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Order isomorphism f:R->R

  1. Jan 4, 2009 #1
    order isomorphism f:R-->R

    let f is order isomorphism from (R,<) to (R,<). why f is continuous ?
    so f is bijection and a<b <--> f(a)<f(b), so what ?
     
  2. jcsd
  3. Jan 6, 2009 #2

    HallsofIvy

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    Re: order isomorphism f:R-->R

    f maps the interval [a, a+ epsilon] one to one and onto the interval [f(a),f(a+ epsilon)]. As epsilong goes to 0, f(a+ epsilon) must go to a, hence continuity.
     
  4. Jan 7, 2009 #3
    Re: order isomorphism f:R-->R

    Another way of proving it is by showing that inverse images of open sets of R (in the order topology) are open. This is easy; testing just elements of a basis for the topology (such as the set of bounded open intervals) suffices. Given an open interval (c, d), c = f(a) and d = f(b) for some a and b, and it's easy to see that f-1((c, d)) = (a, b), which is open.

    (This works equally well for any ordered set.)
     
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