# open set is the disjoint union of a countable collection of open intervals

I don't understand that $$\{I_x\}_{x\in O}$$ is disjoint. For example, $$O = \{(1 , 2)\}$$. Let $$x = 1.5$$. Then, $$a_x = 1$$, and $$b_x = 2$$. Therefore, $$I_x = (1, 2)$$. Let $$y = 1.6$$. Then, similarly, $$I_y = (1, 2)$$. That is, $$I_x$$ and $$I_y$$ are not disjoint, but equal.

Could you explain how $$\{I_x\}_{x \in O}$$ can be disjoint?

• Royden Fitzpatrick? – BCLC Jul 28 '18 at 9:34

The theorem doesn't require $I_x$ and $I_y$ to be disjoint for every $x$ and $y$; only that the collection is mutually disjoint. That is, we require that if $I_x \cap I_y \not = \emptyset$, then $I_x = I_y$. (This is a bit sloppily stated in the proof, though.)