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Let $f(x)$ be a function defined for all $x$ that maps compact intervals to compact intervals: $I$ $compact$ $interval$ $\Rightarrow$ $f(I)$ $compact$ $interval$. Do not assume $f(x)$ is continuous.

(a) Prove that on any compact interval $I$ the function attains its maximum and minimum.

(b) Prove that $f(x)$ has the Intermediate Value Property.

Possible derivative of question here: Prove that the image of a a closed and bounded interval in $\mathbb{R}$ is a a closed and bounded interval in $\mathbb{R}$?

Except I can't assume that $f(x)$ is continuous. I know that I will have to use the Continuous Mapping Theorem since these two properties make up the theorem, but the intuition is pretty backwards to me.

Any help is appreciated! Thank you.

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Hint for a, Recall compact sets in $\mathbb{R}$ are closed and bounded.

For b, if $f(x)<f(y)$ both $f(x)$ and $f(y)$ live in an interval defined by $f(I)$, it is immediate that there is some value in $f(I)$ between them, call it, $f(z)$, as intervals are connected.

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a) Anyway, $f(I)$ is closed and bounded, so $M:=\sup f(I)$ and $m:=\inf f(I)$ both exist and $M,m\in f(I)$.

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