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

i'm studying real analysis with royden, and i've looked up similar qeustions and answers but i couldn't get the exact answer that i need.

i don't need the whole process of proof , and i confused with certain phrase.

First let me show you what i read. royden

i know the procedure in this proof, but i don't know why {I$_x$} is countable.

they explain this in line 12 , but i can't understand it.

especially, i think, when i try to construct {I$_x$} , there exist repeated open intervals for certain interval in open subset $O$ , then the number of open interval of {I$_x$} must be finite.

then how can we say that {I$_x$} is countable?

## with specific example , i understood I$_x$ for [3,10] as a interval of open set $O$ then, I$_7$ and I$_8$ is same , so we write only one of them into {I$_x$} , it means unless $O$ has countable interval, {I$_x$} cannot be countable ..

what do i missed? or misunderstood? let me know

## 1 Answer

Each $I_x$ contains a rational number say $q_x$.

Since the $I_x$ are disjoint, each $q_x$ belongs to exactly one $I_x$.

This produces a bijection between the sets $\{I_x\}$ and $\{q_x\}$, which is a subset of the rationals and hence countable.

• is my confusion came from only the concept between finite and countable? then i understood the answer from which i asked. "countable" that the proposition says, is not meant "infinitely countable" and "countable" means "finite or infinitely countable" and we can conclude that we need at most countable union, but we can say "finite" in certain open set. am i right?
– MOON
Apr 26 '16 at 8:49
• Yes, countable means either finite or countably infinite. So a finite set is countable. Apr 26 '16 at 8:51