Prove that the range of h is the entire R

[drawar]

Homework Statement

The function h: \mathbb{R} \to \mathbb{R} is continuous on \mathbb{R} and let h(\mathbb{R})=\left\{ {h(x):x \in \mathbb{R}} \right\} be the range of h. Prove that if h(\mathbb{R}) is not bounded above and not bounded below, then h(\mathbb{R})=\mathbb{R}

Homework Equations

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 M > 0 s.t |h(x)|> M for all real x.

 

[mfb]

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.

 

[HallsofIvy]

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.

 

[drawar]

Thank you both for the great help. So let me continue from HallsofIvy's hint: By the IVT, there exists a real number x such that h(x)=y. Since y was arbitrary, h(x)=\mathbb{R}. Is that right?

 

[mfb]

Since y was arbitrary, h(x)=\mathbb{R}. Is that right?

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