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Consider the image of an open set $(a,b)$ under the open and continuous mapping $f$. We show, $f$ cannot have any extremum in $(a,b)$.

We know, connected sets are mapped to connected sets under a continuous map. Hence, $f[(a,b)]=(c,d)$ is connected (and open). Suppose, at $\xi \in (a,b)$, $f(\xi)=\sup_{(a,b)} f=M$ (or $\inf_{(a,b)}f=m$). [Being a continuous function, $f$ must attain its supremum/infimum.]

Hence, the image of the set $(a,b)$ under $f$ becomes $(c,M]$ or $[m,d)$, which is a contradiction of the fact that $f$ maps open sets to open sets. Therefore, in any open interval, the function cannot attain glb/lub at an interior point. So, we conclude that the inf and sup are at the end points, i.e. $\sup_{[a,b]}f=f(a)$ or $f(b)$.

Hence the theorem.

Is the proof valid? I am aware of the duplicates. I just want this method verified.

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  • $\begingroup$ Possible duplicate of Every continuous open mapping of $\mathbb{R}$ into $\mathbb{R}$ is monotonic $\endgroup$ – Mees de Vries Jan 28 at 17:21
  • $\begingroup$ i am going to take down my previous question $\endgroup$ – Subhasis Biswas Jan 28 at 17:23
  • $\begingroup$ If the question is essentially the same, you just want to change the presentation, it makes more sense to edit the old question, rather than delete and then immediately re-ask. $\endgroup$ – Mees de Vries Jan 28 at 17:23
  • $\begingroup$ Thank you! will keep that in mind. $\endgroup$ – Subhasis Biswas Jan 28 at 17:24
  • $\begingroup$ Your proof is not correct, first because the extreme value theorem doesn't apply on an open interval, and second because such "global" reasoning doesn't invalidate the possibility of a function whose global extrema on $[a,b]$ are at the endpoints but nevertheless has some local extremum. $\endgroup$ – Ian Jan 28 at 17:25

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