For a continuous function on closed bounded interval is absolute minimum value, absolute maximum values are same as inf, sup of image set I was reading theorem “5.3.9” of “Introduction to real analysis Robert G. Bartle and Donald R. Sherbert”. I didn’t not understand one step in the proof of theorem. In a second line of proof author writes “ by “Maximum and minimum theorem 5.3.4” we have $m,M\in f(I)$. (Where, $m=\inf f(I)$, and $M=\sup f(I)$) How? Can anybody explain?
My opinion: By maximum minimum theorem, as $f$ is continuous on closed and bounded interval $I$ hence, $f$ has absolute minimum and absolute maximum on $I$. Hence by definition of absolute minimum, absolute maximum values we have, there exists points $c_*,c^*\in I$ Such that,
$f(c_*)≤f(x)$ for all $x\in I$ and $f(c^*)≥ f(x)$ for all $x\in I$ .
So how author concluded $m=\inf f(I),M=\sup f(I)\in f(I)$?
Is $m=\inf(f(I))=f(c_*)$ and $M=\sup(f(I))=f(c^*)$ and hence $m,M\in f(I)$?
Am I correct? Please help.

You just need to apply the following fact to the set $A=f[I]$.$\newcommand{\R}{\mathbb R}$
Fact. If $A$ is subset of $\R$ and if $m=\min A$, $M=\max A$ then $m=\inf A$, $M=\sup A$.
This is true for any partially ordered set, but since you are asking this in context of real analysis, I have only mentioned $A\subseteq\R$.
This simply says that if a set has maximum, then it is equal to supremum. (The same is true for minimum and infimum.) Notice that there are sets which have supremum, but do not have maximum - for example, $A=(0,1)$.
Why this claim is true - it depends a bit on how you defined maximum and minimum. But for any reasonable definition you should be able to see that if $M=\max A$, then

*

*$M$ is an upper bond of $A$, i.e., for any $a\in A$ you have $a\le M$.

*$M$ belongs to $A$, and thus for any upper bound $U$ of the set $A$ we have $M\le U$. (We have $a\le U$ for every $a\in A$. Since $M\in A$, the inequality is true for $M$, too.)

