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Countably Many Subdivisions of the Real Line into Open Intervals

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Let [tex]A=\{(a_\alpha,b_\alpha),\alpha\in I\}[/tex] be a family of mutually non-intersecting intervals on the real line. Here [tex]I[/tex] is an arbitrary set of indexes. Prove that this family contains at most countably many elements (intervals).

    2. Relevant equations

    None, other than all of the stuff on countability of a set.

    3. The attempt at a solution

    I've tried to find an enumeration but I haven't gotten anywhere. I'm guessing that this is a property of the real line - it can only be split up into a countable number of open intervals, but I can't see how to prove it. A hint to start may be all I need.

    At least I've been able to get the other five problems on this homework :)
     
  2. jcsd
  3. Feb 10, 2008 #2

    quasar987

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    To show a set is countable, it is sufficient to show there is an injection from the set to some countable set... such as, oh I don't know... the rationals?
     
  4. Feb 10, 2008 #3
    Okay, (correct me if I'm wrong then), I can define an injection [tex]f:A\rightarrow Q[/tex] as [tex]f((a_\alpha,b_\alpha))=\frac{a_\alpha}{b_\alpha}[/tex] (in my class we've used bijections, but I can see that an injection will also work - surjectivity is not necessary for a set that may not be infinite). My only qualm about that is, what if [tex]b_\alpha[/tex] is zero for some alpha? Also, I'm assuming that we can ignore the uniqueness of a rational number (that is, ignore the fact that 1/3 and 9/27 are in the same class). Afterall, we showed in my class that [tex]Q^+ \sim J\times J[/tex], which also ignores differentiation of elements within classes.

    Thanks for your help!
     
  5. Feb 10, 2008 #4

    Dick

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    a0/b0 isn't necessarily rational! And you can't ignore your uniqueness problem. quasar987 is likely suggesting you just pick a rational in each interval and assign it to f((a0,b0)). So your intervals can be placed in 1-1 correspondence with a subset of Q.
     
  6. Feb 10, 2008 #5
    You could start by counting all the intervals of length at least 1/2.
     
  7. Feb 10, 2008 #6

    EnumaElish

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    What if you used [itex]f_\alpha = f((a_\alpha,b_\alpha), k_\alpha)=k_\alpha a_\alpha + (1-k_\alpha)b_\alpha[/itex] for some [itex]0 < k_\alpha < 1[/itex] (e.g. "some middle point" in each interval)?

    A remaining question is, what if either of a or b is irrational for some alpha, so [itex]f_\alpha[/itex] is irrational? What property of the Rationals can you use to find a rational number "near" [itex]f_\alpha[/itex] and within [itex](a_\alpha,b_\alpha)[/itex] for each [itex]\alpha[/itex]?
     
    Last edited: Feb 10, 2008
  8. Feb 10, 2008 #7
    My mistake, I misread it... first map the real line to the interval (0, 1) then count all the intervals whose image under that map have length at least 1/2, of which there are at most 1.
     
  9. Feb 10, 2008 #8

    quasar987

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    The argument is very simple. Every interval of R contains rationals. Create a map that assigns to every alpha in I a rational contained in [tex](a_\alpha,b_\alpha)[/tex]. Since your intervals are disjoint by hypotheses, this map in injective. Done. You've created an injection from I to Q.
     
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