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anonymous
 one year ago
every collection of disjoint open intervals is countable
anonymous
 one year ago
every collection of disjoint open intervals is countable

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we know that rations are dense in the reals

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3This is true for any open set in R. It as most a countable union of open intervals

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3disjoint open intervals*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i need to prove this

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3I am sure if you google it you will find many proofs. Anything I do will be a regurgitation.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3I can help if you have questions on any part.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3if \(O\) 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
 one year ago
Best ResponseYou've already chosen the best response.3If you are more in to set theory then there is a pretty cool proof using that approach.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r is that interval \( I\) open

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3Yes, this is the definition of open set on order topology.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3technically (0,1) is a basis element for the topology, but you got the right idea.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3basically 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.3I would say more like \[0.25\in (0,1)\cup(34,199)\] then we would get \((0,1)\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3aight ill check back later. movie time with the wifey

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3no there is no need for it to be symmetric about x

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.3we want a maxl interval that contains x, not a maxl open set that contains x
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