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.