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Prove that the range of h is the entire R

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data
    The function [itex]h: \mathbb{R} \to \mathbb{R}[/itex] is continuous on [itex]\mathbb{R}[/itex] and let [itex]h(\mathbb{R})=\left\{ {h(x):x \in \mathbb{R}} \right\}[/itex] be the range of [itex]h[/itex]. Prove that if [itex]h(\mathbb{R})[/itex] is not bounded above and not bounded below, then [itex]h(\mathbb{R})=\mathbb{R}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Well, this problem sounds so intuitive I don't know how to prove it. The only thing I can write down here is there exists a [itex]M > 0[/itex] s.t [itex]|h(x)|> M[/itex] for all real [itex]x[/itex].
     
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  3. May 1, 2013 #2

    mfb

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    For some y, can you show that some smaller value and some larger value have to be in the range of h? Afterwards, the solution is just 1 step away.
     
  4. May 1, 2013 #3

    HallsofIvy

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    Let y be any number. Since y is not an upper bound for f, there exist Y1>y such that Y1= f(x1) for some x1 Since y is not a lower bound for f, the exist Y2< y such that Y2= f(x2) for some x2. Now use the "intermediate value property.
     
  5. May 1, 2013 #4
    Thank you both for the great help. So let me continue from HallsofIvy's hint: By the IVT, there exists a real number [itex]x[/itex] such that [itex]h(x)=y[/itex]. Since [itex]y[/itex] was arbitrary, [itex]h(x)=\mathbb{R}[/itex]. Is that right?
     
  6. May 1, 2013 #5

    mfb

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    ##\{h(x)|x \in \mathbb{R}\}=\mathbb{R}##. h(x) is a single number.
    Apart from that: right.
     
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