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anonymous

  • one year ago

every collection of disjoint open intervals is countable

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  1. anonymous
    • one year ago
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    we know that rations are dense in the reals

  2. zzr0ck3r
    • one year ago
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    This is true for any open set in R. It as most a countable union of open intervals

  3. zzr0ck3r
    • one year ago
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    disjoint open intervals*

  4. anonymous
    • one year ago
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    i need to prove this

  5. zzr0ck3r
    • one year ago
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    I am sure if you google it you will find many proofs. Anything I do will be a regurgitation.

  6. zzr0ck3r
    • one year ago
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    I can help if you have questions on any part.

  7. zzr0ck3r
    • one year ago
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    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.

  8. zzr0ck3r
    • one year ago
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    If you are more in to set theory then there is a pretty cool proof using that approach.

  9. anonymous
    • one year ago
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    @zzr0ck3r is that interval \( I\) open

  10. anonymous
    • one year ago
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    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)

  11. anonymous
    • one year ago
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    for any x

  12. zzr0ck3r
    • one year ago
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    Yes, this is the definition of open set on order topology.

  13. zzr0ck3r
    • one year ago
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    technically (0,1) is a basis element for the topology, but you got the right idea.

  14. zzr0ck3r
    • one year ago
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    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.

  15. anonymous
    • one year ago
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    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) ?

  16. zzr0ck3r
    • one year ago
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    I would say more like \[0.25\in (0,1)\cup(34,199)\] then we would get \((0,1)\)

  17. zzr0ck3r
    • one year ago
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    aight ill check back later. movie time with the wifey

  18. anonymous
    • one year ago
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    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.

  19. zzr0ck3r
    • one year ago
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    no there is no need for it to be symmetric about x

  20. zzr0ck3r
    • one year ago
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    we want a maxl interval that contains x, not a maxl open set that contains x

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