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It is a fact from analysis that a continuous and open real-valued function of a real variable is strictly monotonic. The proof I know runs something like this: Suppose $f$ is an open and continuous map but is not strictly monotonic. Consequently, there exist three numbers $a < c < b$ such that either $ f(a) \geq f(c) \leq f(b) \;\;\;\; (1) $ or $ f(a) \leq f(c) \geq f(b) \;\;\;\; (2) $ If $(1)$ holds then the exteme value theorem guarantees that $f$ attains its infimum on $[a,b]$; but by assumption, the infimum is at least as small as $f(c)$ so in fact $f$ attains its infimum on $(a,b)$. Also by assumption, $f$ carries open intervals to open intervals. With this though we have a contradiction since an open interval cannot contain it's own infimum. A similar argument yields considering suprema yields an analagous contradiction. Therefore, $f$ is strictly monotonic.

My question is, Is there a more constructive way to prove this that doesn't involve contradiction? Although I think the proof given is nice, I don't think I could have come up with it own my own because the consequences of $f$ not being strictly monotonic as exhibited in (1) and (2) would not have occurred to me. So, it would be good to see a direct way of proving this.

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    I experienced something similar before with the 'no continuous real function is a 2-to-1 mapping'; everyone around me would say such is 'intuitively clear' and leave me behind in the dust, and while I thought it was *plausible*, it took me lots of beating it to death (in the form of casework) to have confidence that it would work, something they had either swept under the rug or truly did find intuitive. (In which case, I wonder what their brains are made of, and whether it is me who is defective.) I'm still wondering whether there's a cleaner/slicker demonstration of this one too.2016-09-02

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f is open map and continuous, let f:R->R Let the inverse of f is F.so F is continuous from f(R) to R. Now lets assume there exists distinct x,y in R such that f(x)=f(y). WLG lets assume f(x)=f(y)=c for some c in R. F(x) is not continuous at c as x,y are distinct and hence the contradiction.

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    yeah its open that's why inverse of$f$will be continuous and I write inverse of f as F2014-10-09