Prove that the image of a a closed and bounded interval in $\mathbb{R}$ is a a closed and bounded interval in $\mathbb{R}$?

Question

According to the excercise 7.22 of the book Topology by Franzosa:

Combining the Extreme Value Theorem and the Intermediate Value Theorem, prove the following theorem:

Let $[a, b]$ be a closed and bounded interval in $\mathbb{R}$, and assume that $f : [a, b] \rightarrow \mathbb{R}$ is continuous. Then the image of $f$ is a closed and bounded interval in $\mathbb{R}$.

Of course we can prove the theorem by the fact that image of compact domain is compact if the function is continuous. But what about proving it by using only the Extreme Value Theorem and the Intermediate Value Theorem?

Thanks a lot.

Answer 1

How's this:

Left $f:[a,b]\rightarrow\mathbb R$ be continuous. We know that $[a,b]$ is compact, so by the extreme value theorem, $f$ achieves both it's maximum (say, $d$) at a point $y\in [a,b]$ and it's minimum (say, $c$) at a point $x\in[a,b]$. Thus, $f([a,b])\subseteq[c,d]$. Without loss of generality, assume $x<y$. Let $z\in[c,d]$. Then by the intermediate value theorem, there exists $w$ such that $x<w<y$ and such that $f(w)=z$. Thus, $f([a,b])=[c,d].$