Open subset of R written as a countable union of pairwise disjoint open intervals?

[dumbQuestion]

I wasn't sure if I should post this in the analysis or topology forum, but this seems to be closely related to compactness so I thought I'd post it here. When dealing with ℝ, the following theorem seems to be really important:"Every non-empty open set G in ℝ can be uniquely expressed as a finite or countably infinite union of pairwise disjoint open intervals in ℝ"Unfortunately, I have a very difficult time figuring out this proof even though apparently it seems like it's supposed to be pretty obvious. Sadly, the theorem doesn't even make intuitive sense to me. The proofs I see all start at like this:

(I'm just copying from this example document http://www.math.louisville.edu/~lee/RealAnalysis/IntroRealAnal-ch05.pdf , the proof on the top of page 5-7, because this seems to be a common way to tackle this proof)

"Let G be open in R. For x ∈ G let α_x = inf {y | (y, x] ⊂ G} and β_x =
sup {y| [x, y) ⊂ G}. The fact that G is open implies α_x < x < β_x. Define I_x = (α_x, β_x). Then I_x ⊂ G"

Here is where my confusion begins: isn't I_x G itself? We know G is an open interval on ℝ, so G = (a,b) for some a < b. So isn't α_x = a and β_x =
sup {y| [x, y) ⊂ G} = b? I mean just by def of sup and inf? I feel this is where all my confusion stems from and if I could clear up my faulty logic at this step I could digest the remainder of the proof.

 

[jgens]

Not all open sets are intervals in \mathbb{R}. Take the set (-1,0) \cup (0,1) for example.

 

[micromass]

Why do you think G is an open interval?? The theorem merely says that G is an open set. Not all open sets are open intervals! An easy example is (0,1)\cup (2,3).

 

[dumbQuestion]

aaaaah, thanks guys! I'm facepalming at the moment...