Every continuous open mapping from $\mathbb{R}$ into $\mathbb{R}$ is monotonic

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.

• Possible duplicate of Every continuous open mapping of $\mathbb{R}$ into $\mathbb{R}$ is monotonic – Mees de Vries Jan 28 at 17:21
• i am going to take down my previous question – Subhasis Biswas Jan 28 at 17:23
• 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. – Mees de Vries Jan 28 at 17:23
• Thank you! will keep that in mind. – Subhasis Biswas Jan 28 at 17:24
• 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. – Ian Jan 28 at 17:25