,b)\subseteq \mathcal{O},y\in (a,b)\wedge x\in (a,b)$$\exists(a,b)\subseteq \mathcal O, y\in (a,b)\wedge x\in (a,b)$

$\mathrm{\forall }$$\forall$ open sets $B$$B$ contains $x$$x$, $B\cup (a,b)$$B\cup (a,b)$ is also an open set containing $x$$x$ and $y$$y$

$B\mapsto B\cup (a,b)$$B\mapsto B\cup (a,b)$ is a map from $G_x$$G_x$ to $G_y$$G_y$

And since the union of $B\cup (a,b)$$B\cup (a,b)$ is $I_x$$I_x$

Thus $I_x\subseteq I_y$$I_x\subseteq I_y$

By duality, $I_y\subseteq I_x$$I_y\subseteq I_x$, just consider $\mathrm{\forall }$$\forall$ open sets $A$$A$ contains $y$$y$, $A\mapsto A\cup (a,b)$$A\mapsto A\cup (a,b)$

Actually, it naturally define an equivalence relation, $x\sim y\iff \mathrm{\exists }(a,b),x,y\in (a,b)$$x\sim y\Leftrightarrow\exists (a,b),x,y\in (a,b)$

$I_x$$I_x$ is the equivalence class

Proposition. $\sim$$\sim$ is an equivalence relation

$\diamond$$\diamond$ $x\sim x$$x\sim x$

$\diamond$$\diamond$ $x\sim y\iff y\sim x$$x\sim y\Leftrightarrow y\sim x$

is obviously

$\diamond$$\diamond$ $x\sim y,y\sim z$$x\sim y,y\sim z$, then $x\sim z$$x\sim z$

Proof. $x,y\in (a,b),y,z\in (c,d)$$x,y\in(a,b),y,z\in(c,d)$ then $x,z\in (a,b)\cup (c,d)$$x,z\in (a,b)\cup (c,d)$

Since $y\in (a,b)$$y\in (a,b)$ and $y\in (c,d)$$y\in (c,d)$, $(a,b)\cup (c,d)$$(a,b)\cup(c,d)$ is not disjoint union, $(a,b)\cup (c,d)=(a,d)$$(a,b)\cup(c,d)=(a,d)$

Thus $I_x$$I_x$ is a partition of $G$$G$

To see it is countable, consider $\mathbb{Q}$$\mathbb Q$

$r\mapsto I_r$$r\mapsto I_r$ is subjection to $\{x\in \mathcal{O},I_x\}$$\{x\in\mathcal O,I_x\}$, thus $\{x\in \mathcal{O},I_x\}$$\{x\in\mathcal O,I_x\}$ is countable.