anonymous
  • anonymous
every collection of disjoint open intervals is countable
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
we know that rations are dense in the reals
zzr0ck3r
  • zzr0ck3r
This is true for any open set in R. It as most a countable union of open intervals
zzr0ck3r
  • zzr0ck3r
disjoint open intervals*

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anonymous
  • anonymous
i need to prove this
zzr0ck3r
  • zzr0ck3r
I am sure if you google it you will find many proofs. Anything I do will be a regurgitation.
zzr0ck3r
  • zzr0ck3r
http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv
zzr0ck3r
  • zzr0ck3r
I can help if you have questions on any part.
zzr0ck3r
  • zzr0ck3r
if \(O\) is an open set and \(x∈O\) then there exists an interval I such that \(x∈I⊂O\). If there exists one such interval, then there exists one 'largest' interval which contains \(x\) (the union of all such intervals). Denote by \(\{I_α\}\) the family of all such maximal intervals. First all intervals \(I_α\) are pairwise disjoint (otherwise they wouldn't be maximal) and every interval contains a rational number, and therefore there can only be a countable number of intervals in the family.
zzr0ck3r
  • zzr0ck3r
If you are more in to set theory then there is a pretty cool proof using that approach.
anonymous
  • anonymous
@zzr0ck3r is that interval \( I\) open
anonymous
  • anonymous
for example, (0,1) is open because if x is inside (0,1) you can form a 'ball' around x that is contained in (0,1)
anonymous
  • anonymous
for any x
zzr0ck3r
  • zzr0ck3r
Yes, this is the definition of open set on order topology.
zzr0ck3r
  • zzr0ck3r
technically (0,1) is a basis element for the topology, but you got the right idea.
zzr0ck3r
  • zzr0ck3r
basically because any collection of intervals will contain a rational point, we can index the intervals with one such rational point. The rational numbers are countable, so the collection is a countable collection.
anonymous
  • anonymous
so for example the point x= 0.25 in (0,1) the maximal open interval that is still contained in (0,1) would be (0, .5) ?
zzr0ck3r
  • zzr0ck3r
I would say more like \[0.25\in (0,1)\cup(34,199)\] then we would get \((0,1)\)
zzr0ck3r
  • zzr0ck3r
aight ill check back later. movie time with the wifey
anonymous
  • anonymous
I may be missing a condition here, but I was assuming the maximal interval had to be symmetric about x. So for \(x = 0.25 \) and \( O =(0,1)\cup(34,199) \), if you say that the interval \( I\) does not have to be symmetric about \(x\), then wouldn't the whole interval \((0,1)\cup(34,199)\) would be the maximal set that contains \( x\) ? Thanks for discussing this, i am trying to learn this.
zzr0ck3r
  • zzr0ck3r
no there is no need for it to be symmetric about x
zzr0ck3r
  • zzr0ck3r
we want a maxl interval that contains x, not a maxl open set that contains x

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