# If f:RR->RR is continuous over the entire domain Ms is not bounded above or below, show that f(RR)=RR.

If $f:\mathbb{R}\to \mathbb{R}$ is continuous over the entire domain Ms is not bounded above or below, show that $f\left(\mathbb{R}\right)=\mathbb{R}.$

My approach would be to say that since $f$ is continuous on $\mathbb{R}$ it must be continuous on and using the intermediate value theorem this tells us that $f$ must take every value between $f\left(a\right)$ and $f\left(b\right)$. Now I want to argue that we can make this closed bounded interval as large as we want and the result still holds and I want to show that this implies that we can make the values of $f\left(a\right)$ as small as we like and $f\left(b\right)$ as large as we like (or the other way round) since $f$ is neither bounded below or above but I'm not sure if this argument can used to show that the image of $f$ is $\mathbb{R}$ since we switch from closed bounded intervals to an unbounded interval.

Does my argument hold to show the result?
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Sanaa Hudson
A little bit cleaner argument:

Suppose $c\in \mathbb{R}$. We will show that there exists $x\in \mathbb{R}$ with $f\left(x\right)=c$
$f$ is not bounded below, so there exists $a\in \mathbb{R}$ such that $f\left(a\right)
$f$ is not bounded above, so there exists $b\in \mathbb{R}$ such that $f\left(b\right)>c$
Now $f$ is continuous on $\left[a,b\right]$ (or $\left[b,a\right]$ if $b) and $c\in \left(f\left(a\right),f\left(b\right)\right)$, so by intermediate value property there exists $x\in \left(a,b\right)$ (or $\left(b,a\right)$) with $f\left(x\right)=c$