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Topology: Bijection with Open intervals

  1. Oct 31, 2007 #1
    Good Morning,
    I am trying to prove that any 2 open intervals (a,b) and (c,d) are equivalent.
    Show that f(x) = ((d-c)/(b-a))*(x-a)+c is one-to-one and onto (c,d).

    a,b,c,d belong to the set of Real numbers with a<b and c<d.
    Let f: (a,b)->(c,d) be a linear function which i graphed to help me visualize, restricting the domain to (a,b). I know I have to prove that f is a bijection but I am not sure how to do this.
    I know that I am supposed to show that any two open intervals are equivalent, even if the intervals are different lengths, so the open interval (3,4) is equivalent to the open interval (1,8), (3,4) is a subset of (1,9).

    I would appreciate any help with this if you can get me on the right track...thanks so much!:confused:
     
  2. jcsd
  3. Oct 31, 2007 #2

    HallsofIvy

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    You prove that it is a bijection by proving just what you said: that it is one-to-one and onto.

    one-to-one: if f(x)= f(y) then x= y: if ((d-c)/(b-a))*(x-a)+c= ((d-c)/(b-a))*(y-a)+c just do the algebra1

    onto: if y is in (c,d), you must show that there exist an x in (a, b) so that f(x)= y. That is, if y= ((d-c)/(b-a))*(x-a)+c, with c< y< d, solve for x and show that a< x< b.
     
  4. Oct 31, 2007 #3
    Once you've done that, then for fun you can modify that bijection to prove that (a,b) and (c,d] (or variations of that) are equivalent. Just take an infinite countable subset of each (e.g. a rational subset) so that d maps into the subset of (a,b) and then all the elements of the subset get "shifted over" by index, and all the other elements mapped using your bijection.
     
    Last edited: Oct 31, 2007
  5. Nov 1, 2007 #4
    OK Thanks so much. That was very helpful, I understand this now :smile:
     
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