- #1
dumbQuestion
- 124
- 0
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 αx = inf {y | (y, x] ⊂ G} and βx =
sup {y| [x, y) ⊂ G}. The fact that G is open implies αx < x < βx. Define Ix = (αx, βx). Then Ix ⊂ G"
Here is where my confusion begins: isn't Ix G itself? We know G is an open interval on ℝ, so G = (a,b) for some a < b. So isn't αx = a and βx =
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.
(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 αx = inf {y | (y, x] ⊂ G} and βx =
sup {y| [x, y) ⊂ G}. The fact that G is open implies αx < x < βx. Define Ix = (αx, βx). Then Ix ⊂ G"
Here is where my confusion begins: isn't Ix G itself? We know G is an open interval on ℝ, so G = (a,b) for some a < b. So isn't αx = a and βx =
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.