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Analysis, Proof about, f being continuous, bijective

  1. Feb 28, 2010 #1
    Let f: R[tex]\rightarrow[/tex]R be a non-decreasing function. Suppose that f maps Q to Q and f: Q[tex]\rightarrow[/tex]Q bijection. Prove that f: R[tex]\rightarrow[/tex]R is continuous, one to one and onto.

    Hello everyone, I have been staring at this statement for a while now and I just don't understand it, hence I can't even begin to prove it. Can someone explain to me in different words what am I being asked to do. Is f the same same function and Q[tex]\rightarrow[/tex]Q is somehow "inside" R[tex]\rightarrow[/tex]R. I don't understand how f could be the same if R and Q don't have the same cardinality, or in other words-I am lost. Any type of help is greatly appreciated.
     
    Last edited: Feb 28, 2010
  2. jcsd
  3. Feb 28, 2010 #2
    f maps Q to Q means that if q is in Q, f(q) is also in Q.
    I am assuming (unless otherwise noted) that Q stands for the rationals.
     
  4. Feb 28, 2010 #3
    Yes, Q=rationals, R=reals; but how and why is mapping of Q to Q relevant here, since I am later asked about map of function R to R?
     
  5. Feb 28, 2010 #4
    It is relevant because it is a critical hypothesis for the problem. You are also given that f is non decreasing and that f restricted to Q is a bijection. Think about what the map f restricted to Q looks like. Where does f(q) go to?

    Now, where does f map the irrational points to?
     
  6. Feb 28, 2010 #5
    Since I=irrationals are uncountable and so are reals I guess we could create a bijection between the two (not sure if I am correct here), but I suppose I would be mapped to R.
     
  7. Feb 28, 2010 #6
    Ah, I deleted my previous post, what I said was incorrect!. But... let me try to be more helpful: try this outline of a proof (if you like):

    1) show that the restriction of f to Q is continuous (treat Q as a metric space on its own and apply the epsilon delta definition).

    2) show that there exists a unique extension of f:Q->Q to f:R->R such that f is continuous on R (proving this is fairly easy via contradiction).

    3) show that f being non-decreasing on Q and continuous on Q implies it is this candidate function (step 2 isn't entirely necessary, but it should help provide a lot more intuition to make this last leap).

    I'm quite drowsy, so if something sounds confusing tell me.
     
  8. Feb 28, 2010 #7
    I'm sorry, but I'm not familiar with the term "unique extension" what do you mean by that?
     
  9. Feb 28, 2010 #8
    Oh no prob. An extension of a function f:Q->Q to a function g:R->R is g such that for every x in Q, f(x) = g(x).
     
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